\magnification=\magstep1

\newfam\msbfam 
\font\tenmsb=msbm10 \textfont\msbfam=\tenmsb  
\font\sevenmsb=msbm7 \scriptfont\msbfam=\sevenmsb
\def\hexnumber#1{\ifcase#1 0\or1\or2\or3\or4\or5\or6\or7\or8\or9\or
 A\or B\or C\or D\or E\or F\fi}
\edef\msbhx{\hexnumber\msbfam}
\mathchardef\subsetneq="2\msbhx28

\newfam\msbfam 
\font\tenmsb=msbm10 \textfont\msbfam=\tenmsb  
\font\sevenmsb=msbm7 \scriptfont\msbfam=\sevenmsb

\def\Bbb#1{{\fam\msbfam\relax#1}}
\def\N{{\Bbb N}}
\def\Q{{\Bbb Q}}
\def\C{{\Bbb C}}
\def\R{{\Bbb R}}
\def\T{{\Bbb T}}
\def\Z{{\Bbb Z}}

\def\hata{{\widehat{a}}}
\def\hatb{{\widehat{b}}}
\def\hatz{{\widehat{z}}}
\def\hatC{{\widehat{C}}}
\def\hatD{{\widehat{D}}}
\def\hatE{{\widehat{E}}}
\def\hatalpha{{\widehat{\alpha}}}
\def\hatbeta{{\widehat{\beta}}}
\def\hatgamma{{\widehat{\gamma}}}

\def\tildeF{{\widetilde{F}}}
\def\tildeH{{\widetilde{H}}}
\def\tildeS{{\widetilde{S}}}

\def\scra{{\cal A}}
\def\scrb{{\cal B}}
\def\scrc{{\cal C}}
\def\scrd{{\cal D}}
\def\scre{{\cal E}}
\def\scrf{{\cal F}}
\def\scrg{{\cal G}}
\def\scrh{{\cal H}}
\def\scrm{{\cal M}}
\def\scrn{{\cal N}}
\def\scrs{{\cal S}}
\def\scru{{\cal U}}
\def\scrv{{\cal V}}
\def\scrx{{\cal X}}
\def\scrts{{\cal T\cal S}}
\def\scrgg{{\cal G\cal G}}
\def\scrcc{{\cal C\cal C}}
\def\scrfgcc{{\cal F\cal G\cal C\cal C}}
\def\scrbw{{\cal BW}}
\def\scrpc{{\cal PC}}
\def\scrsesag{{\cal SESAG}}
\def\scrbrsx{{\cal BRSX}}
\def\scrsnsx{{\cal SNSX}}
\def\scrsescc{{\cal SESCC}}
\def\scrles{{\cal LES}}
\def\scrbrles{{\cal BRLES}}

\def\tildeC{{\widetilde{C}}}
\def\tildeH{{\widetilde{H}}}
\def\tildeS{{\widetilde{S}}}

\def\barS{{\overline{S}}}
\def\barU{{\overline{U}}}
\def\barPsi{{\overline{\Psi}}}

\def\deg{{\rm deg}}
\def\rank{{\rm rank}}
\def\im{{\rm im}}
\def\id{{\rm id}}
\def\ker{{\rm ker}}
\def\coker{{\rm coker}}
\def\cvx{{\rm cvx}}
\def\Sd{{\rm Sd}}
\def\Cl{{\rm Cl}}
\def\Int{{\rm Int}}
\def\Tor{{\rm Tor}}
\def\Ext{{\rm Ext}}
\def\Hom{{\rm Hom}}
\def\Cell{{\rm Cell}}

\centerline{\bf More notes for Math 8301-8302,
Manifolds and Topology, Fall 2004-Spring 2005}

\centerline{\bf Homology Theory}

\vskip.1in\noindent
{\it Definition.}
A {\bf (chain) complex} consists of
\itemitem{(1)} a bi-infinite sequence $\ldots,C_1,C_0,C_{-1},\ldots$
of additive Abelian groups; and
\itemitem{(2)} for each integer $n$, an additive group homomorphism
$\partial_n:C_n\to C_{n-1}$

\noindent
such that, for all integers $n$, we have $\partial_n\circ\partial_{n+1}=0$.
A chain complex $C_\bullet$ is {\bf nonnegative} if:
for all integers $n<0$, we have $C_n=\{0\}$.

\vskip.1in
There is a category of chain complexes. (What are the arrows?)
An arrow from one chain complex to another is called a {\bf chain map}.

\vskip.1in\noindent
{\it Definition.}
Let $C_\bullet$ be a chain complex. For all integers $n$,
$Z_n(C_\bullet)$ denotes the kernel of $\partial_n:C_n\to C_{n-1}$,
and $B_n(C_\bullet)$ denotes the image of
$\partial_{n+1}:C_{n+1}\to C_n$.
For all integers $n$,
we define $H_n(C_\bullet)=(Z_n(C_\bullet))/(B_n(C_\bullet))$.
For all integers $n$,
$C_n$ is called the {\bf $n$th chain group} of $C_\bullet$,
$B_n(C_\bullet)$ is called the {\bf $n$th boundary group} of $C_\bullet$,
$Z_n(C_\bullet)$ is called the {\bf $n$th cycle group} of $C_\bullet$
and
$H_n(C_\bullet)$ is called the {\bf $n$th homology group} of $C_\bullet$.
For any integer~$n$,
an $n$-cycle is an element of $Z_n(C_\bullet)$ and
an $n$-boundary is an element of $B_n(C_\bullet)$.
An $n$-cycle is said to {\bf bound}
if it is an element of $B_n(C_\bullet)$.

\vskip.1in
Thus $H_n(C_\bullet)$ is the group obtained by modding out
$n$-cycles by those $n$-cycles that bound.

A {\bf graded group} is a bi-infinite sequence of additive Abelian
groups.  What are the arrows in the category of graded groups?

With this terminology, $H_\bullet$ is a functor from
$\{\hbox{chain complexes}\}$ to
$\{\hbox{graded groups}\}$.

For any integer $n\ge0$, let
$\Delta^n:=\{(x_0,\ldots,x_n)\in[0,1]^n\,|\,x_0+\cdots+x_n=1\}
\subseteq\R^{n+1}$.
For any topological space, for any integer $n\ge0$,
a {\bf parametric $n$-simplex} in $X$ is a continuous function
$\Delta^n\to X$.
Thus the set of parametric $n$-simplices in $X$
is $C(\Delta^n,X)$.

Recall, for any set $S$, that $\langle S\rangle$ denotes
the free group on $S$.
Then $\langle\cdot\rangle$ is a functor
from $\{\hbox{sets}\}$ to
$\{\hbox{groups}\}$.
Let $\scrf:\{\hbox{groups}\}\to\{\hbox{sets}\}$
denote the forgetful functor.

\vskip.1in\noindent
{\bf EXERCISE 12C:}
Show that $(\langle\cdot\rangle,\scrf)$ is an adjoint pair.

\vskip.1in
For any set $S$, let $\Z[S]$ denote the group of
formal finite linear combinations of elements of $S$
with coefficients in $\Z$.
This group is called the
{\bf free additive Abelian group generated by $S$}.
Then $\Z[\cdot]$ is a functor from
from $\{\hbox{sets}\}$ to
$\{\hbox{additive Abelian groups}\}$.
Let $\scrf':\{\hbox{additive Abelian groups}\}\to\{\hbox{sets}\}$
denote the forgetful functor.

\vskip.1in\noindent
{\bf EXERCISE 12D:}
Show that $(\Z[\cdot],\scrf')$ is an adjoint pair.

\vskip.1in
For any integer $n\ge1$, for any integer $i\in[0,n]$,
define $\varepsilon_i^n:\Delta^{n-1}\to\Delta^n$
by $\varepsilon_i^n(x_1,\ldots,x_n)=(x_1,\ldots,x_i,0,x_{i+1},\ldots,x_n)$.
For any integer $n\ge1$, for any integer $i\in[0,n]$,
define $\partial_i^n:C(\Delta^n,X)\to C(\Delta^{n-1},X)$
by $\partial_i^n(\sigma)=\sigma\circ\varepsilon_i^n$.

For any topological space $X$,
for any integer $n\ge0$,
let $S_n(X):=\Z[C(\Delta^n,X)]$ be the free additive Abelian group
generated by parametric $n$-simplices in $X$.
For any topological space $X$,
for any integer $n<0$, let $S_n(X):=\{0\}$ be the
trivial additive Abelian group.
For any topological space $X$,
for any integer $n\ge1$,
let $\partial_n:S_n(X)\to S_{n-1}(X)$ be the unique
$\Z$-linear extension of
$$\sigma\qquad\mapsto\qquad
\sum_{i=0}^n\,(-1)^i[\partial_i^n(\sigma)]\qquad:\qquad
C(\Delta^n,X)\qquad\to\qquad S_{n-1}(X).$$
For any topological space $X$,
for any integer $n<1$,
let $\partial_n:S_n(X)\to S_{n-1}(X)$ be the zero map.

\vskip.1in\noindent
{\bf EXERCISE 12E:}
Show that $(S_\bullet(X),\partial_\bullet)$
is a nonnegative chain complex.
That is, for all integers $n$,
show that $\partial_n\circ\partial_{n+1}=0$.

\vskip.1in
For any topological space $X$,
for any integer $n$,
we define $Z_n(X):=Z_n(S_\bullet(X))$,
$B_n(X):=B_n(S_\bullet(X))$
and $H_n(X):=H_n(S_\bullet(X))$.
Then $H_\bullet$ is a functor from the category
$\{\hbox{topological spaces}\}$ to the category
$\{\hbox{graded groups}\}$.

\vskip.1in\noindent
{\it Definition.}
Let $X$ be a topological space, let $n\in Z$
and let $\alpha,\beta\in Z_n(X)$.
We say that $\alpha$ is {\bf homologous} to $\beta$
if $\beta-\alpha$ bounds.

\vskip.1in
Equivalently, the images of $\alpha$ and $\beta$
in $H_n(X)=(Z_n(X))/(B_n(X))$ are equal.

Recall, for any chain complex $C_\bullet$, for any $n\in\Z$, that we
we defined $Z_n(C_\bullet):=\ker(\partial_n:C_n\to C_{n-1})$
and $B_n(C_\bullet):=\im(\partial_{n+1}:C_{n+1}\to C_n)$.
Then $Z_n$ and $B_n$ are functors from $\{\hbox{chain complexes}\}$
to $\{\hbox{additive Abelian groups}\}$.
Then $Z_\bullet$ and $B_\bullet$ are functors from
$\{\hbox{chain complexes}\}$
to $\{\hbox{graded groups}\}$.

\vskip.1in\noindent
{\it Definition.}
Let $C_\bullet$ be a chain complex and let $n\in\Z$.
For any $\alpha\in C_n$, we say that $\alpha$
is a {\bf cycle} if $\alpha\in Z_n(C)$.
For any $\alpha\in Z_n(C_\bullet)$,
we say that $\alpha$ {\bf bounds}
if $\alpha\in B_n(C_\bullet)$.
For any $\alpha,\beta\in Z_n(C_\bullet)$,
we say that $\alpha$ and $\beta$ are {\bf homologous}
if $\beta-\alpha$ bounds.

\vskip.1in
We now turn to the question of computability of homology groups
of topological spaces.
We begin with
$H_0:\{\hbox{topological spaces}\}\to\{\hbox{additive Abelian groups}\}$.

Define a functor $\scrpc$ from $\{\hbox{topological spaces}\}$
to $\{\hbox{sets}\}$ by letting
$\scrpc(X)$ be the set of path-components of $X$.
We leave it to the reader to consider what
$\scrpc$ does to arrows in $\{\hbox{topological spaces}\}$.

\vskip.1in\noindent
{\bf EXERCISE 12F:}
Show that
$H_0:\{\hbox{topological spaces}\}\to\{\hbox{additive Abelian groups}\}$
is equivalent to
$X\mapsto\Z[\scrpc(X)]:
\{\hbox{topological spaces}\}\to\{\hbox{additive Abelian groups}\}$.

\vskip.1in
By the preceding exercise,
if a topological space $X$ has five path-components,
then $H_0(X)$ is isomorphic to $\Z\oplus\Z\oplus\Z\oplus\Z\oplus\Z$.
Computation of $H_0$ is equivalent to counting
connected components.

We now move on to $H_1$.
We begin with the most basic question:

\vskip.1in\noindent
{\it Question:}
What is $H_1(S^1)$?

\vskip.1in\noindent
{\it Definition.}
For any group $G$, let $[G,G]$ denote the subgroup
of $G$ generated by the set $\{xyx^{-1}y^{-1}\,|\,x,y\in G\}$.

\vskip.1in
Let $\scrn$ denote the collection of all normal subgroups $N$
of $G$ such that $G/N$ is Abelian. Then it's not hard to show
that $[G,G]$ is the unique minimal element of $\scrn$.
In particular, $[G,G]$ is a normal subgroup of $G$
and $G/[G,G]$ is Abelian.
The group $G/[G,G]$ is called the {\bf Abelianization} of $G$.

\vskip.1in\noindent
{\bf EXERCISE 13A:}
Let $\scrf:\{\hbox{groups}\}\to\{\hbox{Abelian groups}\}$
be the functor $\scrf(G)=G/[G,G]$ which carries a group
to its Abelianization.
Let $\scrg:\{\hbox{Abelian groups}\}\to\{\hbox{groups}\}$
be inclusion: $\scrg(A)=A$.
Show that $(\scrf,\scrg)$ is an Adjoint pair.

\vskip.1in
Let $I:=[0,1]$.
Let $\kappa:I\to\Delta^1$ be defined by
$\kappa(t)=(1-t,t)$.

\vskip.1in\noindent
{\bf EXERCISE 13B:}
Let $X$ be a topological space, let $x_0\in X$
and let $\sigma\in P_{x_0}^{x_0}(X)$ be a loop at $x_0$ in $X$.
Then $\sigma\circ\kappa^{-1}\in C(\Delta^1,X)\subseteq S_1(X)$.
Show that $\sigma\circ\kappa^{-1}\in Z_1(X)$,
{\it i.e.}, that $\partial_1(\sigma\circ\kappa^{-1})=0$.

\vskip.1in
Let $X$ be a topological space and let $x_0\in X$.
Define a map $\Phi:P_{x_0}^{x_0}(X)\to Z_1(X)$
by $\Phi(\sigma)=\sigma\circ\kappa^{-1}$.
Let $C:P_{x_0}^{x_0}(X)\to\pi_1(X_0,x_0)$
and $D:Z_1(X)\to H_1(X)$ be the canonical maps.
Then we will show later that there is a unique
map $\Psi_{(X,x_0)}:\pi_1(X,x_0)\to H_1(X)$
such that $D\circ\Phi=\Psi_{(X,x_0)}\circ C$.
This map $\Psi_{(X,x_0)}:\pi_1(X,x_0)\to H_1(X)$
is a homomorphism of groups and
is called the {\bf Hurewicz homomorphism}.

We omit the proof of the following interesting result:

\vskip.1in\noindent
{\it Theorem.}
Let $X$ be a path-connected topological space
and let $x_0\in X$.
Then the Hurewicz homomorphism $\Psi_{(X,x_0)}:\pi_1(X,x_0)\to H_1(X)$
is surjective and its kernel is $[\pi_1(X,x_0),\pi_1(X,x_0)]$.

\vskip.1in
Let $\scrc$ be the category
$\{\hbox{pointed path connected topological spaces}\}$.
For any group~$G$, let $A_G:=G/[G,G]$ be the Abelianization of $G$
and let $c_G:G\to A_G$ be the canonical map.
Because of the preceding theorem,
for any pointed path-connected topological space $Z=(X,x_0)$,
there is a unique isomorphism of Abelian groups
$\barPsi_Z:A_{\pi_1(Z)}\to H_1(X)$
such that
$\barPsi_Z\circ c_{\pi_1(Z)}=\Psi_Z$.
Then $\barPsi$ is an equivalence between the functor
$$Z\mapsto A_{\pi_1(Z)}:
\scrc\to\{\hbox{Abelian groups}\}$$
and the functor
$$(X,x_0)\mapsto H_1(X):
\scrc\to\{\hbox{Abelian groups}\}.$$

In particular, for any path-connected topological space~$X$,
for any $x_0\in X$, we see that $H_1(X)$ is the Abelianization
of $\pi_1(X,x_0)$.
This means that, for a person that has facility with the algebraic process
of computing the Abelianization of groups,
one can get $H_1$ from~$\pi_1$.
So $H_1$ is ``no harder than'' $\pi_1$.
In fact, in practice, it often happens that computation of $\pi_1$
is difficult, whereas $H_1$ is easier.

Also, we can answer our question about computing $H_1(S^1)$.
Let $x\in S^1$ be any point.
Then $H_1(S^1)$ is isomorphic to the Abelianization
of $\pi_1(S^1,x)$.
So $H_1(S^1)$ is isomorphic to the Abelianization of $\Z$.
However, $\Z$ is Abelian, and is therefore isomorphic to
its own Abelianization.
Then $H_1(S^1)$ is isomorphic to $\Z$.

Let $X$ be the genus two orientable surface.
Let's next compute $H_1(X)$.
Recall that the fundamental group of $X$ is isomorphic
to $G:=\langle a,b,c,d\,|\,[a,b][c,d]=1\rangle$.
Then $H_1(X)$ is isomorphic to the Abelianization
of $G$, which is isomorphic to
$$\langle a,b,c,d\,|\,
[a,b]=1,[a,c]=1,[a,d]=1,[b,c]=1,[b,d]=1,[c,d]=1,
[a,b][c,d]=1\rangle.$$
The last of these relations ($[a,b][c,d]=1$) is implied by the others,
and so can be omitted.
Then $H_1(X)$ is isomorphic to
$$\langle a,b,c,d\,|\,\
[a,b]=1,[a,c]=1,[a,d]=1,[b,c]=1,[b,d]=1,[c,d]=1\rangle.$$
Up to isomorphism, this is the free Abelian group on $\{a,b,c,d\}$.
Then $H_1(X)$ is isomorphic to $\Z\oplus\Z\oplus\Z\oplus\Z$.

We now take the point of view that we ``know'' $H_0$ and $H_1$.
To continue to talk about $H_n$ for $n\ge2$,
we develop some notation about ``affine'' parametric simplices.

For any integer $m\ge0$, for any $S\subseteq\R^m$,
let $\cvx(S)$ be the set of all $c_0s_0+\cdots+c_ks_k$
such that $k\ge0$ is an integer,
such that $s_0,\ldots,s_k\in S$,
such that $c_0,\ldots,c_k\in[0,1]$
and such that $c_0+\cdots+c_k=1$.
Then $\cvx(S)$ is called the {\bf convex hull} of $S$.

Recall, for any integer $m\ge0$, that
$e_0^k,\ldots,e_k^k$ is the standard basis of $\R^{k+1}$
and that $\Delta^k=\cvx\{e_0^k,\ldots,e_k^k\}\subseteq\R^{k+1}$.

Let $k,m\ge0$ be integers and let $x_0,\ldots,x_k\in\R^m$.
Let $X:=\cvx\{x_0,\ldots,x_k\}$.
Let $[x_0,\ldots,x_k]$ denote the map 
$c_0e_0^k+\cdots+c_ke_k^k\mapsto c_0x_0+\cdots+c_kx_k:
\Delta^k\to X$.
Then $[x_0,\ldots,x_k]\in C(\Delta^k,X)\subseteq S_k(X)$.

Let $k,m\ge0$ be integers, let $X$ be a convex subset of $\R^m$
and let $c=(c_0,\ldots,c_k)\in X^{k+1}$.
We define $[c]:=[c_0,\ldots,c_k]\in S_k(X)$.
For any integer $i\in[0,k]$,
we define $c^i:=(c_0,\ldots,c_{i-1},c_{i+1},\ldots,c_k)\in X^k$.

Let $k,m\ge0$ be integers, let $X$ be a convex subset of $\R^m$
and let $c\in X^{k+1}$.
We leave it as an unassigned exercise to show
that
$\displaystyle{\partial[c]=\sum_{i=0}^k\,(-1)^i[c^i]}$.

For example, we have the following:
Let $m\ge0$ be an integer, let $X$ be a convex subset of $\R^m$
and let $x,y,z\in X$.
Then $\partial[x,y,z]=[y,z]-[x,z]+[x,y]$.

Let $I:=[0,1]$.
Let $p:=(0,0)$, $q:=(1,0)$, $r:=(0,1)$, $s:=(1,1)$.
Then $p,q,r,s$ are the four ``corners'' of the square $I\times I$.
Let $\sigma:=[p,q,r]-[q,r,s]\in S_2(I\times I)$.
One now verifies readily that
$\partial\sigma=
-[r,s]+[p,q]+[q,s]-[p,r]$.

Let $X$ be a topological space and let $x_0\in X$.
Recall that $\kappa:I\to\Delta^1$
is defined by $\kappa(t)=(1-t,t)$,
so $\kappa(0)=(1,0)=e_0^1$ and $\kappa(1)=(0,1)=e_1^1$.
Define $\Phi:P_{x_0}^{x_0}(X)\to Z_1(X)$
by $\Phi(\sigma)=\sigma\circ\kappa^{-1}$.
Let $C:P_{x_0}^{x_0}(X)\to\pi_1(X_0,x_0)$
and $D:Z_1(X)\to H_1(X)$ be the canonical maps.
As promised earlier, we show that there is a unique map
$$\Psi_{(X,x_0)}:\pi_1(X,x_0)\to H_1(X)$$
such that $D\circ\Phi=\Psi_{(X,x_0)}\circ C$.

This statement is equivalent to the following:

\vskip.1in\noindent
{\it Proposition.}
Let $X$ be a topological space and let $x_0\in X$.
Let $\alpha,\beta:I\to X$ be loops at $x_0$ in $X$.
Let $\kappa$ be defined as above.
Assume that $\alpha$ and $\beta$ are endpoint fixed homotopic.
Then $(\beta\circ\kappa^{-1})-(\alpha\circ\kappa^{-1})\in Z_1(X)$
bounds.

\vskip.1in\noindent
{\it Proof:}
Let $H:I\times I\to X$ be
an endpoint fixed homotopy from $\alpha$ to $\beta$.
Let $p$, $q$, $r$, $s$ and~$\sigma$ be as defined above.
We aim to show that $\partial(H_*(\sigma))=
(\beta\circ\kappa^{-1})-(\alpha\circ\kappa^{-1})$.

Since
$\partial\sigma=[q,s]-[p,r]-[r,s]+[p,q]$,
we see that
$$\partial(H_*(\sigma))=
(H_*([r,s]))-(H_*([p,q]))-(H_*([q,s]))+(H_*([p,r])).$$
We have $H_*([r,s])=(H(1,\cdot))\circ\kappa^{-1}=\beta\circ\kappa^{-1}$.
Similarly, $H_*([p,q])=\alpha\circ\kappa^{-1}$.
It remains to show that $H_*([q,s])=H_*([p,r])$.

Let $c:I\to X$ be the constant map at $x_0$,
defined by $c(t)=x_0$.
Then $H(\cdot,0)=H(\cdot,1)=c$.
Then $H_*([q,s])=(H(\cdot,1))\circ\kappa^{-1}=c\circ\kappa^{-1}$.
Similarly, $H_*([p,r])=c\circ\kappa^{-1}$.
Then $H_*([q,s])=H_*([p,r])$.
{\bf QED}

\vskip.1in\noindent
{\it Proposition.}
Let $X$ and $Y$ be topological spaces
and let $f,g:X\to Y$ be continuous maps.
Assume that $f$ and $g$ are homotopic.
Then $f_*:H_1(X)\to H_1(Y)$ is equal to
$g_*:H_1(X)\to H_1(Y)$.

\vskip.1in\noindent
{\it Proof:}
Let $h:I\times X\to Y$ be a continuous map such
that $h(0,\cdot)=f$ and $h(1,\cdot)=g$.

Let $p,q,r,s\in I\times\Delta^1$ be defined by
$p:=(0,e_0^1)$, $q:=(0,e_1^1)$,
$r:=(1,e_0^1)$ and $s:=(1,e_1^1)$.
We leave it as an unassigned exercise to show that
there exists $\sigma\in S_2(I\times\Delta^1)$
such that $\partial\sigma=[q,s]-[p,r]-[r,s]+[p,q]$.

For all $\omega\in C(\Delta^1,X)$,
define $R_\omega:I\times\Delta^1\to Y$
by $R_\omega(u,v)=h(u,\omega(v))$.
Define $\phi:S_1(X)\to S_2(Y)$ by:
for all $\omega\in C(\Delta^1,X)$,
$\phi(\omega)=(R_\omega)_*(\sigma)$.

Recall that $\Delta^0=\{1\}\subseteq\R$.
Let $a,b\in I\times\Delta^0$ be defined by
$a:=(0,1)$ and $b:=(1,1)$.
Let $\rho:=[a,b]$.
Then $\partial\rho=[b]-[a]$.

For all $\omega\in C(\Delta^0,X)$
define $L_\omega:I\times\Delta^0\to Y$
by $L_\omega(u,v)=h(u,\omega(v))$.
Define $\psi:S_0(X)\to S_1(Y)$ by:
for all $\omega\in C(\Delta^1,X)$,
$\psi(\omega)=(L_\omega)_*(\rho)$.

\vskip.1in\noindent
{\bf EXERCISE 13C:}
Show, for all $\omega\in C(\Delta^1,X)$,
that $\partial_2(\phi(\omega))=
-[g_*(\omega)]+[f_*(\omega)]-[\psi(\partial_1(\omega))]$.

\vskip.1in
Let $\omega_0\in H_1(X)$.
We wish to show that $f_*(\omega_0)=g_*(\omega_0)$.

Let $c:Z_1(X)\to H_1(X)$ be the canonical map.

Choose $\omega\in Z_1(X)$ such that $c(\omega)=\omega_0$.
Since $\omega\in Z_1(X)=\ker(\partial_1:S_1(X)\to S_0(X))$,
we conclude that $\partial_1(\omega)=0$.
Then, by Exercise 13C,
we see that $\partial_2(\phi(\omega))=
[g_*(\omega)]-[f_*(\omega)]$.

Then
$[g_*(\omega)]-[f_*(\omega)]\in\im(\partial_2:S_2(X)\to S_1(X))=B_1(X)$.
So, as $B_1(X)=\ker(c)$, we get
$c([g_*(\omega)]-[f_*(\omega)])=0$.
So, as $c(g_*(\omega))=g_*(\omega_0)$
and as $c(f_*(\omega))=f_*(\omega_0)$,
we conclude that $f_*(\omega_0)=g_*(\omega_0)$, as desired.
{\bf QED}

\vskip.1in\noindent
{\bf EXERCISE 14A:}
Let $k$ and $m$ be integers and let $U\subseteq\R^m$ be star-shaped.
Assume that $k\ne0$.
Show, for all $\omega\in Z_k(U)$,
that $\omega$ bounds.
That is, show that $H_k(U)=0$.

\vskip.1in
We now extend the preceding theorem slightly,
and, at the same time, reorganizing the notation a bit.

\vskip.1in\noindent
{\it Theorem.}
Let $X$ and $Y$ be topological spaces
and let $f,g:X\to Y$ be continuous maps.
Assume that $f$ is homotopic to $g$.
Then $f_*:H_1(X)\to H_1(Y)$
is equal to $g_*:H_1(X)\to H_1(Y)$
and $f_*:H_2(X)\to H_2(Y)$
is equal to $g_*:H_2(X)\to H_2(Y)$.

\vskip.1in\noindent
{\it Proof:}
Choose a continuous map $h:I\times X\to Y$
such that $h(0,\cdot)=f$ and such that $h(1,\cdot)=g$.

We make the following definitions:
\itemitem{(A)}
For all $\omega\in C(\Delta^0,X)$,
define $L_\omega:I\times\Delta^0\to Y$
by $L_\omega(u,v)=h(u,\omega(v))$.
\itemitem{(B)}
For all $\omega\in C(\Delta^1,X)$,
define $R_\omega:I\times\Delta^1\to Y$
by $R_\omega(u,v)=h(u,\omega(v))$.
\itemitem{(C)}
For all $\omega\in C(\Delta^2,X)$,
define $P_\omega:I\times\Delta^2\to Y$
by $P_\omega(u,v)=h(u,\omega(v))$.

\noindent
Above, ``L'' stands for ``line segment'',
``R'' stands for ``rectangle'' and ``P'' stands for prism.
Note that $I\times\Delta^0$ is a line segment,
$I\times\Delta^1$ is a rectangle and
$I\times\Delta^2$ is a prism.

Fix $\rho\in S_1(I\times\Delta^0)$
and $\sigma\in S_2(I\times\Delta^1)$
as in the preceding proof.
Also, as in the preceding proof,
define $\psi:S_0(X)\to S_1(Y)$
and $\phi:S_1(X)\to S_2(Y)$ by:
\itemitem{$(A')$} for all $\omega\in C(\Delta^0,X)$,
$\psi(\omega)=(L_\omega)_*(\rho)$; and
\itemitem{$(B')$} for all $\omega\in C(\Delta^1,X)$,
$\phi(\omega)=(R_\omega)_*(\sigma)$.

\noindent
For any $\tau\in S_3(I\times\Delta^2)$,
define $\chi_\tau:S_2(X)\to S_3(Y)$ by
\itemitem{$(C')$} for all $\omega\in C(\Delta^2,X)$,
$\chi_\tau(\omega)=(P_\omega)_*(\tau)$.

According to Exercise 13C, for all $\omega\in C(\Delta^1,X)$,
we have
\itemitem{(1)} $(g_*-f_*)(\omega)=
(\partial_2\circ\phi+\psi\circ\partial_1)(\omega))$.

\noindent
Then, by linearity, (1) holds for all $\omega\in S_1(X)$.

\vskip.1in\noindent
{\bf EXERCISE 14B:}
Show that there exists $\tau\in S_3(I\times\Delta^2)$
such that, for all $\omega\in C(\Delta^2,X)$, we have
\itemitem{(2)} $(g_*-f_*)(\omega)=
(\partial_3\circ\chi_\tau+\phi\circ\partial_2)(\omega))$.

\noindent
(Hint: For all integers $i\in[0,2]$, let
$\delta_i:=\id\times\varepsilon_i^2:
I\times\Delta^1\to I\times\Delta^2$.
Define $l,r:\Delta^2\to I\times\Delta^2$
by $l(u)=(0,u)$ and $r(u)=(1,u)$.
Let $\tau_0:=r-l+
(\delta_0)_*(\sigma)-(\delta_1)_*(\sigma)+(\delta_2)_*(\sigma)\in
S_2(I\times\Delta^2)$.
Show that $\partial_2\tau_0=0$.
Then argue from Exercise 14A that $\tau_0$ bounds.
Choose $\tau\in S_3(I\times\Delta^2)$ such that $\partial_3\tau=\tau_0$.
Then show that (2) above holds.)

\vskip.1in
By linearity, since (2) holds for all $\omega\in C(\Delta^2,X)$,
it follows that (2) holds for all $\omega\in S_2(X)$.

By (1), for all $\omega\in Z_1(X)$,
$(g_*-f_*)(\omega)=(\partial_2\circ\phi)(\omega)$,
so $(g_*-f_*)(\omega)$ bounds.
It follows, for all $\omega\in H_1(X)$,
that $(g_*-f_*)(\omega)=0$.

By (2), for all $\omega\in Z_2(X)$,
$(g_*-f_*)(\omega)=(\partial_3\circ\chi_\tau)(\omega))$,
so $(g_*-f_*)(\omega)$ bounds.
It follows, for all $\omega\in H_2(X)$,
that $(g_*-f_*)(\omega)=0$.
{\bf QED}

\vskip.1in
For any $k\in\Z$,
define a functor $C_\bullet\mapsto C_{\bullet+k}:
\{\hbox{chain complexes}\}\to\{\hbox{chain complexes}\}$
by $(C_{\bullet+k})_i=C_{i+k}$.

The preceding two proofs motivate us to formulate the
following definition:

\vskip.1in\noindent
{\it Definition.}
Let $C_\bullet$ and $D_\bullet$ be chain complexes.
Let $f,g:C_\bullet\to D_\bullet$ be chain maps.
A {\bf chain homotopy} from $f$ to $g$ is a chain
map $h:C_\bullet\to D_{\bullet+1}$
such that $g-f=\partial\circ h+ h\circ\partial$.
If a chain homotopy from $f$ to $g$ exists,
then we say that $f$ and $g$ are {\bf chain homotopic}.

\vskip.1in
The maps $\psi:S_0(X)\to S_1(Y)$,
$\phi:S_1(X)\to S_2(Y)$ and
$\chi_\tau:S_2(X)\to S_3(Y)$
of the preceding proof
are the beginnings of a map
$S_\bullet(X)\to S_{\bullet+1}(Y)$.
The formulas (1) and (2) of the preceding
proof are the beginnings of the assertion that
this map is a chain homotopy.

We leave it to the interested reader to
continue the sequence $\psi,\phi,\chi_\tau,\ldots$
and the formulas (1), (2), \dots, and prove:

\vskip.1in\noindent
{\it Theorem.}
Let $X$ and $Y$ be topological spaces
and let $f,g:X\to Y$ be continuous maps.
Assume that $f$ and $g$ are homotopic.
Then $f_*:S_\bullet(X)\to S_\bullet(Y)$
and $g_*:S_\bullet(X)\to S_\bullet(Y)$ are chain homotopic.

\vskip.1in
For any $k\in\Z$, define functors
$B_k,Z_k:\{\hbox{chain complexes}\}\to\{\hbox{additive Abelian groups}\}$
by $B_k(C_\bullet):=\partial_{k+1}(C_{k+1})$
and $Z_k(C_\bullet):=\ker(\partial_k:C_k\to C_{k-1})$.

We have an easy algebraic result:

\vskip.1in\noindent
{\it Theorem.}
Let $C_\bullet$ and $D_\bullet$ be chain complexes
and let $f,g:C_\bullet\to D_\bullet$ be chain maps.
Assume that $f$ and $g$ are chain homotopic.
Then $f_*:H_\bullet(C_\bullet)\to H_\bullet(D_\bullet)$
and $g_*:H_\bullet(C_\bullet)\to H_\bullet(D_\bullet)$
are equal.

\vskip.1in\noindent
{\it Proof:}
Let $k\in\Z$. We wish to show that
$f_*:H_k(C_\bullet)\to H_k(D_\bullet)$
and $g_*:H_k(C_\bullet)\to H_k(D_\bullet)$
are equal.
Let $\omega\in Z_k(C_\bullet)$.
We wish to show that $(g(\omega))-(f(\omega))\in B_k(D_\bullet)$.

Let $h:H_\bullet(C_\bullet)\to H_{\bullet+1}(D_\bullet)$
be a chain homotopy.
Then
$g-f=\partial\circ h+ h\circ\partial$.
Then
$[g(\omega)]-[f(\omega)]=[\partial_{k+1}(h(\omega))]+ [h(\partial_k(\omega))]$.
Since $\omega\in Z_k(C_\bullet)$,
we get $\partial_k(\omega)=0$.
Then
$[g(\omega)]-[f(\omega)]=
\partial_{k+1}(h(\omega))\in\partial_{k+1}(D_{k+1})=B_k(D_\bullet)$.
{\bf QED}

\vskip.1in
Combining the two preceding theorems, we get

\vskip.1in\noindent
{\it Corollary.}
Let $X$ and $Y$ be topological spaces
and let $f,g:X\to Y$ be continuous maps.
Assume that $f$ and $g$ are homotopic.
Then $f_*:H_\bullet(X)\to H_\bullet(Y)$
and $g_*:H_\bullet(X)\to H_\bullet(Y)$ are equal.

\vskip.1in\noindent
{\it Corollary.}
Let $X$ and $Y$ be homotopy equivalent topological spaces.
Then $H_\bullet(X)\cong H_\bullet(Y)$.

\vskip.1in\noindent
{\it Corollary.}
Let $X$ be a contractible topological space,
and let $*$ be a topological space with only one point.
Then $H_\bullet(X)\cong H_\bullet(*)$.

\vskip.1in
Note that $H_0(*)\cong\Z$ and that, for all integers $k\ne0$,
$H_k(*)$ is isomorphic to the trivial additive Abelian group $\{0\}$.

\vskip.1in\noindent
{\it Definition.}
Let $C_\bullet$ be a nonnegative chain complex.
An {\bf augmentation} of $C_\bullet$ is a surjective homomorphism
$\varepsilon:C_0\to\Z$ such that $\varepsilon\circ\partial_1:C_1\to\Z$
is equal to zero.
An {\bf augmented chain complex} consists of a nonnegative
chain complex, together with an augmentation.

\vskip.1in\noindent
{\it Example.}
Let $X$ be a nonempty topological space,
and define $\varepsilon:S_0(X)\to\Z$ by
$\displaystyle{\varepsilon\left({\mathop\sum_i}\,c_ix_i\right)=
{\mathop\sum_i}\,c_i}$.
Then $\tildeS_\bullet(X):=(S_\bullet,\varepsilon)$
is an augmented chain complex.

\vskip.1in
With the notation of the preceding example, $\tildeS_\bullet$ is a
functor from $\{\hbox{topological spaces}\}$ to
$\{\hbox{augmented chain complexes}\}$.

Given an augmented chain complex $(C_\bullet,\varepsilon)$, let
$C_\bullet^*$ be the chain complex
$$\cdots\to C_2\to C_1\to C_0\to\Z\to0\to0\to\cdots$$
where all the maps are boundary maps from $C_\bullet$,
except for the map $C_0\to\Z$ which is $\varepsilon$.
Note, for all integers $n\ne-1$, that $C_n^*=C_n$
and that $C_{-1}^*=\Z$.
Then $(C_\bullet,\varepsilon)\mapsto C_\bullet^*$
is a functor $\scrf$ from
$\{\hbox{augmented chain complexes}\}$
to $\{\hbox{chain complexes}\}$

With all this notation, we define a functor
$$\tildeH_\bullet:=H_\bullet\circ\scrf:
\{\hbox{augmented chain complexes}\}\to
\{\hbox{graded groups}\}.$$
Precomposing this functor with the functor $\tildeS_\bullet$,
we obtain a functor
$$\{\hbox{topological spaces}\}\to\{\hbox{graded groups}\}.$$
This last functor is also denoted by $\tildeH_\bullet$,
and the reader must pick up from context which of the two
functors $\tildeH_\bullet$ is under discussion at a given time.

The {\bf reduced homology} groups of a topological space $X$
are the groups $\tildeH_\bullet(X)$.

We leave it as an unassigned exercise to show, for any
augmented chain complex $\tildeC_\bullet=(C_\bullet,\varepsilon)$,
for any integer $k\ne0$,
that $H_k(C_\bullet)=\tildeH_k(\tildeC_\bullet)$.
We leave it as an unassigned exercise to show, for any
chain complex $C_\bullet$,
that $H_0(C_\bullet)=\Z\oplus[\tildeH_0(\tildeC_\bullet)]$.
Thus ``augmenting the chain complex reduces the rank of homology
by one in degree zero, and has no effect in other degrees''.

For any topological space $X$,
for any integer $k\ne0$,
we have that $H_k(X)=\tildeH_k(X)$.
Moreover, for any topological space $X$,
we have that $H_0(X)=\Z\oplus[\tildeH_k(X)]$.
We leave it as an unassigned exercise to prove
that, if $*$ denotes a topological space with exactly
one point, then $\tildeH_\bullet(*)=0$.

\vskip.1in\noindent
{\it Definition.}
Let $\scrd$ be a diagram in the category of additive
Abelian groups and let $A$ be an object in $\scrd$.
We will say that $\scrd$ is {\bf exact} at $A$ if:
\itemitem{(1)} there is exactly one arrow $f:X\to A$ in $\scrd$ whose
target is $A$;
\itemitem{(2)} there is exactly one arrow $g:A\to Y$ in $\scrd$ whose
domain is $A$; and
\itemitem{(3)} $\im(f)=\ker(g)$, {\it i.e.},
$\{f(x)\,|\,x\in X\}=\{a\in A\,|\,g(a)=0\}$.

\vskip.1in\noindent
{\it Definition.}
A diagram
$$0\to A'\to A\to A''\to0$$
in $\{\hbox{additive Abelian groups}\}$
is called a {\bf short exact sequence}
if it is exact at $A'$, at $A$ and at $A''$.

\vskip.1in
For any arrow $f:C_\bullet\to D_\bullet$ in $\{\hbox{chain complexes}\}$
or $\{\hbox{graded groups}\}$,
we define $\im(f),\ker(f)\in\{\hbox{chain complexes}\}$
by $(\im(f))_n=\im(f_n:C_n\to D_n)$ and
$(\ker(f))_n:=\ker(f_n:C_n\to D_n)$.
Then $\ker$ and $\im$ are functors from the arrow category of
the category $\{\hbox{chain complexes}\}$
to the category $\{\hbox{chain complexes}\}$.

\vskip.1in\noindent
{\it Definition.}
Let $\scrd$ be a diagram in the category of chain complexes
or of graded groups.
Let $A_\bullet$ be an object in $\scrd$.
We will say that $\scrd$ is {\bf exact} at $A_\bullet$ if:
\itemitem{(1)} there is exactly one arrow $f:X_\bullet\to A_\bullet$
in $\scrd$ whose target is $A_\bullet$;
\itemitem{(2)} there is exactly one arrow $g:A_\bullet\to Y_\bullet$
in $\scrd$ whose domain is $A_\bullet$; and
\itemitem{(3)} $\im(f)=\ker(g)$.

\vskip.1in\noindent
{\it Definition.}
A diagram
$$0\to A'_\bullet\to A_\bullet\to A''_\bullet\to0$$
in $\{\hbox{chain complexes}\}$
is called a {\bf short exact sequence}
if it is exact at $A'_\bullet$, at $A_\bullet$ and at $A''_\bullet$.

\vskip.1in\noindent
{\it Definition.}
Let $A_\bullet$, $B_\bullet$ be chain complexes.
We say that $A_\bullet$ is a {\bf subcomplex} of $B_\bullet$ if:
for all integers $n$, $A_n$ is a subgroup of $B_n$.

\vskip.1in
For any chain complex $B_\bullet$, for any subcomplex $A_\bullet$,
we define the chain complex $B_\bullet/A_\bullet$ by
$(B_\bullet/A_\bullet)_n=B_n/A_n$.

For any arrow $f:A\to B$ in $\{\hbox{additive Abelian groups}\}$,
we define $\coker(f):=B/(\im(f))$; note that, with this definition,
$$0\to\ker(f)\to A\to B\to\coker(f)\to0$$
is exact at $\ker(f)$, at $A$, at $B$ and at $\coker(f)$.

For any arrow $f:A_\bullet\to B_\bullet$ in
$\{\hbox{additive Abelian groups}\}$,
we define $\coker(f):=B_\bullet/(\im(f))$;
note that, with this definition,
$$0\to\ker(f)\to A_\bullet\to B_\bullet\to\coker(f)\to0$$
is exact at $\ker(f)$, at $A_\bullet$, at $B_\bullet$ and at $\coker(f)$.

\vskip.1in\noindent
{\it Definition.}
A diagram in $\{\hbox{additive Abelian groups}\}$
of the form
$$0\to A\to B\to C\quad\quad\quad D\to E\to F\to0$$
is called a {\bf broken six} if it is exact
at $A$, $B$, $E$ and $F$.
A diagram in $\{\hbox{additive Abelian groups}\}$
of the form
$$0\to A\to B\to C\to D\to E\to F\to0$$
is called a {\bf snake six} if it is exact
at $A$, $B$, $C$, $D$, $E$ and $F$.

\vskip.1in
To get into the spirit of the terminology, reorganize the preceding
diagram so that $D$ lies under $A$, $E$ lies under $B$ and $F$ lies
under $C$.  Then the arrow from $C$ to $D$ has to follow a path that
gives it something of a snake shape.

Let $\scrsesag:=\{\hbox{short exact sequences of additive Abelian groups}\}$.
Let $\scrbrsx:=\{\hbox{broken sixes}\}$.
Let $\scrsnsx:=\{\hbox{snake sixes}\}$.
Let $\scra$ be the arrow category of $\scrsesag$.
Define a functor $\scrg:\scra\to\scrbrsx$ by:
for all $\phi=(f',f,f''):(A',A,A'')\to(B',B,B'')$ in $\scrsesag$,
let $\scrg(\phi)$ be the diagram
$$0\to\ker(f')\to\ker(f)\to\ker(f'')\qquad\qquad\qquad
\coker(f')\to\coker(f)\to\coker(f'')\to0.$$
Let $\scrf:\scrsnsx\to\scrbrsx$ be the forgetful functor.

\vskip.1in\noindent
{\it Snake Lemma.}
There is a functor $\scrh:\scra\to\scrsnsx$
such that $\scrf\circ\scrh=\scrg$.

\vskip.1in\noindent
{\it Proof:}
See the opening scene in the movie ``It's My Turn''
with Jill Clayburg and Michael Douglas. Alternatively,
spend quite some time chasing diagrams.
{\bf QED}

\vskip.1in\noindent
{\it Definition.}
A diagram in $\{\hbox{graded groups}\}$ of the form
$$A'_\bullet\to A_\bullet\to A''_\bullet$$
is said to be a {\bf broken long exact sequence}
if it is exact at $A_\bullet$.
It is said to be a {\bf broken complex}
if the composite arrow $A'_\bullet\to A''_\bullet$ is zero.

\vskip.1in\noindent
{\it Definition.}
A {\bf long exact sequence} consists of
\itemitem{(1)} a broken long exact sequence
$A'_\bullet\to A_\bullet\to A''_\bullet$; and
\itemitem{(2)} a {\bf connecting homomorphism}
$A''_\bullet\to A'_{\bullet-1}$

\noindent
such that
$$A''_{\bullet+1}\to A'_\bullet\to A_\bullet\to A''_\bullet\to A'_{\bullet-1}$$
is exact at $A'_\bullet$, at $A_\bullet$ and at $A''_\bullet$.

\vskip.1in
In the preceding diagram, the arrow $A''_{\bullet+1}\to A'_\bullet$
is obtained by applying the functor $C_\bullet\mapsto C_{\bullet+1}$
to the connecting homomorphism $A''_\bullet\to A'_{\bullet-1}$.

Let $\scrsescc$ be the category of short exact sequences of
chain complexes.
Let $\scrles$ be the category of long exact sequences.
Let $\scrbrles$ be the category of broken long exact sequences.
Define a functor $\scrg_0:\scrsescc\to\scrbrles$ by:
for any short exact sequence
$0\to A'_\bullet\to A_\bullet\to A''_\bullet$
of chain complexes, we define
$\scrg_0(0\to A'_\bullet\to A_\bullet\to A''_\bullet)$
to be the broken long exact sequence:
$$H_\bullet(A'_\bullet)\to H_\bullet(A_\bullet)\to H_\bullet(A''_\bullet).$$
Let $\scrf_0:\scrles\to\scrbrles$ be the forgetful functor.

\vskip.1in\noindent
{\it Snake Corollary.}
There is a functor $\scrh_0:\scrsescc\to\scrles$
such that $\scrf_0\circ\scrh_0=\scrg_0$.

\vskip.1in\noindent
{\it Proof:}
We leave it as an exercise to show that this follows from
the Snake Lemma.
Alternatively, spend quite some time chasing diagrams.
{\bf QED}

\vskip.1in\noindent
{\it Definition.}
Let $(X,A)$ be a topological pair.
We define $H_\bullet(X,A):=H_\bullet[(S_\bullet(X))/(S_\bullet(A))]$.

\vskip.1in
The functor $H_\bullet$ defined above goes from the category
of topological pairs to the category of graded groups.
We make the convention that $\Z[\emptyset]=\{0\}$
so that $S_\bullet(\emptyset)=0$.
Then, for any topological space $X$,
we have $H_\bullet(X,\emptyset)\cong H_\bullet(X)$.
More precisely,
$H_\bullet(\cdot,\emptyset)$ and $H_\bullet$
are equivalent functors.

For any topological pair $(X,B)$, we have a short
exact sequence of chain complexes:
$$0\to S_\bullet(B)\to S_\bullet(X)\to(S_\bullet(X))/(S_\bullet(B))\to0,$$
and, applying the Snake Corollary, we get a long exact sequence
whose broken form is
$$H_\bullet(B)\to H_\bullet(X)\to H_\bullet(X,B).$$

Slightly more generally, if $X$ is a topological space and
if $A\subseteq B\subseteq X$,
then we have a short exact sequence of chain complexes:
$$0\to(S_\bullet(B))/(S_\bullet(A))\to(S_\bullet(X))/(S_\bullet(A))
\to(S_\bullet(X))/(S_\bullet(B))\to0,$$
and, applying the Snake Corollary, we get a long exact sequence
whose broken form is
$$H_\bullet(B,A)\to H_\bullet(X,A)\to H_\bullet(X,B).$$

\vskip.1in\noindent
{\it Example.}
Let $X:=\R^2$ and let $A:=\R^2\backslash\{(0,0)\}$.
Then, as part of the above long exact sequence we get a diagram
in the category of Abelian groups
$$H_2(X)\to H_2(X,A)\to H_1(A)\to H_1(X)$$
which is exact at $H_2(X,A)$ and at $H_1(A)$.
Since $X$ is contractible, we see that $H_2(X)$ and $H_1(X)$ are
both trivial, and so exactness implies that the map
$H_2(X,A)\to H_1(A)$ is an isomorphism.
Since $A$ is homotopy equivalent to the circle $S^1$
and since $H_1(S^1)\cong\Z$,
we conclude that $H_2(X,A)\cong\Z$.

\vskip.1in
To clarify the importance of this last example,
we need to have some additional terminology.

\vskip.1in\noindent
{\it Definition.}
Let $X$ be a topological space, let $A\subseteq X$
and let $n\in\Z$.
We define $Z_n(X,A):=
\{\gamma\in S_n(X)\,|\,\partial_n\gamma\in S_{n-1}(A)\}$.
We define $B_n(X,A):=(B_n(X))+(S_n(A))$.

\vskip.1in
It is an unassigned exercise to show that the functor
$(X,A)\mapsto(Z_\bullet(X,A))/(B_\bullet(X,A))$
is equivalent to the functor $(X,A)\mapsto H_\bullet(X,A)$.

\vskip.1in\noindent
{\it Definition.}
Let $(X,A)$ be a topological pair and let $n\in\Z$.
For any $\alpha\in S_n(X)$, we say that $\alpha$
is a {\bf cycle mod $A$} if $\alpha\in Z_n(X,A)$.
For any $\alpha\in Z_n(X,A)$, we say that
$\alpha$ {\bf bounds mod $A$} if $\alpha\in B_n(X,A)$.
For any $\alpha,\beta\in Z_n(X,A)$, we say that
$\alpha$ and $\beta$ are {\bf homologous mod $A$}
if $\beta-\alpha$ bounds mod $A$.

\vskip.1in
Let $X:=\R^2$ and let $A:=\R^2\backslash\{(0,0)\}$.
Let $\sigma\in C(\Delta^2,X)$
and assume that $\partial_2(\sigma)\in S_1(A)$.
Relative homology allows us to define the ``number of times''
that $\sigma$ covers $(0,0)$, as follows:
We know that $\sigma$ is a cycle mod $A$,
{\it i.e.}, that $\sigma\in Z_2(X,A)$.
Following the preceding example,
fix an isomorphism $H_2(X,A)\to\Z$.
Such an isomorphism is called an {\bf orientation} of $\R^2$
at $(0,0)$.
Since $H_2(X,A)\cong(Z_2(X,A))/(B_2(X,A))$,
we obtain a composite homomorphism
$Z_2(X,A)\to(Z_2(X,A))/(B_2(X,A))\cong H_2(X,A)\cong\Z$.
The image of $\sigma$ under this map is what we will call
the ``number of times'' that $\sigma$ covers $(0,0)$.
Note that this number might be negative.
Intuitively, each time that $\sigma$ covers $(0,0)$,
it either covers it in an orientation-preserving way
or in an orientation reversing-way, and, to know which
is which require fixing an orientation at $(0,0)$
on $X$.
Each orientation-preserving covering of $(0,0)$ counts for $+1$,
while orientation-reversing covering of $(0,0)$ counts for $-1$.

\vskip.1in\noindent
{\it Excision Theorem.}
Let $X$ be a topological space.
For all $S\subseteq X$, let $\barS:=\Cl_X(S)$
and let $S^\circ:=\Int_X(S)$.
Let $U,A\subseteq X$
and assume that $\barU\subseteq A^\circ$.
Let $i:(X\backslash U,A\backslash U)\to(X,A)$ be the inclusion map.
Then $i_*:H_\bullet(X\backslash U,A\backslash U)\to H_\bullet(X,A)$
is an isomorphism.

\vskip.1in
We will sketch part of the proof in a moment, but first
let's mention an application.
Let $X:=S^2$ be the two-sphere.
Let $x,y\in X$ be distinct points.
(Imagine $x$ as the north pole, $y$ as the south pole.)
Let $U:=\{y\}$ and let $A:=X\backslash\{x\}$.
Then, by the Excision Theorem,
we ahve $H_2(X,A)\cong H_2(X\backslash U,A\backslash U)$.
Stereographic projection gives an isomorphism,
in the category $\{\hbox{topological pairs}\}$,
betwen $(X\backslash U,A\backslash U)$ and $(\R^2,\R^2\backslash\{0\})$.
Then, by the preceding example,
$H_2(X,A)\cong H_2(X\backslash U,A\backslash U)\cong\Z$.
An isomorphism $H_2(X,A)\to\Z$ is called
an {\bf orientation} of $S^2$ at $x$,
and allows us to count the number of times
that a given parametric $2$-simplex $\sigma$ in $S^2$ covers the
point $x$, counting $+1$ for orientation-preserving coverings
of $x$ and $-1$ for orientation-reversing coverings of $x$.
This is all provided that 
$\partial_2(\sigma)\in S_1(A)$.

Given a topological space $X$ and $\scru\subseteq\{\hbox{subsets of }X\}$,
of $X$, we define, for each $n\in\Z$,
$\displaystyle{S_n^\scru(X):=\sum_{U\in\scru}\,S_n(U)}$;
for any $\alpha\in S_n(X)$, we say that $\alpha$ is
{\bf subordinate} to $\scru$ if $\alpha\in S_n^\scru(X)$.

\vskip.1in\noindent
{\bf EXERCISE 14C:}
Recall that $e_0^1,e_1^1$ is the standard basis of $\R^2$.
Let $p:=e_0^1$, $q:=e_1^1$.
Recall that $\Delta^1:=\cvx\{p,q\}$.
Let $r:=(p+q)/2$.
Show that $[p,q]-([p,r]+[r,q])$ bounds.

\vskip.1in
Let $p,q,r$ be as in Exercise 14C.
Let $X$ be a topological space and let $\alpha\in C(\Delta^1,X)$.
We define the {\bf subdivision} of $\alpha$
to be $\Sd(\alpha):=\alpha_*([p,r]+[r,q])\in S_1(X)$
and note, by Exercise 14C,
that $\alpha-(\Sd(\alpha))$ bounds.
The subdivison map extends by $\Z$-linearity to a map
$\Sd:S_1(X)\to S_1(X)$ and this map has the property that,
for any $\alpha\in S_1(X)$, we have that
$\alpha-(\Sd(\alpha))$ bounds.

\vskip.1in\noindent
{\it Sketch of proof that, in the Excision Theorem,
the map $i_*:H_1(X\backslash U,A\backslash U)\to H_1(X,A)$
is surjective:}
Let $\alpha\in Z_1(X,A)$.
We wish to show that $\alpha$ is homologous mod $A$
to an element of $Z_1(X\backslash U,A\backslash U)$.

Let $\scrv:=\{X\backslash\barU,A^\circ\}$.
Then $\scrv$ is an open cover of $X$.
By Exercise 14C, if we subdivide~$\alpha$,
then we obtain a chain whose difference with $\alpha$ bounds.
Via a Lebesgue number argument, if we subdivide $\alpha$
sufficiently, we will obtain a chain $\beta$
such that $\beta-\alpha$ bounds and
such that $\beta$ that is subordinate to $\scrv$.
Then $\beta\in S_1^\scrv(X)=(S_1(X\backslash\barU))+(S_1(A^\circ))$.
Choose $\gamma\in S_1(X\backslash\barU)$
and $\delta\in S_1(A^\circ)$ such that $\beta=\gamma+\delta$.
Then $\beta$ is homologous to $\gamma$ mod $A$.
Then $\alpha$ is homologous to $\gamma$ mod $A$.
It suffices to show that $\gamma\in Z_1(X\backslash U,A\backslash U)$.

Since $\alpha$ is a cycle mod $A$,
it follows that $\gamma$ is a cycle mod $A$.

Then $\gamma\in(S_1(X\backslash\barU))\cap(Z_1(X,A))
\subseteq Z_1(X\backslash U,A\backslash U)$.
{\bf QED}

\vskip.1in\noindent
{\it Definition.}
A {\bf long exact ladder} is an arrow in the category
$\{\hbox{long exact sequences}\}$.

\vskip.1in
For any long exact sequence $L$,
let $A^L_\bullet\to B^L_\bullet\to C^L_\bullet$
denote its broken form,
and let $f^L:A^L_\bullet\to B^L_\bullet$ and $g^L:B^L_\bullet\to C^L_\bullet$
be the arrows in this broken form.
For any long exact sequence $L$,
let $h^L_\bullet:C^L_\bullet\to A^L_{\bullet-1}$ be the connecting
homomorphism of $L$.

For any long exact ladder $p:L\to M$, let
$\alpha^p:A^L_\bullet\to A^M_\bullet$,
$\beta^p:B^L_\bullet\to B^M_\bullet$ and
$\gamma^p:C^L_\bullet\to C^M_\bullet$
be the underlying arrows in $\{\hbox{graded groups}\}$.

A long exact ladder $p:L\to M$ is said to be {\bf right isomorphic}
if $\gamma^p:C^L_\bullet\to C^M_\bullet$ is an isomorphism in the
category $\{\hbox{graded groups}\}$.
Let $\scrc$ be the category
$\{\hbox{right isomorphic long exact ladders}\}$.

Let $\scra:\{\hbox{long exact ladders}\}\to\{\hbox{broken complexes}\}$
be defined by: for any long exact ladder $p:L\to M$,
$\scra(p)$ is the broken complex
$$A^L_\bullet\to B^L_\bullet\oplus A^M_\bullet\to B^M_\bullet,$$
where the first arrow is
$$a\mapsto(f^L(a),\alpha^p(a))
:A^L_\bullet\to B^L_\bullet\oplus A^M_\bullet,$$
and where the second arrow is
$$(b,a)\mapsto(\beta^p(b))-(f^M(a)):
B^L_\bullet\oplus A^M_\bullet\to B^M_\bullet.$$

\vskip.1in\noindent
{\it Proposition.}
Let $\scra:\{\hbox{long exact ladders}\}\to\{\hbox{broken complexes}\}$
be as described above.
Let $\scrf:\{\hbox{long exact sequences}\}\to
\{\hbox{broken long exact sequences}\}$ be the forgetful functor.
There is a functor $\scrbw:\scrc\to\{\hbox{long exact sequences}\}$
such that $\scra|\scrc=\scrf\circ\scrbw$.

\vskip.1in
The functor $\scrbw$ described in the last proposition
is the {\bf Barratt-Whitehead functor}.

\vskip.1in\noindent
{\it Definition.}
A {\bf Mayer-Vietoris triple} or {\bf MV triple} consists of
\itemitem{(1)} a topological space $X$; and
\itemitem{(2)} two subsets $A,B\subseteq X$

\noindent
such that, if $A^\circ$ and $B^\circ$ denote the
interiors of $A$ and $B$ in $X$, then $A^\circ\cup B^\circ=X$.

\vskip.1in
Let $\scrd$ denote the arrow category of
$\{\hbox{short exact sequences of chain complexes}\}$.
Recall that the Snake Corollary gives a functor
$$\{\hbox{short exact sequences of chain complexes}\}
\to\{\hbox{long exact sequences}\},$$
and this functor induces a functor
$\scrs:\scrd\to\{\hbox{long exact ladders}\}$.

We now define a functor $\scrm:\{\hbox{MV triples}\}\to\scrd$,
as follows: Let $(X,A,B)\in\scrm$.
Let $S_\bullet(A\cap B)\to S_\bullet(A)$ be induced by the
inclusion $A\cap B\to A$.
Let
$$S_\bullet(A)\to(S_\bullet(A))/(S_\bullet(A\cap B))$$
be the canonical map. Then
$$0\to S_\bullet(A\cap B)\to S_\bullet(A)\to
(S_\bullet(A))/(S_\bullet(A\cap B))\to0$$
is a short exact sequence of chain complexes, which we denote by $E$.
Similary, inclusion and canonical maps give a short exact sequence
$$0\to S_\bullet(B)\to S_\bullet(X)\to(S_\bullet(X))/(S_\bullet(B))\to0,$$
which we denote by $F$.
The inclusions $A\cap B\to B$ and $A\to X$
induce arrows $i:S_\bullet(A\cap B)\to S_\bullet(B)$
and $j:S_\bullet(A)\to S_\bullet(X)$.
Then a short diagram chase shows that there is a unique
arrow $k:(S_\bullet(A))/(S_\bullet(A\cap B))\to(S_\bullet(X))/(S_\bullet(B))$
such that $(i,j,k):E\to F$ is an object in $\scrd$.
We define $\scrm(X,A,B)=(i,j,k)$.

\vskip.1in\noindent
{\it Theorem (Mayer-Vietoris).}
For any MV triple $(X,A,B)$,
we have $\scrs(\scrm(X,A,B))\in\scrc$.

\vskip.1in\noindent
{\it Proof:}
Let $p:=\scrs(\scrm(X,A,B))$.
Let $L$ be the domain of $p$ and let $M$ be the target of $p$.
We wish to show that $\gamma^p:C^L_\bullet\to C^M_\bullet$ is an isomorphism.

We have $C^L_\bullet=H_\bullet(A,A\cap B)$ and $C^M_\bullet=H_\bullet(X,B)$.
Moreover, the map
$$\gamma^p:
H_\bullet(A,A\cap B)\to H_\bullet(X,B)$$
is induced by the inclusion $(A,A\cap B)\to(X,B)$.
Let $S:=X\backslash A$.
Then the closure in $X$ of $S$ is contained in the interior
in $X$ of $B$, therefore, by the Excision Theorem,
the inclusion ($X\backslash S,B\backslash S)\to(X,B)$
induces an isomorphism in homology.
However, $(A,A\cap B)=(X\backslash S,B\backslash S)$,
so we are done.
{\bf QED}

\vskip.1in\noindent
{\it Corollary (Mayer-Vietoris).}
Let $(X,A,B)$ be a MV triple.
Let
$$i:A\cap B\to A,\qquad j:A\cap B\to B,\qquad
v:A\to X,\qquad w:B\to X$$
be inclusion maps.
Define $f:H_\bullet(A\cap B)\to(H_\bullet(A))\oplus(H_\bullet(B))$
by $f(\sigma)=(i_*(\sigma),j_*(\sigma))$.
Define $g:(H_\bullet(A))\oplus(H_\bullet(B))\to H_\bullet(X)$
by $g(\sigma,\tau)=(v_*(\sigma))-(w_*(\tau))$.
Then there is a long exact sequence whose broken form is
$$H_\bullet(A\cap B)\to(H_\bullet(A))\oplus(H_\bullet(B))\to H_\bullet(X),$$
where the first arrow is $f$ and the second is $g$.

\vskip.1in\noindent
{\it Proof:}
Let $\scrbw$ be the Barratt-Whitehead functor.
Then the required long exact seqence is
$\scrbw(\scrs(\scrm(X,A,B)))$.
Let $\scrf:\{\hbox{long exact sequences}\}\to
\{\hbox{broken long exact sequences}\}$
be the forgetful functor.
We must show that $\scrf(\scrbw(\scrs(\scrm(X,A,B))))$
is the broken long exact sequence described in the statement
of the theorem.

By the preceding proposition,
$\scrf(\scrbw(\scrs(\scrm(X,A,B))))=
\scra(\scrs(\scrm(X,A,B)))$.
The result then follows directly from the definitions
of $\scra$, $\scrs$ and $\scrm$.
{\bf QED}

\vskip.1in\noindent
{\it Example.}
Let $X:=S^2$.
Let $x,y\in X$ be distinct.
(Think of the north pole and the south pole.)
Let $A:=X\backslash\{y\}$ and let $B:=X\backslash\{x\}$.
Then, by the Mayer-Vietoris corollary we have a
long exact sequence part of which reads
$$(H_2(A))\oplus(H_2(B))\to H_2(X)\to H_1(A\cap B)\to
(H_1(A)\oplus H_1(B)).$$
So, as $A$ and $B$ are contractible, we conclude that
$H_2(X)$ is isomoprhic to $H_1(A\cap B)$.
However $A\cap B$ is homotopy equivalent to $S^1$,
and we deduce from this that $H_1(A\cap B)\cong\Z$.
Then $H_2(S^2)\cong\Z$.

\vskip.1in
We comment that all the Mayer-Vietoris work that was done
with homology $H_\bullet:\{\hbox{topological spaces}\}\to
\{\hbox{graded groups}\}$ could as easily have been
done with reduced homology
$\tildeH_\bullet:\{\hbox{topological spaces}\}\to
\{\hbox{graded groups}\}$.
We would then have the result:

\vskip.1in\noindent
{\it Corollary (Mayer-Vietoris).}
Let $(X,A,B)$ be a MV triple,
and assume that $A\cap B\ne\emptyset$.
Let
$$i:A\cap B\to A,\qquad j:A\cap B\to B,\qquad
v:A\to X,\qquad w:B\to X$$
be inclusion maps.
Define $f:\tildeH_\bullet(A\cap B)\to
(\tildeH_\bullet(A))\oplus(\tildeH_\bullet(B))$
by $f(\sigma)=(i_*(\sigma),j_*(\sigma))$.
Define $g:(\tildeH_\bullet(A))\oplus(\tildeH_\bullet(B))\to\tildeH_\bullet(X)$
by $g(\sigma,\tau)=(v_*(\sigma))-(w_*(\tau))$.
Then there is a long exact sequence whose broken form is
$$\tildeH_\bullet(A\cap B)\to(\tildeH_\bullet(A))\oplus(\tildeH_\bullet(B))\to
\tildeH_\bullet(X),$$
where the first arrow is $f$ and the second is $g$.

\vskip.1in\noindent
{\bf EXERCISE DUE AT THE FINAL:}
Let $d\ge0$ be an integer.
Let $C$ be a closed subset of $\R^d$.
Prove, for any $n\in\Z$, that
$$\tildeH_n(\R^d\backslash C)\cong
\tildeH_{n+1}(\R^{d+1}\backslash(C\times\{0\})).$$
({\it Hint:} Find open contractible subsets $A,B\subseteq\R^{d+1}$
such that $A\cup B=\R^{d+1}\backslash(C\times\{0\})$
and such that $A\cap B=(\R^d\backslash C)\times\R.$)

\vskip.1in\noindent
{\it Corollary.}
Let $d\ge0$ be an integer.
Let $C$ and $D$ be closed subsets of $\R^d$
and that $C\ne\R^d\ne D$.
Assume that $C$ and $D$ are homeomorphic.
Then, for all $n\in\Z$,
we have $H_n(\R^d\backslash C)\cong H_n(\R^d\backslash D)$.

\vskip.1in
That is, homology cannot detect the difference
between the complements of two homeomorphic closed proper subsets
of Euclidean space.
Thus, while it is known that ``a knot is determined by its complement'',
this fact cannot be verified using homology, since any
two knot complements have the same homology.

One might think that the only use of the last corollary
is in showing how weak homology is, but, amazingly, we will
soon see that it eventually yields a very strong consequence:
an open mapping theorem for maps between equidimensional
topological manifolds.

\vskip.1in\noindent
{\it Proof of the last corollary:}
It suffices to show that
$\tildeH_n(\R^d\backslash C)\cong\tildeH_n(\R^d\backslash D)$.
Let $C':=C\times\{0\}^d\subseteq\R^{2d}$
and let $D':=D\times\{0\}^d\subseteq\R^{2d}$.
Let $f:C\to D$ be a homeomorphism.
Let $\Gamma:=\{(c,f(c))\,|\,c\in C\}$ be the graph of $f$.
We leave it as an unassigned exercise to verify,
via The Tietze Extension Theorem that
$\R^{2d}\backslash C'$ is homeomorphic to $\R^{2d}\backslash\Gamma$.
A similar argument shows that
$\R^{2d}\backslash D'$ is homeomorphic to $\R^{2d}\backslash\Gamma$.
Then $\tildeH_{d+n}(\R^{2d}\backslash C')\cong
\tildeH_{d+n}(\R^{2d}\backslash D')$.

By $d$ applications of the preceding exercise,
we have
$\tildeH_{d+n}(\R^{2d}\backslash C')\cong\tildeH_n(\R^{2d}\backslash C)$.
Similarly, we have
$\tildeH_{d+n}(\R^{2d}\backslash D')\cong\tildeH_n(\R^{2d}\backslash D)$.
The result follows.
{\bf QED}

\vskip.1in
Recall that a topological space is a {\bf simple closed curve}
if it is homeomorphic to the circle $S^1$.

\vskip.1in\noindent
{\it The Jordan Curve Theorem.}
Let $C$ be a simple closed curve in $\R^2$.
Then $\R^2\backslash C$ has two connected components.

\vskip.1in\noindent
{\it Proof:}
By the preceding corollary, we have
$H_0(\R^2\backslash C)\cong H_0(\R^2\backslash S^1)$,
so $H_0(\R^2\backslash C)\cong\Z\oplus\Z$.
Then $\R^2\backslash C$ has two connected components.
{\bf QED}

\vskip.1in
An important issue that often comes up in topology is when one
can conclude that a continuous bijection is a homeomorphism.
This is not always true ({\it e.g.}, the map
$t\mapsto(\cos(t),\sin(t)):[0,2\pi)\to S^1$),
but there are a number of ``open mapping'' theorems in
mathematics that clarify important cases where it {\it is} true.

For example, if $K$ is a compact topological space,
and if $g:K\to L$ is continuous, then,
because all closed subsets of $K$ are compact,
it follows that $g$ is a closed map.
Consequently, with the additional assumption
that $g:K\to L$ is bijective
(which implies that the image of the complement
is the complement of the image),
we may conclude that $g:K\to L$ is open.
A continuous open bijection is a homeomorphism,
and so we conclude that:
If $K$ is compact,
and if $g:K\to L$ is a continuous bijection,
then $g$ is a homeomorphism.
This then has the following consequence:

\vskip.1in\noindent
{\it Fact.}
Let $X$ and $Y$ be topological spaces
and let $f:X\to Y$ be a continuous bijection.
Then, for any compact subset $K\subseteq X$,
we have: $K$ and $f(K)$ are homeomorphic.

\vskip.1in
We aim for another example, proved via homology,
that many continuous bijections are open.
This is a surprising argument, because
it might seem difficult to verify openness of a set
via as weak an invariant as homology.
In fact, an open disk in $\R^2$ is homotopy equivalent
to a closed disk, and therefore the two have the same homology.
So, using only homology, it might seem difficult
to distinguish between open sets and closed sets.
However, the following clever observation
belies this impression.

\vskip.1in\noindent
{\it Remark.}
Let $X$ be a locally path-connected topological space
and let $P,Q\subseteq X$.
Assume that $P$ is closed in $X$
and that $P\cap Q=\emptyset$.
Assume the following:
\itemitem{(1)} $H_0(X\backslash P)\not\cong\Z$;
\itemitem{(2)} $H_0(Q)\cong\Z$; and
\itemitem{(3)} $H_0(X\backslash(P\cup Q))\cong\Z$.

\noindent
Then $Q$ is open in $X$.

\vskip.1in\noindent
{\it Proof:}
Let $Y:=X\backslash P$.
Then, as $X$ is locally path-connected, $Y$ is as well.
Then the path-components of $Y$ are open in $Y$,
and therefore are open in $X$.

By (2), $Q$ is path-connected.
Let $A$ denote the path-component of $Y$ containing $Q$.
It suffices to show that $A=Q$.

By (1), $H_0(Y)\ne\Z$, so $Y$ is not path-connected,
so $A\subsetneq Y$, and, in particular, $Y\not\subseteq A$.
Then $Q\cup(Y\backslash Q)=Y\not\subseteq A$,
so, since $Q\subseteq A$,
we conclude that $Y\backslash Q\not\subseteq A$.

By (3), we have $H_0(Y\backslash Q)\cong\Z$,
so $Y\backslash Q$ is path-connected.
So, because $Y\backslash Q\not\subseteq A$
and because $A$ is a path-component of $Y$,
we conclude that $(Y\backslash Q)\cap A=\emptyset$,
or, equivalently, $A\subseteq Q$.
Then, as $Q\subseteq A$, we get $A=Q$.
{\bf QED}

\vskip.1in\noindent
{\it Proposition.}
Let $d\ge0$ be an integer.
Let $f:\R^d\to\R^d$ be a continuous, injective map.
Then $f:\R^d\to\R^d$ is open.

\vskip.1in\noindent
{\it Proof:}
Let $B$ be an open ball in $\R^d$ and let $Q:=f(B)$.
We wish to show that $Q$ is open in $\R^d$.

Let $\partial B$ denote the boundary of $B$ in $\R^d$.
Let $P:=f(\partial B)$.
As $\partial B$ and $B\cup\partial B$ are both compact subsets of $\R^d$,
the preceding Fact yields both
that $P$ is homeomorphic to $\partial B$ and
that $P\cup Q$ is homeomorphic to $B\cup\partial B$.

Then, by the preceding corollary,
we get both that
$H_0(\R^d\backslash P)\cong H_0(\R^d\backslash\partial B)\cong\Z\oplus\Z$
and that
$H_0(\R^d\backslash(P\cup Q))\cong
H_0(\R^d\backslash(B\cup\partial B))\cong\Z$.
Finally, since $B$ is path-connected,
$Q$~is as well, and so $H_0(Q)\cong\Z$.
Then, by the last remark, with $X:=\R^d$,
we see that $Q$~is open in $\R^d$.
{\bf QED}

\vskip.1in
Again, remarkably, although homology seems too
weak to prove openness of anything,
and although preceding corollary highlights
the weakness of homology (it cannot distinguish
between the complements of homeomorphic proper closed sets
of Euclidean space),
nevertheless, the preceding corollary was a primary tool
in proving an open mapping theorem.

We can generalize this last proposition,
once we have the notion of a topological manifold:

\vskip.1in\noindent
{\it Definition.}
Let $d\ge0$ be an integer.
A topological space is said to be a {\bf topological $d$-manifold}
if every point has an open neighborhood that is homeomorphic to $\R^d$.

\vskip.1in
We leave it as an unassigned exercise to verify that, in a topological
$d$-manifold, every point has arbitrarily small open neighborhoods
that are homeomorphic to $\R^d$. That is:

\vskip.1in\noindent
{\it Remark.}
Let $d\ge0$ be an integer.
Let $X$ be a topological $d$-manifold.
Let $x\in X$ and let $V$ be an open neighborhood of $x$ in $X$.
Then there is an open neighborhood $X_0$ of $x$ in $X$
such that $X_0\subseteq V$ and
such that $X_0$ is homeomorphic to $\R^d$.

\vskip.1in
Note that because of the preceding remark,
any open subset of a topological $d$-manifold
is again a topological $d$-manifold.

\vskip.1in\noindent
{\it Theorem.}
Let $d\ge0$ be an integer.
Let $X$ and $Y$ be topological $d$-manifolds.
Let $f:X\to Y$ be a continuous, injective map.
Then $f:X\to Y$ is open.

\vskip.1in\noindent
{\it Proof:}
Let $U$ be an open subset of $X$.
We wish to show that $f(U)$ is open in $Y$.
Let $y\in f(U)$.
We wish to show that there is an open neighborhood $N$ of $y$ in $Y$
such that $N\subseteq f(U)$.

Let $Y_0$ be an open neighborhood of $y$ in $Y$
such that $Y$ is homeomorphic to $\R^d$.
Fix $x\in U$ such that $f(x)=y$.
By the preceding remark, let $X_0$ be an open neighborhood of $x$ in $X$
such that $X_0\subseteq U\cap(f^{-1}(Y_0))$ and
such that $X_0$ is homeomorphic to $\R^d$.

Then $f|X_0:X_0\to Y_0$ is continuous and injective,
so, by the preceding proposition, $f|X_0:X_0\to Y_0$ is open.
Then $N:=f(X_0)$ is an open subset of $Y$.
Moreover, we have $y=f(x)\in f(X_0)=N$.
Finally, we have $N=f(X_0)\subseteq f(U)$.
{\bf QED}

\vskip.1in\noindent
{\it Corollary.}
Let $d\ge0$ be an integer.
Let $X$ and $Y$ be topological $d$-manifolds.
Let $f:X\to Y$ be a continuous, bijective map.
Then $f:X\to Y$ is a homeomorphism.

\vskip.1in\noindent
{\it Corollary.}
Let $d\ge0$ be an integer.
Let $U,S\subseteq\R^d$.
Assume that $U$ is open in $\R^d$.
Assume that $U$ and $S$ are homeomorphic.
Then $S$ is open in $\R^d$.

\vskip.1in\noindent
{\it Proof:}
Let $f:U\to S$ be a homeomorphism.
By the preceding theorem,
$f:U\to\R^d$ is open,
so $f(U)$ is open in $\R^d$.
That is, $S$ is open in $\R^d$.
{\bf QED}

\vskip.1in
The preceding corollary is sometimes called
{\bf Invariance of Domain}.

A classical question is whether a nonempty open subset of
Euclidean space can be homeomorphic to a nonempty open subset
of another. The fact that the answer is negative is
also sometimes called {\bf Invariance of Domain},
since a connected open subset
of Euclidean space is sometimes called a ``domain'',
and since such a fact asserts that dimension is a
topological invariant of domains.

In fact, we can now prove something {\it a priori} stronger:

\vskip.1in\noindent
{\it Corollary.}
Let $d,k\ge0$ be integers and assume that $d\ne k$.
Let $X$ be a topological $k$-manifold
and let $Y$ be a topological $d$-manifold.
Then $X$ is not homeomorphic to $Y$.

\vskip.1in\noindent
{\it Proof:}
Assume, for definiteness, that $d<k$; the proof is otherwise similar.
Let $f:X\to Y$ be a homeomorphism, and we aim for a contradiction.

Choose an open subseteq $X_0$ of $X$
such that $X_0$ is homeomorphic to $\R^k$.
Choose a subset $Z$ of $X_0$
such that $Z$ is homeomorphic to $\R^d$
and such that $Z$ is not open in $X_0$.
Then $Z$ is not open in $X$.

The map $f|Z:Z\to Y$ is a continuous injection, so, by the
preceding thorem, $f(Z)$ is open in $Y$.
Then, as $f:X\to Y$ is a homeomorphism,
we conclude that $Z$ is open $X$, a contradiction.
{\bf QED}

\vskip.1in
A topological space is a {\bf topological manifold}
if there exists some integer $d\ge0$ such that it
is a topological $d$-manifold.

The preceding corollary tells us that, if $X$ is
a topological manifold, then its {\bf dimension}, $\dim(X)$,
is well-defined, via: $\dim(X)$ is the unique integer
$d\ge0$ such that $X$ is a topological $d$-manifold.

Here is an unassigned, but interesting exercise to ponder:
Let $X$ and $Y$ be topological manifolds.
Let $f:X\to Y$ be a continuous bijection.
Show that $f:X\to Y$ is a homeomorphism.

\vskip.1in\noindent
{\bf START OF WINTER SEMESTER:}

\vskip.1in\noindent
{\it Definition.}
Let $m\ge0$ be an integer and let $P:\R^m\to\R$.
We say that $P$ is {\bf homogeneous quadratic}
if there exist $(c_{ij})\in\R^{m\times m}$
such that, for all $(x_1,\ldots,x_m)\in\R^m$,
we have
$$P(x_1,\ldots,x_m)=\sum_{i,j=1}^m\,c_{ij}x_ix_j.$$

\vskip.1in
Note that the matrix $(c_{ij})$ described above is
not unique. However, if we require it to be symmetric
then it is unique. Alternatively, if we require it to be
upper triangular, then it is unique as well.

\vskip.1in\noindent
{\it Definition.}
Let $m,n\ge0$ be integers and let $P:\R^m\to\R^n$.
For all integers $k\in[1,n]$, let $q_k:\R^n\to\R$
be the $k$th coordinate projection,
$q_k(y_1,\ldots,y_n)=y_k$.
Then we say that $P$ is {\bf homogeneous quadratic}
if, for all integers $k\in[1,n]$,
we have that $q_k\circ P:\R^m\to\R$ is homogeneous quadratic.

\vskip.1in
By {\bf vector space} we will always mean {\it real} vector
space, unless otherwise specified. By $\dim(V)$ we will always
mean the real dimension of the vector space $V$, unless otherwise
specified.

\vskip.1in\noindent
{\it Definition.}
Let $V$ and $W$ be vector spaces and let $Q:V\to W$.
We say that $Q$ is {\bf homogenous quadratic} if
there exist integers $m,n\ge0$ and
isomorphsims $f:\R^m\to V$ and $g:W\to\R^n$
such that $g\circ Q\circ f:\R^m\to\R^n$ is homogeneous quadratic.

\vskip.1in In freshman calculus, functions $f:\R\to\R$ are
approximated by inhomogeneous linear functions $x\mapsto a+bx$, and
then, later, by quadratic Taylor polynomials $x\mapsto a+bx+cx^2$.  In
higher dimensions, a function $f:\R^m\to\R^n$ is approximated by the
sum of a constant and a homogeneous linear map $L:\R^m\to\R^n$.
The analysis of $L$ involves developing an entirely new branch
of mathematics called ``linear algebra''.

One also wants to do quadratic Taylor polynomials, in which
$f:\R^m\to\R^n$ is approximated by a sum of a constant,
a homogeneous linear function $L:\R^m\to\R^n$, and a
homogeneous quadratic function $Q:\R^m\to\R^n$.

Question: Why is there no subject called ``quadratic algebra''?

Answer: For any vector spaces $V$ and $W$, for any homogeneous
quadratic $Q:V\to W$, there exists a unique symmetric bilinear
$B:V\times V\to W$ such that, for all $v\in V$,
we have $Q(v)=B(v,v)$.

\vskip.1in\noindent
{\bf EXERCISE 15A:}
Let $V:=\R^2$, $W:=\R$ and define $Q$ by $Q(x,y)=x^2+2xy+3y^2$.
Find $B:V\times V\to W$.

\vskip.1in
Thus questions about quadratic algebra can be recast as questions
about symmetric bilinear algebra. So quadratic algebra is subsumed
in bilinear algebra. (In a similar way, cubic algebra is
subsumed in trilinear algebra, {\it etc.})
But this just begs the:

Question: Why is there no subject called ``bilinear algebra''?

\vskip.1in\noindent
{\it Definition.}
Let $V$, $W$ and $X$ be vector spaces and let $B:V\times W\to X$
be bilinear. Then we say that $B$ is {\bf universal}
if, for any vector space $Y$, for any bilinear map $C:V\times W\to Y$,
there is a unique linear map $L:X\to Y$ such that $C=L\circ B$.

\vskip.1in
Answer: For any vector spaces $V$ and $W$,
there is a vector space $X$ and
a universal bilinear map $B:V\times W\to X$.

We will explain a construction of $X$ and $B$ below,
but if we can find such $X$ and $B$,
then any question about bilinear maps
from $V\times W$ to another vector space $Y$ can be recast
as a question about linear maps from $X$ to $Y$.
Thus the subject of bilinear algebra is subsumed in
linear algebra.

\vskip.1in\noindent
{\bf EXERCISE 15B:}
Let $V$, $W$, $X$ and $X'$ be vector spaces. Let
$$B:V\times W\to X\qquad\hbox{and}\qquad B':V\times W\to X'$$
be universal bilinear maps.
Prove that there is a vector space isomorphism $L:X\to X'$
such that $L\circ B=B'$.

\vskip.1in
Exercise 15B asserts that, given $V$ and $W$,
up to simply renaming the elements of $X$,
there is a unique $X$ and universal map $V\times W\to X$.

Let's now address existence.
Given $V$ and $W$, we will construct a vector space
denoted $V\otimes_\R W$ and a universal bilinear
$V\times W\to V\otimes_\R W$.

Let $X_0:=\R[V\times W]$ be the
vector space of formal finite $\R$-linear combinations
of elements of the set $V\times W$.
Let $K$ be the vector subspace of $X_0$ spanned by the union of the
following four sets:
$$\eqalign{
\{&a(v,w)-(av,w)\,|\,a\in\R,v\in V,w\in W\},\cr
\{&a(v,w)-(v,aw)\,|\,a\in\R,v\in V,w\in W\},\cr
\{&(v+v',w)-(v,w)-(v',w)\,|\,v,v'\in V,w\in W\},\cr
\{&(v,w+w')-(v,w)-(v,w')\,|\,v\in V,w,w'\in W\}.}$$
Let $V\otimes_\R W:=X_0/K$.

Let $i:V\times W\to X_0$ be the inclusion
and let $p:X_0\to V\otimes_\R W$ be the canonical map.
The standard notation for the image of $(p\circ i)(v,w)$ is $v\otimes w$.
Then the map
$$p\circ i=(v,w)\mapsto v\otimes w:V\times W\to V\otimes_\R W$$
will be called the {\bf canonical bilinear map},
and we leave it as an unassigned exercise to verify
that it is a universal bilinear map.

\vskip.1in\noindent
{\bf EXERCISE 15C:}
Show that the matrix multiplication map
$R^{3\times1}\times\R^{1\times4}\to\R^{3\times4}$
is universal.

\vskip.1in\noindent
{\bf EXERCISE 15D:}
Compute $\dim(\R^3\otimes_\R\R^4)$.

\vskip.1in
All rings will be assumed to have multiplicative unity $1$.

In fact, in all of the preceding remarks, we may replace $\R$
by any commutative ring~$R$, as long as we use ``$R$-module''
as a replacement for ``vector space''. (Note that an $\R$-module
is the same as a vector space.)

Note that any additive Abelian group $A$ may be thought of as a $\Z$-module,
in which, for all $n\in\Z$, for all $a\in A$,
$na$ is defined to be the $n$-fold sum $a+\cdots+a$.
Moreover any $\Z$-module has an underlying additive Abelian group.
Thus the category of $\Z$-modules is isomorphic to the category
of additive Abelian groups.

A $\Z$-linear map $A\to B$ is the same as an additive group
homomorphism. A $\Z$-bilinear map $A\times B\to C$ will
sometimes be called a {\bf bihomomorphism}.

Then, for any additive Abelian groups $A$ and $B$,
we construct an additive Abelian group $A\otimes_\Z B$
and a bihomomorphism $S:A\times B\to A\otimes_\Z B$
which is {\bf universal} in the sense that,
for any additive Abelian group $C$,
for any bihomomorphism $T:A\times B\to C$
there is a homomorphism $H:A\otimes_\Z B\to C$
such that $T=H\circ S$.

We will abbreviate $\otimes_\Z$ as $\otimes$.
For any integer $n$ we abbreviate $\Z/(n\Z)$ as $Z/n$.

As with Exercise 15B, for any universal bihomomorphism
$A\times B\to D$, we have that:
$D$ is isomorphic to $A\otimes B$.

\vskip.1in\noindent
{\bf EXERCISE 15E:}
Let $m,n\in\Z$ and assume that $(m,n)\ne(0,0)$.
Let $g$ be the gcd of $m$ and $n$.
Let $\phi:\Z/m\to\Z/g$ and $\psi:\Z/n\to\Z/g$
be canonical maps.
Show that the map
$$(x,y)\quad\mapsto\quad[\phi(x)][\psi(y)]\qquad:\qquad
(\Z/m)\times(\Z/n)\quad\to\quad\Z/g$$
is universal.

\vskip.1in
We conclude, from Exercise 15E, that $(\Z/m)\otimes(\Z/n)$
is isomorphic to $\Z/g$.

\vskip.1in\noindent
{\bf EXERCISE 15F:}
Show, for any additive Abelian group $A$,
that $(n,a)\mapsto na:\Z\times A\to A$
is universal.

\vskip.1in
We conclude, from Exercise 15F, that $\Z\otimes A$
is isomorphic to $A$. In fact, one may show that
this isomorphism is ``natural'' in the sense that
the functor
$$\Z\otimes\cdot\qquad:\qquad\{\hbox{additive Abelian groups}\}
\to\{\hbox{additive Abelian groups}\}$$
is equivalent to the identity functor.

\vskip.1in\noindent
{\bf EXERCISE 15G:}
Show, for any additive Abelian groups $A$, $B$ and $C$,
that
$$((a,b),c)\mapsto(a\otimes c,b\otimes c)\qquad:\qquad
(A\oplus B)\times C\to(A\otimes C)\oplus(B\otimes C)$$
is universal.

\vskip.1in
We conclude, from Exercise 15F, that
$(A\oplus B)\otimes C$
is isomorphic to $(A\otimes C)\oplus(B\otimes C)$.
A similar argument shows that
$A\otimes(B\oplus C)$
is isomorphic to $(A\otimes B)\oplus(A\otimes C)$.

\vskip.1in\noindent
{\bf EXERCISE 15H:}
Compute $[(\Z/3)\oplus(\Z/27)\oplus(\Z/54)]
\otimes[(\Z/2)\oplus(\Z/4)\oplus(\Z/12)]$
up to isomorphism.
Express your answer in the form
$(\Z/a)\oplus(\Z/b)\oplus(\Z/c)\oplus\cdots$
such that $a|b|c|\cdots$.

\vskip.1in\noindent
{\it Definition.}
Let $A$, $B$, $C$ and $D$ be additive Abelian groups.
Let $f:A\to B$ and $g:C\to D$ be homomorphisms.
Then we define $f\otimes g:A\otimes C\to B\otimes D$
by $(f\otimes g)(a\otimes c)=(f(a))\otimes(g(c))$.

\vskip.1in
We remark that it is an exercise to show that this
is well-defined. (Note that every element of $A\otimes C$
is a finite sum of elements from $\{a\otimes c\,|\,a\in A,c\in C\}$,
however a given element of $A\otimes C$ may be expressed
as such as sum in two very different ways.)

To see that the map is well-defined, one needs only
observe that
$$(a,c)\mapsto(f(a))\otimes(g(c)):A\times C\to B\otimes D$$
is a bihomomorphism and therefore determines a homomorphism
$$H:A\otimes C\to B\otimes D$$
such that the composite
$$A\times C\to A\otimes C\to B\otimes D$$
is
$(a,c)\mapsto(f(a))\otimes(g(c)):A\times C\to B\otimes D$.
Thus the homomorphism $H$ has the property that
$H(a\otimes c)=(f(a))\otimes(g(c))$.
Then $f\otimes g=H$, so, as $H$ is well-defined by construction,
we see that $f\otimes g$ is, as well.

\vskip.1in
For any additive Abelian groups $A$ and $B$, we set
$A^B:=A\otimes B$.
Then, for any $B$, we have a functor
$A\mapsto A^B:\{\hbox{additive Abelian groups}\}
\to\{\hbox{additive Abelian groups}\}$
whose effect on arrows is defined by
$f^B:=f\otimes\id_B$.

\vskip.1in\noindent
{\it Fact.}
If $0\to A'\to A\to A''\to0$ is exact, then
$(A')^B\to A^B\to(A'')^B\to0$ is exact.

\vskip.1in
In short, we say that $A\mapsto A^B$ is ``right-exact''.

\vskip.1in\noindent
{\bf EXERCISE 15I:}
Let $i:2\Z\to\Z$ be the inclusion map and let $B:=\Z/2$.
show that $i^B$ is not injective.

\vskip.1in
For all additive Abelian groups $A$, let
$\phi_A:\Z[A]\to A$ denote the extension
by $\Z$-linearity of the identity map $A\to A$,
and let $K_A$ denote the kernel of $\phi_A$.
Let $\iota_A:K_A\to\Z[A]$ denote the injection map.

\vskip.1in\noindent
{\it definition.}
For any additive Abelian groups $A$ and $B$,
let $\Tor(A,B)$ denote the kernel of
$(\iota_A)^B:(K_A)^B\to(\Z[A])^B$.
This is called the {\bf torsion product} of $A$ and $B$
and is, alternatively, denoted $A*B$.

\vskip.1in
An additive Abelian group $F$ is said to be {\bf free Abelian}
if there exists a set $S$ such that $F$ is isomorphic to $\Z[S]$.
It is a standard fact, which we will assume without proof,
that any subgroup of a free Abelian group is again free Abelian.

A subset $S$ of an additive Abelian group $A$ is said to be a
{\bf generating set} if the identity map $s\to s$ extends,
by $\Z$-linearity, to a surjection $\Z[S]\to A$.
It is said to be a {\bf free generating set}
if the identity map $s\to s$ extends,
by $\Z$-linearity, to an isomorphism $\Z[S]\to A$.
Note that an additive Abelian group is free Abelian
iff it admits a free generating set.

Let $A$ and $B$ be additive Abelian groups
and let $f:A\to B$ be a surjection.
A {\bf splitting} of $F$ is
a homomorphism $g:B\to A$ such that $f\circ g=\id_B$.
We say that $f$ {\bf splits} if $f$ admits a splitting.
A {\bf split surjection} is a surjection which splits.

\vskip.1in\noindent
{\bf EXERCISE 15J:}
Let $A$ and $B$ be additive Abelian groups
and let $f:A\to B$ be a split surjection.
Let $K$ be the kernel of $f$.
Show that there is an isomorphism $h:K\oplus B\to A$
such that, for all $k\in K$, for all $b\in B$
we have $f(h(k,b))=b$.

\vskip.1in
That is, in the category of
``additive Abelian groups with a homomorphism to $B$''
the pair $(A,f)$ is isomorphic to $(B\oplus K,\pi)$,
where $\pi:K\oplus B\to B$ is projection onto the second factor.

\vskip.1in\noindent
{\it Remark.}
Let $A$ be an additive Abelian group
and let $F$ be a free Abelian group.
Then any surjection $f:A\to F$ is split.

\vskip.1in\noindent
{\it Proof:}
Let $S$ be a free generating set for $F$.
For all $s\in s$, choose $a_s\in A$ such that $f(a_s)=s$.
Extend $s\mapsto a_s:S\to A$ by $\Z$-linearity to a map $g:F\to A$.
Then $f\circ g=\id_F$.
{\bf QED}

\vskip.1in\noindent
{\bf EXERCISE 15K:}
Let $A$ be an additive Abelian group
and let $F$ and $F'$ be free Abelian groups.
Let $0\to F'\to F\to A\to0$ be a short exact sequence.
show that $A*B$ is isomorphic to the kernel of $(F')^B\to F^B$.

\vskip.1in\noindent
{\bf EXERCISE 15L:}
Let $m,n\in\Z\backslash\{0\}$.
Compute $(\Z/m)*(\Z/n)$.

\vskip.1in\noindent
{\bf EXERCISE 15M:}
For any additive Abelian group $A$, show that $A*\Z=0$.

\vskip.1in\noindent
{\bf EXERCISE 15N:}
For any additive Abelian group $A$, show that $\Z*A=0$.

\vskip.1in\noindent
{\bf EXERCISE 15O:}
For any additive Abelian groups $A$, $B$ and $C$,
show that $A*(B\oplus C)=(A*B)\oplus(A*C)$.

\vskip.1in\noindent
{\bf EXERCISE 15P:}
For any additive Abelian groups $A$, $B$ and $C$,
show that $(A\oplus B)*C=(A*C)\oplus(B*C)$.

\vskip.1in
In the preceding exercises, do not use the fact
that $A*B$ is isomorphic to $B*A$.
We will assume this fact from here on out.
Note that Exercise 15M and Exercise 15N both guarantee that $A*B$
is not the same as $A\otimes B$.

\vskip.1in\noindent
{\bf EXERCISE 15Q:}
Compute
$[(\Z/3)\oplus(\Z/27)\oplus(\Z/54)]
*[(\Z/2)\oplus(\Z/4)\oplus(\Z/12)]$,
up to isomorphism.
Put your answer in the form
$(\Z/a)\oplus(\Z/b)\oplus(\Z/c)\oplus\cdots$,
with $a|b|c|\cdots$.

\vskip.1in
For any integer $n\ge1$, for any integer $i\in[0,n]$,
define $\varepsilon_i^n:\Delta^{n-1}\to\Delta^n$
by $\varepsilon_i^n(x_1,\ldots,x_n)=(x_1,\ldots,x_i,0,x_{i+1},\ldots,x_n)$.
For any integer $n\ge1$, for any integer $i\in[0,n]$,
define $\partial_i^n:C(\Delta^n,X)\to C(\Delta^{n-1},X)$
by $\partial_i^n(\sigma)=\sigma\circ\varepsilon_i^n$.

For any topological space $X$,
for any integer $n\ge0$,
let $S_n(X;A):=A[C(\Delta^n,X)]$ be the additive
Abelian group of formal finite $A$-linear combinations
of elements of $C(\Delta^n,X)$.
For any topological space $X$,
for any integer $n<0$, let $S_n(X;A):=\{0\}$ be the
trivial additive Abelian group.
For any topological space $X$,
for any integer $n\ge1$,
let $\partial_n:S_n(X;A)\to S_{n-1}(X;A)$ be
the unique homomorphism extending
$$a\sigma\qquad\mapsto\qquad
\sum_{i=0}^n\,(-1)^ia[\partial_i^n(\sigma)].$$
For any topological space $X$,
for any integer $n<1$,
let $\partial_n:S_n(X;A)\to S_{n-1}(X;A)$ be the zero map.

\vskip.1in\noindent
{\bf EXERCISE 16A:}
Show, for all $n\in\Z$, that $S_n(X;A)$
is isomorphic to $(S_n(X))\otimes A$.

\vskip.1in
In addition to Exercise 15R,
we remark that, for all $n\in Z$,
the map $\partial_n:S_n(X;A)\to S_{n-1}(X;A)$
is the result of applying the functor
$B\mapsto B^A:
\{\hbox{additive Abelian groups}\}\to
\{\hbox{additive Abelian groups}\}$ to the map
$\partial_n:S_n(X)\to S_{n-1}(X)$.

Note that $B\mapsto B^A$ defines a functor
$\{\hbox{chain complexes}\}\to\{\hbox{chain complexes}\}$.
With this notation, we summarize the preceding remars as:
$S_\bullet(X;A)\cong(S_\bullet(X))^A$.

We now pose a

\vskip.1in\noindent
{\it Question:}
If we know, for a topological space $X$,
how to compute $H_\bullet(X)$,
can we then compute $H_\bullet(X;A)$.

\vskip.1in
The philosophy of the Universal Coefficient Theorem
is that the answer is ``yes'', provided that we also
know how to compute $\otimes$ and $*$,
{\it i.e.}, how to compute tensor and torsion products.

\vskip.1in\noindent
{\it Definition.}
A chain complex $C_\bullet$ is {\bf free},
if, for all $n\in\Z$, we have that $C_n$ is free.

\vskip.1in
Note, for any topological space $X$, that $S_\bullet(X)$ is free.

\vskip.1in\noindent
{\it Universal Coefficient Theorem.}
Let $C_\bullet$ be a free chain complex,
let $A$ be an additive Abelian group.
Then, for all $n\in Z$, $H_n((C_\bullet)^A)$ is isomorphic
to $[(H_nC_\bullet)^A]\oplus[(H_{n-1}C_\bullet)*A]$.

\vskip.1in\noindent
{\it Corollary.}
Let $X$ be a topological space,
let $A$ be an additive Abelian group and
let $n\in\Z$.
Then $H_n(X;A)$ is isomorphic to
$[(H_nX)^A]\oplus[(H_{n-1}X)*A]$.

\vskip.1in
For the next exercise, you may assume, for any integer $n\ge0$,
the following facts:
\itemitem{(1)} We have $H_0(\R P^n)=\Z$;
\itemitem{(2)} For any even integer $k\in[1,n-1]$,
we have $H_k(\R P^n)=0$;
\itemitem{(3)} For any odd integer $k\in[1,n-1]$,
we have $H_k(\R P^n)=\Z/2$;
\itemitem{(4)} If $n$ is even, then $H_n(\R P^n)=0$;
\itemitem{(5)} If $n$ is odd, then $H_n(\R P^n)=\Z$;
\itemitem{(6)} For any integer $k\in(-\infty,-1]\cup[n+1,\infty)$,
we have $H_k(\R P^n)=0$.

\vskip.1in\noindent
{\bf EXERCISE 16B:}
Compute $H_\bullet(\R P^3;\Z/2)$.

\vskip.1in\noindent
{\it Proof of the Universal Coefficient Thoerem:}
For any $n\in\Z$,
let $Z_n:=Z_n(C_\bullet)$,
let $B_n:=B_n(C_\bullet)$,
let $H_n:=H_n(C_\bullet)=Z_n/B_n$.
We wish to show that
$$H_n((C_\bullet)^A)\cong[(H_n)^A]\oplus[H_{n-1}*A].$$

For any $n\in\Z$,
as $B_{n-1}$ is a subgroup of the free Abelian group $C_{n-1}$,
we see that $B_{n-1}$ is free Abelian, and so the surjection
$\partial_n:C_n\to B_{n-1}$ splits.
Then, by Exercise 15J, we obtain, for all $n\in\Z$,
an isomorphism $C_n\cong Z_n\oplus B_{n-1}$.

For all $n\in\Z$, let $i_n:B_n\to Z_n$ be the inclusion map.
Then, in the category of chain complexes,
we conclude that
$$\cdots\to C_{n+1}\to C_n
\to C_{n-1}\to\cdots$$
is isomorphic to
$$\cdots\quad\to\quad Z_{n+1}\oplus B_n\quad\to\quad Z_n\oplus B_{n-1}
\quad\to\quad Z_{n-1}\oplus B_{n-2}\quad\to\quad\cdots,$$
where the boundary maps in the latter are
$$\eqalign{\ldots,\qquad
\alpha_n:=(x,y)\mapsto(i_n(y),0)\quad&:
\quad Z_{n+1}\oplus B_n\to Z_n\oplus B_{n-1}
\qquad,\qquad\cr
\alpha_{n-1}:=(y,z)\mapsto(i_{n-1}(z),0)\quad&:\quad
Z_n\oplus B_{n-1}\to Z_{n-1}\oplus B_{n-2}\qquad,\ldots.}$$

Applying the functor $B\mapsto B^A$, we obtain a chain complex
$$\cdots\,\to\,(Z_{n+1})^A\oplus(B_n)^A
\,\to\,(Z_n)^A\oplus(B_{n-1})^A
\,\to\,(Z_{n-1})^A\oplus(B_{n-2})^A\,\to\,\cdots,$$
where the boundary maps are
$$\eqalign{\ldots,\quad
(\alpha_n)^A=(x,y)\mapsto(i_n^A(y),0)\quad&:\quad
(Z_{n+1})^A\oplus(B_n)^A\to(Z_n)^A\oplus(B_{n-1})^A
\qquad,\qquad\cr
(\alpha_{n-1})^A=(y,z)\mapsto(i_{n-1}^A(z),0)\quad&:\quad
(Z_n)^A\oplus(B_{n-1})^A\to(Z_{n-1})^A\oplus(B_{n-2})^A
\quad,\ldots.}$$

Now fix $n\in\Z$.
We wish to show that $H_n$ applied to this chain complex
yields an additive Abelian group isomorphic to
$[(H_n)^A]\oplus[H_{n-1}*A]$.
That is, we wish to show that
$${{\ker[(\alpha_{n-1})^A]}\over{\im[(\alpha_n)^A]}}\cong
[(H_n)^A]\oplus[H_{n-1}*A].$$

Because $(\alpha_{n-1})^A(x,y)=((i_{n-1})^A(y),0)$
and $(\alpha_n)^A(y,z)=((i_n)^A(z),0)$,
it follows that
$$\eqalign{\ker[(\alpha_{n-1})^A]&=[(Z_n)^A]\oplus[\ker(i_{n-1})^A],\cr
\im[(\alpha_n)^A]&=[\im(i_n)^A]\oplus\{0\}.}$$
It therefore suffices to show both
that the cokernel $(i_n)^A:(B_n)^A\to(Z_n)^A$
is isomorphic to $(H_n)^A$
and that the kernel of $(i_{n-1})^A:(B_{n-1})^A\to(Z_{n-1})^A$
is isomorphic to $H_{n-1}*A$.

We have short exact sequences
$$\eqalign{0\to B_{n-1}&\to Z_{n-1}\to H_{n-1}\to 0,\cr
0\to B_n&\to Z_n\to H_n\to 0,}$$
where $B_{n-1}$, $Z_{n-1}$, $B_n$ and $Z_n$ are all free Abelian.
If we apply $B\mapsto B^A$, we get two lines of exact sequences
$$\eqalign{(B_{n-1})^A&\to(Z_{n-1})^A\to(H_{n-1})^A\to 0,\cr
(B_n)^A&\to(Z_n)^A\to(H_n)^A\to 0,}$$
where the leftmost maps are
$(i_{n-1})^A:(B_{n-1})^A\to(Z_{n-1})^A$ and
$(i_n)^A:(B_n)^A\to(Z_n)^A$.

From the first line, by definition of torsion product,
the kernel of 
$(i_{n-1})^A:(B_{n-1})^A\to(Z_{n-1})^A$
is isomorphic to $H_{n-1}*A$.
Exactness on the second line implies that the cokernel of
$(i_n)^A:(B_n)^A\to(Z_n)^A$ is isomorphic to $(H_n)^A$.

This is what was required.
{\bf QED}

\vskip.1in
There is some naturality in Universal Coefficients, which yields a
slightly stronger result:

\vskip.1in\noindent
{\it Naturality in the Universal Coefficients Theorem.}
Let $A$ be an additive Abelian group and let $n\in\Z$.
Let $\scrb:=\{\hbox{chain complexes}\}$
and let $\scrc:=\{\hbox{additive Abelian groups}\}$.
Let $\scrf,\scrg:\scrb\to\scrc$
be defined by $\scrf(C_\bullet)=(H_nC_\bullet)\otimes A$
and $\scrg(C_\bullet)=H_n(C_\bullet\otimes A)$.
For all $C_\bullet\in\scrb$,
let $T_{C_\bullet}:=\Tor(H_{n-1}(C_\bullet),A)$.
There exists a natural transformation $\tau:\scrf\to\scrg$
such that: for all $C_\bullet\in\scrb$,
there is a map
$\iota:T_{C_\bullet}\to\scrg(C_\bullet)$
such that
$$(x,y)\qquad\mapsto\qquad(\tau_{C_\bullet}(x))+(\iota(y))\qquad:\qquad
(\scrf(C_\bullet))\oplus T_{C_\bullet}\qquad\to\qquad\scrg(C_\bullet)$$
is an isomorphism of additive Abelian groups.

\vskip.1in
In particular, for all $C_\bullet\in\scrc$, we have
that $\tau_{C_\bullet}:\scrf(C_\bullet)\to\scrg(C_\bullet)$
is injective.
Bear in mind that $\iota$ is not natural in $C_\bullet$.
We omit a proof of the preceding naturality theorem.

We now begin a new topic: The study of finite CW-complexes and celluar
cohomology via spectral sequences.

Let $k\ge0$ and $d\ge0$ be integers.
Let $B^d$ denote the closed unit ball in $\R^d$;
its boundary in $B^d$ is $S^{d-1}$.
Let $kB^d:=\{1,\ldots,k\}\times B^d$
and $kS^{d-1}:=\{1,\ldots,k\}\times S^{d-1}$.
Given a topological space $X$ and a continuous map $f:kS^{d-1}\to X$,
we define $A(X,f):=(X\coprod kB^d)/f$,
and we let $i_{X,f}:X\to A(X,f)$ be the composite of the inclusion
and the canonical map, as shown here:
$$X\qquad\subseteq\qquad X\coprod kB^d
\qquad\to\qquad\left(X\coprod kB^d\right)\,\bigg/\,f\qquad=\qquad A(X,f).$$

\vskip.1in\noindent
{\it Definition.}
Let $k\ge0$, $d\ge0$ be integers,
let $Y$ be a topological space, let $X\subseteq Y$ and
let $f:kS^{d-1}\to X$ be continuous.
We say that $Y$ is {\bf obtained} from $X$ via $f$
if there is a homeomorphism $\psi:Y\to A(X,f)$
such that $\psi|X=i_{X,f}$.

\vskip.1in
Note that, in this case,
the map $\phi:kB^d\to Y$ obtained by composing
$$kB^d\qquad\subseteq\qquad
X\coprod kB^d\qquad\to\qquad A(X,f)\qquad\to\qquad Y$$
has the property that $\phi|(kS^{d-1})=f$.
We will say that such a map $\phi:kB^d\to Y$
is {\bf associated} to $(Y,X,f)$.
Note that different choices of $\psi$ will
yield different possibilities for $\phi$,
all associated to $(Y,X,f)$.

\vskip.1in\noindent
{\it Definition.}
Let $k\ge0$, $d\ge0$ be integers,
let $Y$ be a topological space and let $X\subseteq Y$.
We say that $Y$ is {\bf obtained} from $X$ via $k$ $d$-cells
if there is a continouous map $f:kS^{d-1}\to X$
such that $Y$ is obtained from $X$ via $f$.

\vskip.1in\noindent
{\it Definition.}
Let $d\ge0$ be integers,
let $Y$ be a topological space and let $X\subseteq Y$.
We say that $Y$ is {\bf obtained} from $X$ via $d$-cells
if there is some integer $k\ge0$ such that
$Y$ is obtained from $X$ via $k$ $d$-cells.

\vskip.1in\noindent
{\it Definition.}
Let $X$ be a topological space and let $A\subseteq X$.
We say that $(X^0,\ldots,X^d)$ is a ($d$-dimensional)
{\bf skeletal filtration} from $A$ to $X$ if
$$A\subseteq X^0\subseteq X^1\subseteq\cdots\subseteq X^d=X,$$
if $X^0$ is obtained from $A$ via $0$-cells and if,
for any integer $i\in[1,d]$,
$X^i$ is obtained from $X^{i-1}$ via $d$-cells.

\vskip.1in\noindent
{\it Definition.}
Let $X$ be a topological space.
A {\bf skeletal filtration} of $X$ is a skeletal filtration
from $\emptyset$ to $X$.

\vskip.1in\noindent
{\it Definition.}
Let $X$ be a topological space and let $A\subseteq X$.
We say that $(X,A)$ is a {\bf compact CW-pair}
if $X$ is a compact topological space,
if $A$ is a closed subset of $X$ and
if there exists a skeletal filtration from $A$ to $X$.

\vskip.1in\noindent
{\it Definition.}
We say that $X$ is a {\bf finite CW-complex}
if $(X,\emptyset)$ is a compact CW-pair,
{\it i.e.}, if $X$ admits a skeletal filtration.

\vskip.1in
We note that many interesting topological spaces
either are finite CW-complexes or are homotopy equivalent
to finite CW-complexes.
For example, the fundamenal result of Morse theory
is that any compact smooth manifold is homotopy
equivalent to a finite CW-complex.
(The term ``smooth manifold'' will be defined precisely
later in this course.)
For another example, the realization of any
abstract finite simplicial complex is a finite CW-complex.

Let $\scrc:=\{\hbox{finite CW-complexes with skeletal filtration}\}$,
$\scrts:=\{\hbox{topological spaces}\}$.
Note that there is a forgetful functor $\scrf:\scrc\to\scrts$,
in which one forgets the skeletal filtration.

A chain complex $C_\bullet$ is {\bf finitely generated} if both
$\{n\in\Z\,|\,C_n\ne0\}$ is finite and, for all $n\in\Z$,
$C_n$ is a finitely generated Abelian group.
Let $\scrcc:=\{\hbox{chain complexes}\}$
and $\scrfgcc:=\{\hbox{finitely generated chain complexes}\}$.
Finally, let $\scrgg:=\{\hbox{graded groups}\}$.

We take the point of view that the homology functor
$H_\bullet:\scrcc\to\scrgg$
is a computational nightmare, but that its
restriction $H_\bullet:\scrfgcc\to\scrgg$
is much more feasible since it amounts to
computing kernels, images and cokernels of 
finitely many matrices of finite size.

Our goal in this part of the notes is to develop a functor
$\Cell_\bullet:\scrc\to\scrfgcc$ such that,
for all $Z\in\scrc$,
we have $H_\bullet(\Cell_\bullet(Z))\cong H_\bullet(\scrf(Z))$.
(More precisely, we will choose $\Cell_\bullet$
such that $H_\bullet\circ\Cell_\bullet:\scrc\to\scrgg$
is equivalent to $H_\bullet\circ\scrf:\scrc\to\scrgg$.)
Then, given a finite CW-complex $X$,
to compute $H_\bullet(X)$,
we first choose a skeletal filtration of $X$,
{\it i.e.}, we first choose $Z\in\scrc$ such that $\scrf(Z)=X$,
and then we compute $H_\bullet(X)$ by instead doing
the more feasible computation of $H_\bullet(\Cell_\bullet(Z))$.

\vskip.1in\noindent
{\it Fact.}
Let $Y$ be a topological space and let $X\subseteq Y$.
Let $k,d\in\Z$. Assume $k\ge0$ and $d\ge1$.
Let $f:kS^{d-1}\to X$ be continuous.
Assume that $Y$ is obtaind from $X$ via $f$
and let $\phi:kB^d\to Y$ be associated to $(Y,X,f)$.
Then $(\phi,f)_*:H_\bullet(kB^d,kS^{d-1})\to H_\bullet(Y,X)$
is an isomorphism.

\vskip.1in
We omit the proof and refer the reader intead to Vick's book
on ``Homology'' or to Rotman's book on algebraic topology.
(Look up ``relative homeomorphism'' and read the nearby results.)

\vskip.1in\noindent
{\it Remark.}
Let $q,k,d\in\Z$.\
Assume that $k\ge0$ and $d\ge1$.
If $d\ne q$, then $H_q(kB^d,kS^d)\cong\{0\}$.
If $d=q$, then $H_q(kB^d,kS^d)\cong\Z^k\cong\Z^{k\times1}$.

\vskip.1in
We leave the proof of this as an unassigned exercise for the
interested reader. (Hint: Consider the long exact sequence
associated to $(kB^d,kS^d)$.)

We now fix a topological space $X$ and a collection of
subsets $X^0,\ldots,X^d$ of $X$
such that $X^0\subseteq\cdots X^d=X$.
Let $\scrx:=(X^0,\ldots,X^d)$ and let $Z:=(X,\scrx)$.
For all integers $i<0$, we define $X^i:=\emptyset$.
For all integers $i>d$, we define $X^i:=X$.

If $\scrx$ is a skeletal filtration,
then, for all $q\in\Z$, we define
$$\Cell_q\qquad:=\qquad\Cell_q(Z)\qquad:=\qquad H_q(X^q,X^{q-1}).$$

For all $n\in\Z$, the long exact sequence of $(X^n,X^{n-1})$
has the broken form
$$H_\bullet(X^{n-1})\qquad\to\qquad H_\bullet(X^n)
\qquad\to\qquad H_\bullet(X^n,X^{n-1}).$$
With $n=q$, between the $q$ degree and the $q-1$ degree,
this long exact sequence has a connecting homomorphism
$\Cell_q\to H_{q-1}(X^{q-1})$.
With $n=q-1$, in the $q-1$ degree, this broken long exact sequence
has a map $H_{q-1}(X^{q-1})\to\Cell_{q-1}$.
For all $q\in\Z$, let $\partial_q:\Cell_q\to\Cell_{q-1}$
denote the composition of these two maps:
$$\Cell_q\qquad\to\qquad H_{q-1}(X^{q-1})\qquad\to\qquad\Cell_{q-1}.$$

\vskip.1in\noindent
{\bf EXERCISE 16C:}
Show, for all $q\in\Z$, that $\partial_{q-1}\circ\partial_q=0$.

\vskip.1in\noindent
{\it Example.}
Let $s,t,u$ be three distinct points in $\R^2$.
Let $A$ be the closed line segment from $s$ to $t$.
Let $B$ be the closed line segment from $t$ to $u$.
Let $C$ be the closed line segment from $u$ to $s$.
Let $T:=A\cup B\cup C$.
Note that $H_1(T)\cong\Z$, since $T$ is homeomorphic to $S^1$.
Consider now the case where $d=1$, where $X^0=\{s,t,u\}$ and
where $X=X^1=T$.

Note that $X^0$ is obtained from $\emptyset$ by attaching $3$ $0$-cells,
and that $X^1$ is obtained from $X^0$ by attaching $3$ $1$-cells.
Choose attaching maps to make the last sentence true.
Then the preceding Fact and Remark show,
for all $q\in\Z\backslash\{0,1\}$, that $\Cell_q=0$.
They also show that there are isomorphisms $\Cell_0\cong\Z^{3\times1}$
and $\Cell_1\cong\Z^{3\times1}$.
Choose such isomorphisms and use them to identify
$\partial_1:\Cell_1\to\Cell_0$
with a $3\times 3$ matrix $B:\Z^{3\times1}\to\Z^{3\times1}$.

\vskip.1in\noindent
{\bf EXERCISE 16D:}
Explictly lay out all of the choices mentioned in the
preceding example, and then compute $B\in\Z^{3\times3}$.

\vskip.1in\noindent
{\bf EXERCISE 16E:}
In the preceding example,
show that $H_1(\Cell_\bullet)\cong H_1(X)$.

\vskip.1in
We aim to show, in complete generality,
that if $\scrx$ is a skeletal filtration of $X$,
then we have $H_\bullet(\Cell_\bullet)\cong H_\bullet(X)$.

\vskip.1in\noindent
{\it Example.}
Fix an integer $n\ge0$.
Let $h:S^n\to\R P^n$
and
$h_1:S^{n+1}\to\R P^{n+1}$
be the canonnical maps.
Let $i:S^n\to S^{n+1}$
be defined by $i(p)=(p,0)$
and let $j:\R P^n\to\R P^{n+1}$
be defined by $j(h(p))=h_1(i(p))$;
note that $j$ is well-defined and continuous.
Let $u:B^{n+1}\to S^{n+1}$ be defined by
$u(x)=(x,\sqrt{1-\|x\|^2})$.

Then $j\coprod(h_1\circ u):
\R P^n\coprod B^{n+1}\to\R P^{n+1}$
is surjective and has the same fibers as the canonical map
$$\R P^n\coprod B^{n+1}\qquad\to\qquad
\left(\R P^n\coprod B^{n+1}\right)\,\bigg/\,h
\qquad=\qquad A(\R P^n,h).$$
Then $j\coprod(h_1\circ u)$
factors to a continuous bijection
$$\rho\qquad:\qquad A(\R P^n,h)\qquad\to\qquad\R P^{n+1};$$
by compactness, this map is closed, therefore open,
therefore a homeomorphism, showing that
$\R P^{n+1}$ is obtained from $j(\R P^n)$
via one $(n+1)$-cell.

The above remarks, for varying $n$,
give injective continuous maps (denoted above by $j$) of the form:
$$\R P^0\qquad\to\qquad\R P^1\qquad\to\qquad\R P^2\qquad\to\qquad\cdots.$$
Now consider the case where $d=2$, where $X=X^2=\R P^2$,
where $X^1$ be the image of $\R P^1$ in $\R P^2$
and where $X^0$ be the image of $\R P^0$ in $\R P^2$.
The above remarks then show that,
for each $k\in\{1,2\}$,
$X^k$ is obtained from $X^{k-1}$ by attaching one $k$-cell.
Thus $(X^0,X^1,X^2)$ is a skeletal filtration of
$\R P^2$, showing that $\R P^2$ is a finite CW-complex.

\vskip.1in\noindent
{\bf EXERCISE 16F:}
Compute the cellular chain complex of the skeletal filtration
of $\R P^2$ described above.
Compute its homology (in all degrees).

\vskip.1in
Recall that we are considering the case of a
{\bf filtered topological space}, {\it i.e.},
a topological space $X$ together with a finite
increasing sequence of subsets
$X^0\subseteq X^1\subseteq X^2\subseteq\cdots\subseteq X^d=X$.
We have defined $\scrx:=(X^0,\ldots,X^d)$.
Recall that we have defined,
for all $p<0$, $X^p:=\emptyset$.
We have also defined, for all $p>d$, $X^p:=X$.

Let's pose the question in complete generality:
Can we compute $H_\bullet(X)$
if we know, for all $p$, $H_\bullet(X^p,X^{p-1})$?
This is somehow the topology analogue to the
question of whether one can reconstruct a group
from the quotients of a composition series.

For topology, the answer, developed below, is,
``sometimes yes'', and the basic tool for
handling this question is what are called spectral sequences.
The ``sometimes'' is, of course, disturbing,
but we note that in the case of a {\it skeletal} filtration
(in which each skeleton is obtained by the preceding one
from addition of cells), the answer is better:
The answer is: ``Yes, provided you can calculate
the homology of the cellular chain complex
associated to the filtration''. We have
already assigned exercises to demonstrate the
feasibility of that kind of computation.

\vskip.1in\noindent
{\it Definition.}
A {\bf bigraded group} is a map
$\Z\times\Z\to\{\hbox{additive Abelian groups}\}$.

\vskip.1in
We will now adopt the standard notation that the
category of graded groups is denoted
$\{\hbox{additive Abelian groups}\}^\Z$,
and the category of bigraded groups is denoted
$$\{\hbox{additive Abelian groups}\}^{\Z\times\Z}.$$

For all $p,q\in\Z$, define $D^p_q:=H_q(X^p)$
$E^p_q:=H_q(X^p,X^{p-1})$.
Then $D^\bullet_\bullet,E^\bullet_\bullet\in
\{\hbox{additive Abelian groups}\}^{\Z\times\Z}$,
{\it i.e.}, $D^\bullet_\bullet$ and $E^\bullet_\bullet$
are both examples of bigraded groups.
I advise ``picturing'' a bigraded group $G^\bullet_\bullet$
by placing, at the integer lattice point $(s,t)\in\Z\times\Z$,
{\it neither} the group $G^s_t$,
{\it nor} the group $G^t_s$,
but rather the group $G^{s+t}_s$.
Then $G^1_1$ appears over the point $(1,0)$,
and $G^1_0$ appears over the point $(0,1)$.
In what follows, discussion of ``origin'', ``up'', ``high'', ``low'',
``down'', ``left'', ``right'', ``horizontal'', ``vertical'',
``northeast'', ``southwest'', ``northwest'', ``southeast''
and other points-of-compass will assume this pictorial organization.

This might be odd at first and does require some care to implement.
If you read the literature, you will notice
that the groups we define as $E^p_q$
are in other places defined as $E^1_{p,q-p}$,
and, following that notation, horizontal and
vertical take on more traditional meanings.
Our choice here means that we have the simple formulas:
$D^p_q:=H_q(X^p)$ and $E^p_q:=H_q(X^p,X^{p-1})$,
so we have one advantage in formulas, but one disadvantage
in picturing our bigraded groups.
Let me mention {\it a propos} of this that the groups
we define below as $(rE)^p_q$
are in other expositions defined as $E^{r+1}_{p,q-p}$.

We take the point of view that our goal is to
compute $H_\bullet:=H_\bullet(X)$, which means computing,
for each $q$, for some $p\ge d$, the group $D^p_q$.
That is, we want to compute the ``high'' $D$s.
Note that we already know the ``low'' $D$s to be $0$,
because, for all $p,q\in\Z$, if $p<0$,
then $D^p_q=H_q(X^p)=H_q(\emptyset)=0$.

We recall that,
in the case where $\scrx$ is skeletal,
we have, for all $p\ne q$, $E^p_q=0$,
and we have, for all $q$, $E^q_q=\Cell_q$.
We are therefore taking the point of view
that $E^\bullet_\bullet$ is known, and that,
if $\scrx$ is cellular, then it is $0$
except along the horizontal line through
the origin.
Along that horizontal line, we see the terms
of the cellular chain complex.

Note that, for all $p,q\in\Z$,
we have a the long exact sequence associated to
the pair $(X^p,X^{p-1})$ a part of which appears as:
$$\cdots\qquad\to\qquad D^{p-1}_q\qquad\to\qquad D^p_q\qquad\to\qquad E^p_q$$
which is followed by a connecting homomorphism (denoted $\gamma$) to
$$D^{p-1}_{q-1}\qquad\to\qquad D^p_{q-1}\qquad\to\qquad E^p_{q-1}$$
which is followed by a connecting homomorphism (denoted $\gamma$) to
$$D^{p-1}_{q-2}\qquad\to\qquad D^p_{q-2}\qquad\to\qquad E^p_{q-2}
\qquad\to\qquad\cdots.$$

Thus we obtain bigraded group homomorphisms
$\alpha:D^\bullet_\bullet\to D^{\bullet+1}_\bullet$,
$\beta:D^\bullet_\bullet\to E^\bullet_\bullet$
and $\gamma:E^\bullet_\bullet\to D^{\bullet-1}_{\bullet-1}$,
with the property that
$\im\,\alpha=\ker\,\beta$,
$\im\,\beta=\ker\,\gamma$
and $\im\,\gamma=\ker\,\alpha$.
We should clarify the definitions of $\im$ and $\ker$ appearing here:
For example, we define $\im\,\alpha$ to be the bigraded group
defined by $(\im\,\alpha)^p_q=\im(\alpha:D^{p-1}_q\to D^p_q)$.
On the other hand, $\ker\,\alpha$ is defined by
$(\ker\,\alpha)^p_q=\ker(\alpha:D^p_q\to D^{p+1}_q)$.

We now pause from this discussion to define, quite generally:

\vskip.25in\noindent
{\it Definition.}
A {\bf$\left(\matrix{a&b&c\cr x&y&z}\right)$-exact couple}
consists of
\itemitem{(1)} two bigraded groups $D^\bullet_\bullet$,
$E^\bullet_\bullet$; and
\itemitem{(2)} three maps
$\alpha:D^\bullet_\bullet\to D^{\bullet+a}_{\bullet+x}$,
$\beta:D^\bullet_\bullet\to E^{\bullet+b}_{\bullet+y}$
and $\gamma:E^\bullet_\bullet\to D^{\bullet+c}_{\bullet+z}$

\noindent
such that $\im\,\alpha=\ker\,\beta$,
such that $\im\,\beta=\ker\,\gamma$
and such that $\im\,\gamma=\ker\,\alpha$.

\vskip.25in
Note that we have $\gamma\beta=0$ but
$\beta\gamma:
E^\bullet_\bullet\to E^{\bullet+b+c}_{\bullet+y+z}$
may be nonzero.
However $(\beta\gamma)^2=\beta(\gamma\beta)\gamma=
\beta\cdot0\cdot\gamma=0$.
Thus $(E^\bullet_\bullet,\beta\gamma)$
is simply a family of chain complexes.

With the preceding definition,
we see that a filtration $\scrx$ of a topological space $X$
yields a $\left(\matrix{1&0&-1\cr0&0&-1}\right)$-exact couple,
which basically encodes a family of long exact sequences,
one for each pair of consecutive terms in the filtration of $X$.
In this case, $\beta\gamma$ has bidegree
$\left(\matrix{-1\cr-1}\right)$
and so $(E^\bullet_\bullet,\beta\gamma)$
is pictured a collection of horizontal chain complexes
with all arrows westbound ({\it i.e.}, horizontal and to the left).

Recall that, in the case where $\scrx$ is skeletal,
$E^\bullet_\bullet$ is zero except on the horizontal line
through the origin. In this case, $\beta\gamma$
appears as a family of westbound arrows,
and, along the horizontal line through
the origin, we see the cellular chain complex
$(\Cell_\bullet,\partial)$.
(Observe, for all $p\in\Z$, that $\partial_p=(\beta\gamma)^p_p$.)
We define $H_\bullet^C:=H_\bullet(\Cell_\bullet)$
and we take the point of view that that is known.
That is, we know certain subquotients of the
various groups $E^p_q$.

Recall that we wish to compute the high values of $D^\bullet_\bullet$,
which are the same as $H_\bullet$.
Thus, phrased broadly, our task is to use known information about
$(E^\bullet_\bullet,\beta\gamma)$
to find out information about high values of
$D^\bullet_\bullet$.
(We may also use that the low $D^\bullet_\bullet$ are all zero.)

We further adopt the point of view that any time we
can declare that some group appears as a subquotient
of some $D^p_q$, then we have made progress,
particularly when $p\ge d$, in which case $H_q=D^p_q$.

In fact, for each $q$, if we can get enough subquotients of $H_q$,
then perhaps they will fit together to give the
quotient groups of a composition series for $H_q$.
Finally, if we're very lucky,
all of those groups but one will be nonzero,
and we can then declare that $H_q$ is isomorphic
to that group.

Amazingly, in the case where $\scrx$ is skeletal,
this is exactly what happens.
Moreover, we should point out that in many other
applications of spectral sequences, this
is exactly what happens.

We now pause from the specific discussion about filtered topological
spaces to make a general

\vskip.25in\noindent
{\it Definition.}
Let $C:=(D^\bullet_\bullet,E^\bullet_\bullet,\alpha,\beta,\gamma)$
be a $\left(\matrix{a&b&c\cr x&y&z}\right)$-exact couple.
Then the {\bf derived couple} of $C$
is the $\left(\matrix{a&b-a&c\cr x&y-x&z}\right)$-exact couple
$\hatC=(\hatD^\bullet_\bullet,\hatE^\bullet_\bullet,
\hatalpha,\hatbeta,\hatgamma)$, defined as follows:
Let $\hatD^\bullet_\bullet:=\im\,\alpha$ and
let $\hatE^\bullet_\bullet:=(\ker\,\beta\gamma)/(\im\,\beta\gamma)$.
Define $\hatalpha:\hatD^\bullet_\bullet\to
\hatD^{\bullet+a}_{\bullet+x}$ by
$\hatalpha:=\alpha|(\hatD^\bullet_\bullet)$.
Define $\hatbeta:\hatD^\bullet_\bullet\to
\hatE^{\bullet-a+b}_{\bullet-x+y}$ by
$\hatbeta(\alpha(d))=[\beta(d)]$,
where $[\cdot]:\ker\,\beta\gamma\to
\hatE^\bullet_\bullet$ is the quotient map.
Define $\hatgamma:\hatE^\bullet_\bullet\to
\hatD^{\bullet+c}_{\bullet+z}$ by
$\hatgamma([e])=\gamma(e)$.

\vskip.25in\noindent
{\bf EXERCISE 16G:}
Show that $\hatbeta$ and $\hatgamma$ are well-defined
in the preceding definition.

\vskip.1in\noindent
{\bf EXERCISE 16H:}
In the preceding definition,
show that $\im\,\hatalpha=\ker\,\hatbeta$,
that $\im\,\hatbeta=\ker\,\hatgamma$
and that $\im\,\hatgamma=\ker\,\hatalpha$.

\vskip.1in
We now return to our discussion in which
we have an exact couple obtained from a filtration $\scrx$
of a topological space $X$.
Note, in this case that the
derived couple described above is
a $\left(\matrix{1&-1&-1\cr0&0&-1}\right)$-exact couple.
In particular $\hatbeta\hatgamma$ has bidegree
$\left(\matrix{-2\cr-1}\right)$ and is pictured
as a family of arrows that move in a southwesterly direction.
Note also that, in $(\hatE^\bullet_\bullet,\hatbeta\hatgamma)$,
the terms on the horizontal line
through the origin are exactly $H_\bullet^C$,
which we are taking as known quantities.
As before, $(\hatE^\bullet_\bullet,\hatbeta\hatgamma)$
is a family of chain complexes,
but these chain complexes have the nice property that,
for each one, all but one term is zero.
Thus these are chain complexes with the property
that passing to homology does nothing (up to isomorphism)!
So, later on, when we discuss ${{\widehat{\hatE}}}^\bullet_\bullet$,
we will have ${\widehat{\hatE}}^\bullet_\bullet\cong\hatE^\bullet_\bullet$,
and is therefore known, given that we have computed
the homology of the cellular chain complex of $\scrx$.

Moreover, as we continue and take further and further derived exact
couples, we can determine that all these exact couples have the
property that their $E$-term is known and agrees, up to isomorphism,
with $\hatE^\bullet_\bullet$.

Again, let us recall that, speaking very broadly,
we wish to figure out, from all this known
information, the subquotients of $D^p_q$, for $p\ge d$.  That is, we
wish to determine as many subquotients as possible of the ``high''
values of $D^\bullet_\bullet$.

We now define
$0C:=(D^\bullet_\bullet,E^\bullet_\bullet,\alpha,\beta,\gamma)$,
and, for all integers $r\ge0$,
we recursively define $(r+1)C=\widehat{rC}$.
For all integers $r\ge0$ define
$(rD)^\bullet_\bullet$, $(rE)^\bullet_\bullet$,
$r\alpha$, $r\beta$ and $r\gamma$
to be the components of $rC$,
so that we have
$rC=((rD)^\bullet_\bullet,(rE)^\bullet_\bullet,r\alpha,r\beta,r\gamma)$.

By the preceding remarks, if the original exact couple $C$
is obtained from a skeletal filtration of a topological space,
then we have that $(2E)^\bullet_\bullet$
is isomorphic to $\hatE^\bullet_\bullet$.
Moreover, the bidegree of $(2\beta)(2\gamma)$
is $\left(\matrix{-3\cr-1}\right)$ and is pictured
as a family of arrows that move in a south-by-southwesterly direction.

So, in $((2E)^\bullet_\bullet,(2\beta)(2\gamma))$,
the terms on the horizontal line
through the origin are, up to isomorphism, $H_\bullet^C$,
which we are taking as known quantities.
As before, $((2E)^\bullet_\bullet,(2\beta)(2\gamma))$
is a family of chain complexes,
and these chain complexes again have the nice property that,
for each one, all but one term is zero.
Thus these are chain complexes with the property
that passing to homology does nothing (up to isomorphism)!
Then $(3E)^\bullet_\bullet$ is isomorphic (in the category
of bigraded groups) to $(2E)^\bullet_\bullet$,
which, in turn is isomorphic to $\hatE^\bullet_\bullet$.

Still assuming that $C$ is obtained from a skeltal filtration,
continuing in this manner, we see, for all integers $r\ge1$,
that $(rE)^\bullet_\bullet$ is isomorphic (in the category
of bigraded groups) to $\hatE^\bullet_\bullet$.
That is, as we pass to higher and higher derived couples,
nothing happens to the $E$-term after $r=1$.
So, for all integers $r\ge1$, for all $p,q\in\Z$,
if $p\ne q$, then $(rE)^p_q=0$.
Also, all integers $r\ge1$, for all $q\in\Z$,
we have $(rE)^q_q=H^C_q$.
Recall, for all $p,q\in\Z$, if $p\ge d$, then $D^p_q=H_q$.
We therefore wish to show, for all $q\in\Z$,
that there are some integer $r\ge1$ and some integer $p\ge d$
such that $(rE)^q_q\cong D^p_q$.

More precisely, we will show
for all $q\in\Z$,
that there are some integer $r\ge1$ and some integer $p\ge d$
and some composition series of $D^p_q$ such that every
quotient of the composition series is $0$ except for one,
which is isomorphic to $(rE)^q_q$.

It therefore behooves us to find conditions on exact couples
under which, a given $(rE)^p_q$ can be seen as a subquotient
of some $D^{p'}_{q'}$.

We now return to the discussion of a general exact couple.
First, note, for all $p,q,r\in\Z$, that if $r\ge1$, then
$(rD)^p_q$ is a subgroup of $((r-1)D)^p_q$
and that $(rE)^p_q$ is a subquotient of $((r-1)E)^p_q$.
It follows that:
for all $p,q,r\in\Z$, that if $r\ge0$, then
$(rD)^p_q$ is a subgroup of $D^p_q$
and that $(rE)^p_q$ is a subquotient of $E^p_q$.
Given $p,q\in\Z$, we now attempt to identify these
subgroups (as $r$ varies) of $D^p_q$
and subqotients (as $r$ varies) of $E^p_q$.

Recall that $\hatD^\bullet_\bullet=\im\,\alpha=\alpha(D^\bullet_\bullet)$
and that $\hatalpha:=\alpha|\hatD^\bullet_\bullet$.
Then $(2D)^\bullet_\bullet=\hatalpha(\hatD^\bullet_\bullet)=
(\alpha|\hatD^\bullet_\bullet)(\hatD^\bullet_\bullet)=
\alpha(\hatD^\bullet_\bullet)=\alpha(\alpha(D^\bullet_\bullet))=
\im(\alpha^2)$.
We leave it as an unassigned exercise to continue this argument
to show, for all integers $r\ge0$,
that $(rD)^\bullet_\bullet=\im(\alpha^r)$.
That is, for all $p,q,r\in\Z$, if $r\ge0$,
then $(rD)^p_q=\alpha^r(D^{p-ra}_{q-rb})$.
This shows $(rD)^p_q$ as a subgroup of $D^p_q$.

We now turn to the $E$-terms.
First, note that
$$\hatE^\bullet_\bullet=
{{\ker(\beta\gamma)}
\over
{\im(\beta\gamma)}}=
{{\gamma^{-1}(\ker\,\beta)}
\over
{\beta(\im\,\gamma)}}=
{{\gamma^{-1}(\im\,\alpha)}
\over
{\beta(\ker\,\alpha)}}.$$
Thus we see $\hatE^\bullet_\bullet$
as a subquotient of $E^\bullet_\bullet$.

A similar argument (just put a hat on each step of
the preceding paragraph) shows
$$(2E)^\bullet_\bullet=
{{\hatgamma^{-1}(\im\,\hatalpha)}
\over
{\hatbeta(\ker\,\hatalpha)}}.$$
Thus we see $(2E)^\bullet_\bullet$
as a subquotient of $\hatE^\bullet_\bullet$.
Under the canonical surjection
$\ker(\beta\gamma)\to(\hatE)^\bullet_\bullet$,
pull back the numerator and denominator
(namely, pull back $\hatgamma^{-1}(\im\,\hatalpha)$
and $\hatbeta(\ker\,\hatalpha)$)
in the above displayed equation.
The quotient of the resulting pullbacks
is then isomorphic to $(2E)^\bullet_\bullet$
and shows $(2E)^\bullet_\bullet$
as a subquotient of $E^\bullet_\bullet$.
This gives enough hints to make the following

\vskip.1in\noindent
{\bf EXERCISE 17A:}
Show that
$$(2E)^\bullet_\bullet\cong
{{\gamma^{-1}(\im\,\alpha^2)}
\over
{\beta(\ker\,\alpha^2)}}.$$

\vskip.1in
Generalizing Exercise 17A,
it is an unassigned exercise to see,
for all integers $r\ge1$,
$$(rE)^\bullet_\bullet\cong
{{\gamma^{-1}(\im\,\alpha^r)}
\over
{\beta(\ker\,\alpha^r)}}.$$
This shows $(rE)^\bullet_\bullet$
as a subquotient of $E^\bullet_\bullet$.

To be precise about indices, we need to do
some calculations, given the bidegrees
of $\alpha$, $\beta$ and $\gamma$.
For all $p,q,r\in\Z$, if $r\ge1$, then
$$(rE)^p_q\cong
\left({{\gamma^{-1}(\im\,\alpha^r)}
\over
{\beta(\ker\,\alpha^r)}}\right)^p_q=
{{\gamma^{-1}(\im(\alpha^r:D^{p+c-ra}_{q+z-rx}\to D^{p+c}_{q+z}))}
\over
{\beta(\ker(\alpha^r:D^{p-b}_{q-y}\to D^{p-b+ra}_{q-y+rx}))}}.$$

Now recall, in the skeletal filtered case, that, for all
$p,q\in\Z$, if $p<0$, then $D^p_q=0$.

For the general case, fix $p,q,r\in\Z$ and assume
that $r\ge1$ and that $D^{p+c-ra}_{q+z-rx}=0$.
In this case, we have
$\im(\alpha^r:D^{p+c-ra}_{q+z-rx}\to D^{p+c}_{q+z})=0$
and so the numerator in the preceding displayed equation
is just $(\ker\,\gamma)^p_q$, which is equal to
$(\im\,\beta)^p_q$. So, in this case, we see that
$$(rE)^p_q\cong\left({{\im\,\beta}\over
{\beta(\ker\,\alpha^r)}}\right)^p_q.$$

\vskip.1in\noindent
{\bf EXERCISE 17B:}
Let $A$, $B$ and $C$ be groups,
and let $f:A\to B$ and $g:A\to C$ be homomorphisms.
Show that
$${{\im\,f}\over{f(\ker\,g)}}\cong{{\im\,g}\over{g(\ker\,f)}}.$$

\vskip.25in
Again, assume that
$r\ge1$ and that $D^{p+c-ra}_{q+z-rx}=0$.
We apply Exercise 17B to the preceding expression for $(rE)^p_q$,
with $f:A\to B$ replaced by
$\beta:D^{p-b}_{q-y}\to E^p_q$
and with $g:A\to C$ replaced by
$\alpha^r:D^{p-b}_{q-y}\to D^{p-b+ra}_{q-y+rx}$.
Taking care with our indices, we conclude
$$(rE)^p_q\cong\left({{\im\,\alpha^r}\over
{\alpha^r(\ker\,\beta)}}\right)^{p-b+ra}_{q-y+rx}.$$
Since $\ker\,\beta=\im\,\alpha$, we then have
$$(rE)^p_q\cong\left({{\im\,\alpha^r}\over
{\im\,\alpha^{r+1}}}\right)^{p-b+ra}_{q-y+rx}.$$
Or, equivalently, we have
$$(rE)^p_q\cong
{{\alpha^r(D^{p-b}_{q-y})}\over{\alpha^{r+1}(D^{p-b-a}_{q-y-x})}}.$$

Since all powers of $\alpha$ map $D$-terms to $D$-terms,
we finally see $(rE)^p_q$ as a subquotient of
a term in $D^\bullet_\bullet$.
Specifially, we see $(rE)^p_q$ as a subquotient of
$D^{p-b+ra}_{q-y+rx}$.

In the case where our original exact couple $C$ is obtained
from a skeletal filtration,
we  have $(a,x)=(1,0)$ and $(b,y)=(0,0)$.
So, as $r\to\infty$ this $D$-term, 
$D^{p-b+ra}_{q-y+rx}$, moves ``higher and higher''.
Since the high $D$-terms give the homology of $X$,
this is exactly as we would wish.

We now complete the details,
and we assume from here on out that the exact couple $C$
is obtained from a skeletal filtration.
Given $p\in\Z$,
letting $r_0:=\max\{1,p\}$, we have, for all integers $r\ge r_0$,
that $X^{p-r}=\emptyset$,
so, for all $q\in\Z$, we get
$D^{p+c-ra}_{q+z-rx}=D^{p-r}_q=H_q(X^{p-r})=H_q(\emptyset)=0$,
which, by earlier remarks, then implies that
$$(rE)^p_q\cong
{{\alpha^r(D^p_q)}\over{\alpha^{r+1}(D^{p-1}_q)}}\cong
{{\im(H_q\,X^p\to H_q\,X^{p+r})}\over
{\im(H_q\,X^{p-1}\to H_q\,X^{p+r})}}.$$

Now fix $q\in\Z$.
We wish to show that $H_q^C=H_q$.

Recall that $H_q=H_q\,X$.
For all $p\in\Z$,
let $F^p:=\im(H_q\,X^p\to H_q)$.
Note that, for all integers $p<0$, we have $F^p=0$.
Note that, for all integers $p\ge d$, we have $F^p=H_q$.
Then, as $p$ varies, $\{F^p\}$ is a composition series of $H_q$.
We wish to show that, as $p$ varies,
the quotients $\{F^p/F^{p-1}\}$ are all $0$
except for one, which is isomorphic to $H_q^C$.

For all $p\in\Z$,
there exists $r_0\in\Z$
such that, for all $r\ge r_0$,
we have $(rE)^p_q\cong F^p/F^{p-1}$.
For all integers $r\ge1$,
we recall that $(rE)^\bullet_\bullet$
is isomorphic to $\hatE^\bullet_\bullet$
and that this bigraded group is zero,
except on the line through the origin,
where one sees the cellular homology of $X$.
Specifically, for all $p\in\Z$, for all integers $r\ge1$,
if $p\ne q$, then $(rE)^p_q\cong\hatE^p_q\cong0$.
Also, for all $r\ge1$, we have $(rE)^q_q\cong\hatE^q_q\cong H_q^C$.

Putting all this together, we see, for all $p\in\Z$,
there exists $r_0\in\Z$ such that, for all $r\ge r_0$,
we have
$$F^p/F^{p-1}\cong(rE)^p_q\cong
\cases{
0&if $p\ne q$\cr
H_q^C&if $p=q$}.$$
So we get what was required:
As $p$ varies, all but one of $\{F^p/F^{p-1}\}$ is $0$
and that one is isomorphic to $H_q^C$.

This completes the proof. QED

\vskip.1in
Recall that a chain complex is a graded group with a
differential of degree $-1$.
(A {\bf differential} is a map whose square is zero.)

\vskip.1in\noindent
{\it Definition.}
A {\bf cochain complex} is a graded additive Abelian group with a
diffential of degree $+1$.
That is, it is a bi-infinite sequence $\ldots,C^{-1},C^0,C^1,\ldots$
of additive Abelian groups,
together with a map $d:C^\bullet\to C^{\bullet+1}$
such that $d^2:C^\bullet\to C^{\bullet+2}$ is equal to zero.
This map $d$ is called the {\bf coboundary map}.

\vskip.1in
Traditionally, one uses subscripts to index chain complexes
and superscripts to index cochain complexes.
We will use the indexing convention that,
if $(C^\bullet,d)$ is a cochain complex,
then, for all $k\in\Z$, we have a map $d^k:C^k\to C^{k+1}$.

Let $A$ be a fixed additive Abelian group.
Then, for all additive Abelian groups $B$,
$\Hom(B,A)$ is an additive Abelian group
under the addition $(f+g)(b)=(f(b))+(g(b))$.
Then $\scrh:=\Hom(\cdot,A):
\{\hbox{additive Abelian groups}\}\to\{\hbox{additive Abelian groups}\}$
is a contravariant functor.
That is, if $\phi:B\to C$ is an arrow in $\{\hbox{additive Abelian groups}\}$
then we obtain a map $\phi^*=\scrh(\phi):\scrh(C)\to\scrh(B)$
defined by $\phi^*(f)=f\circ\phi$.

For any chain complex $(C_\bullet,\partial)$,
we define a graded group $\scrh(C_\bullet)$
by $(\scrh(C_\bullet))_n=\scrh(C_n)$,
and we then have a cochain complex
$(\scrh(C_\bullet),\scrh(\partial))$.
Thus we get a (contravariant) functor
$$(C_\bullet,\partial)\mapsto(\scrh(C_\bullet),\scrh(\partial)):
\{\hbox{chain complexes}\}\to\{\hbox{cochain complexes}\}.$$
This functor is also denoted $\Hom(\cdot,A)$.

\vskip.1in\noindent
{\it Fact.}
Let $0\to B'\to B\to B''\to 0$ be an exact sequence
of additive Abelian groups.
Let $A$ be an additive Abelian group
and let $\scrh:=\Hom(\cdot,A)$.
Then $0\to\scrh(B')\to\scrh(B)\to\scrh(B'')$
is exact.

\vskip.1in
That is, $\Hom(\cdot,A)$ is a ``left exact'' functor.
We leave it as an unassigned exercise to show that it is not fully
exact, {\it i.e.}, that $\scrh(B)\to\scrh(B'')$ is not
necessarily surjective.

Recall, that for any additive Abelian group $C$,
we define $K_C$ to be the kernel of the map $\phi_C:\Z[C]\to C$
which extends, by $\Z$-linearity, the identity map $C\to C$.
Recall that $\iota_C:K_C\to\Z[C]$ denotes the injection map.

Recall that the {\bf cokernel} of a map $\sigma:S\to T$ is the group
$T/(\sigma(S))$, so the cokernel of $\sigma$ measures how close
$\sigma$ is to being surjective.

\vskip.1in\noindent
{\it Definition.}
Let $A$ and $C$ be additive Abelian groups.
Let $\scrh:=\Hom(\cdot,A)$.
We define $\Ext(C,A)$ to be the
cokernel of $\scrh(\iota_C):\scrh(\Z[C])\to\scrh(K_C)$.
That is,
$$\Ext(C,A):=
{{\scrh(K_C)}\over{\im[\scrh(\iota_C):\scrh(\Z[C])\to\scrh(K_C)]}}.$$

\vskip.1in\noindent
{\it Remark.}
Let $A$ and $C$ be additive Abelian groups
and let $F'$ and $F$ be free Abelian groups.
Let $0\to F'\to F\to C\to0$ be an exact sequence.
Then $\Ext(C,A)$ is isomorphic to
the cokernel of the map $\Hom(F,A)\to\Hom(F',A)$
induced by $F'\to F$.

\vskip.1in\noindent
{\bf EXERCISE 17C:}
Let $m,n\in\Z\backslash\{0\}$.
Compute $\Ext(\Z/m,\Z/n)$.

\vskip.1in\noindent
{\bf EXERCISE 17D:}
Show, for any additive Abelian group $A$,
that $\Ext(\Z,A)=0$.

\vskip.1in\noindent
{\bf EXERCISE 17E:}
Let $m\in\Z\backslash\{0\}$.
Compute $\Ext(\Z/m,\Z)$.

\vskip.1in
Note, based on your answers to Exercise 17D and to Exercise 17E,
that $\Ext(A,B)$ is different from $\Ext(B,A)$.

\vskip.1in\noindent
{\it Fact.}
Let $A$, $B$ and $C$ be additive Abelian groups.
Then
$$\eqalign{
\Ext(A\oplus B,C)&\cong
\Ext(A,C)\oplus\Ext(B,C)\cr
\Ext(A,B\oplus C)&\cong
\Ext(A,B)\oplus\Ext(A,C).}$$

\vskip.1in
Let $A$ be an additive Abelian group.

We note that $\Ext(\cdot,A)$ is sometimes called
the ``derived functor'' of $\Hom(\cdot,A)$.
One can also prove that $\Ext(A,\cdot)$
is (equivalent to) the derived functor of
the covariant functor $\Hom(A,\cdot)$.

Similarly, $\Tor(\cdot,A)$ is the derived functor
of $\cdot\otimes A$ and one may prove that
$\Tor(A,\cdot)$ is (equivalent to)
the derived functor of $A\otimes\cdot$.
So, since $A\otimes\cdot$ and $\cdot\otimes A$
are equvalent, it follows that
$\Tor(A,\cdot)$ and $\Tor(\cdot,A)$ are equivalent,
which implies, for all additive Abelian groups $B$,
that $\Tor(A,B)$ is isomorphic to $\Tor(B,A)$.

On the other hand, we leave it as an unassigned exercise to
show that there exist choices of $A$ and $B$
for which $\Hom(A,B)$ is not isomorphic to $\Hom(B,A)$.
Thus $\Hom(A,\cdot)$ is not equivalent to $\Hom(\cdot,A)$.
Similarly, $\Ext(A,\cdot)$ is not equivalent to $\Ext(\cdot,A)$.

\vskip.1in\noindent
{\bf EXERCISE 17F:}
Compute
$\Ext([(\Z/3)\oplus(\Z/27)\oplus(\Z/54)],
[(\Z/2)\oplus(\Z/4)\oplus(\Z/12)])$,
up to isomorphism.
Put your answer in the form
$(\Z/a)\oplus(\Z/b)\oplus(\Z/c)\oplus\cdots$,
with $a|b|c|\cdots$.

\vskip.1in
Note that your answers to 
Exercise 15H, Exercise 15Q and Exercise 17F
are all the same.
Moreover, it would be the same if you were to compute
$\Hom([(\Z/3)\oplus(\Z/27)\oplus(\Z/54)],
[(\Z/2)\oplus(\Z/4)\oplus(\Z/12)])$.

\vskip.1in
In fact, we have four ``bifunctors''
($\otimes$, $\Tor$, $\Hom$ and $\Ext$)
on additive Abelian groups,
and they all agree when restricted to
finite additive Abelian groups.
We leave it as an unassigned exercise to verify
that they are all different when considering
infinite additive Abelian groups.

Given a cochain complex $(C^\bullet,d)$,
we define $Z^\bullet(C^\bullet):=\ker\,d$,
we define $B^\bullet(C^\bullet):=\im\,d$,
and we define its {\bf cohomology} to be
$H^\bullet(C^\bullet):=(\ker\,d)/(\im\,d)$.
Specifically, for all $n\in\Z$, we have
$$\eqalign{
Z^n(C^\bullet)&=\ker(d:C^n\to C^{n+1})\cr
B^n(C^\bullet)&=\im(d:C^{n-1}\to C^n)\cr
H^n(C^\bullet)&={{Z^n(C^\bullet)}\over{B^n(C^\bullet)}}.
}$$

We also have a {\bf Universal Coefficients Theorem}
for cohomology:

\vskip.1in\noindent
{\it Theorem.}
Let $(C_\bullet,d)$ be a free chain complex.
Then $H^n(\Hom(C_\bullet,A))$ is isomorphic to
$$\Hom(H^nC_\bullet,A)\qquad\oplus\qquad\Ext(H_{n-1}C_\bullet,A).$$

\vskip.1in
If $X$ is a topological space, then we define
the {\bf cohomology of $X$} as
$H^\bullet(X):=H^\bullet(\Hom(S_\bullet X,\Z))$.
If, in addition, $A$ is an additive Abelian group,
then we define the
{\bf cohomology of $X$ with coefficients in $A$}
as $H^\bullet(X;A):=H^\bullet(\Hom(S_\bullet X,A))$.
Then the preceding theorem implies that $H^\bullet(X;A)$ is isomorphic to
$$\Hom(H_n(X),A))\qquad\oplus\qquad\Ext(H_{n-1}(X),A).$$

\vskip.1in\noindent
{\bf EXERCISE 17G:}
Compute $H^\bullet(\R P^3)$.

\vskip.1in\noindent
{\bf EXERCISE 17H:}
Compute $H^\bullet(\R P^3;\Z/2)$.

\vskip.1in
One might ask what value there is in cohomology and cohomology
with coefficients (as well as homology with coefficients),
given that these are all computable directly from homology.

First, sometimes these other groups are more easily computable
than is homology, and they do provide invariants.
Moreover, sometimes we gain information through Universal
Coefficients about the homology groups, by first discovering
facts about homology or cohomology with coefficients.

Second, there are times when working with $\Z/2$ coefficients
is preferable to working with $\Z$ coefficients.
In fact, a chain with $\Z/2$
coefficients is just the same as a set of parametric simplices,
since the only (non-zero) possibility for a coefficient on a
parametric simplex is $1\in\Z/2$. So, when one ``pictures'' a
chain with $\Z/2$ coefficients, one is simply picturing a
collection of parametric simplices. This is particularly
helpful in dealing with non-orientable manifolds (defined later)
where one can't know whether, when a parametric simplex covers
a point in the space, the covering should count for $+1$ or $-1$.
With $\Z/2$ coefficients, one doesn't even ask the question;
one only asks whether the point is covered or not.

Third, there is a ring structure (to be defined below) in cohomology.
Don Kahn wrote me a note showing that $\R P^3$ has the same homology
groups and cohomology groups as $\R P^2\vee S^3$, but that their
cohomology rings are different, so that cohomology with the ring
structure can sometimes distinguishes spaces that are
indistinguishable using just cohomology groups.
(Note: If $X$ and $Y$ are connected topological manifolds,
then $X\vee Y$ is the topological space obtained by choosing
points $x\in X$ and $y\in Y$ and letting
$X\vee Y$ the space obtained by identifying $x$ with $y$
in $X\coprod Y$.
More precisely, it is the quotient
of the disjoint union $X\coprod Y$ by the
smallest equivalence relation on $X\coprod Y$
such that $x$ is equivalent to $y$.
One can show that, up to homeomorphism, the result is
independent of the choices of $x$ and $y$.
Note, by Mayer-Vietoris, that, for all integers $k\ge1$,
we have $H_k(X\vee Y)=(H_k(X))\oplus(H_k(Y))$.)

\vskip.1in
We now turn to the problem of computing $H_\bullet(X\times Y)$,
given knowledge of $H_\bullet(X)$ and $H_\bullet(Y)$.
This is part of a broader problem of finding the homology of
the total space of a fibration given the homology of its
fiber and base. That broader problem involves something
called the Leray-Serre Spectral Sequence, and we will not
present it here.

For the simplest fibrations, where one simply takes the
product of two spaces, computation breaks down into two
parts, one of which (called the Eilenberg-Zilber Theorem)
is topological and one of which (called the Kunneth Theorem)
is algebraic. To state these results, we need the following
definitions:

\vskip.1in\noindent
{\it Definition.}
Let $C_\bullet$ and $D_\bullet$ be two graded groups.
We then define the graded group
$C_\bullet\otimes D_\bullet$ by
$$(C_\bullet\otimes D_\bullet)_n\qquad:=\qquad
{\mathop\bigoplus_{p\in\Z}}\,C_p\otimes D_{n-p}.$$

\vskip.1in\noindent
{\it Definition.}
Let $C_\bullet$ and $D_\bullet$ be
two chain complexes.
Then $C_\bullet\otimes D_\bullet$ is a chain complex
via the diffential defined by, for all $p,q\in\Z$,
for all $\sigma\in C_p$, for all $\tau\in D_q$, we have:
$$\partial(\sigma\otimes\tau)\quad=\quad
[(\partial\sigma)\otimes\tau]\quad+\quad
[(-1)^p][\sigma\otimes(\partial\tau)].$$

\vskip.1in\noindent
{\it Eilenberg-Zilber Theorem.}
Let $X$ and $Y$ be topological spaces.
Then
$H_\bullet(X\times Y)$ is naturally isomorphic
to $H_\bullet((S_\bullet X)\otimes(S_\bullet Y))$.

\vskip.1in
By ``naturally isomorphic'', we mean that the
functor
$$(X,Y)\mapsto H_\bullet(X\times Y):
\{\hbox{ordered pairs of topological spaces}\}\to
\{\hbox{graded groups}\}$$
is equivalent to the functor
$$(X,Y)\mapsto H_\bullet((S_\bullet X)\otimes(S_\bullet Y)):
\{\hbox{ordered pairs of topological spaces}\}\to
\{\hbox{graded groups}\}.$$

We will comment on the proof later.

\vskip.1in\noindent
{\it Definition.}
Let $A_\bullet$ and $B_\bullet$ be graded groups.
Then we define $\Tor(A_\bullet,B_\bullet)$ to be
the graded group defined by
$$(\Tor(A_\bullet,B_\bullet))_k\qquad:=\qquad
{\mathop\bigoplus_{p\in\Z}}\,\Tor(A_p,B_{k-p}).$$

\vskip.1in\noindent
{\it Kunneth Theorem.}
Let $C_\bullet$ and $D_\bullet$ be chain complexes and let $n\in\Z$.
Then $H_n(C_\bullet\otimes D_\bullet)$ is isomorphic to
$$[(H_\bullet C_\bullet)\otimes(H_\bullet D_\bullet)]_n\qquad\oplus\qquad
[\Tor(H_\bullet C_\bullet,H_\bullet D_\bullet)]_{n-1}.$$

\vskip.1in
You may note some similarity with the Universal Coefficients Theorems
appearing above, and we comment that, in fact, the first of those
two theorems can be used to prove the Kunneth Theorem, see either
Vick's book ``Homology'' or Rotman's book on algebraic topology.
Beyond this, we will not comment on the proof of the Kunneth Theorem.

Putting the two preceding theorems together, for any
two topological spaces $X$ and $Y$, for any $n\in\Z$,
we have that $H_n(X\times Y)$ is isomorphic to
$$\bigg[{\mathop\bigoplus_{p\in\Z}}\,H_p(X)\otimes H_{n-p}(Y)\bigg]
\qquad\bigoplus\qquad
\bigg[{\mathop\bigoplus_{p\in\Z}}\,\Tor(H_p(X),H_{n-1-p}(Y))\bigg].$$

\vskip.1in\noindent
{\bf EXERCISE 17I:}
Compute $H_\bullet(\R P^3\times\R P^2)$.

\vskip.1in
Some comments on the proof of the Eilenberg-Zilber Theorem:
We will not prove this theorem fully. Instead we will
state another theorem called the Acyclic Models Theorem,
and we will show that it implies two results:
\itemitem{(1)} The fact that if $X$ and $Y$ are topological
spaces, if $f:X\to Y$ and $g:X\to Y$ are continuous maps and if
$f$ is homotopic to $g$, then $f_*:H_\bullet(X)\to H_\bullet(Y)$
and $g_*:H_\bullet(X)\to H_\bullet(Y)$ are equal; and
\itemitem{(2)} The Eilenberg-Zilber Theorem.

\vskip.1in
Recall that we have already proved (1). I will take the point of view
that Acyclic Models is simply a ``straightforward'' generalization of
(1), and that a person with a sufficient understanding of the proof of
(1) and of the proof that Acyclic Models implies (1) will be able to
figure out the proof of Acyclic Models. Since we will show how
Acyclic Models implies Eilenberg-Zilber, this will give a quasi-proof
of Eilenberg-Zilber.

Before stating Acycic Models, we need some definitions.

\vskip.1in\noindent
{\it Definition.}
Let $A$ be an additive group and let $S\subseteq A$.
We say that $S$ is a {\bf basis} of $A$ if the $\Z$-linear
extension $\Z[S]\to A$ of the inclusion $S\to A$ is
an isomorphism.

\vskip.1in
Note that $A$ is free Abelian iff $A$ admits a basis.

\vskip.1in\noindent
{\it Definition.}
Let $\scrc$ be a category. Then a {\bf model system} on $\scrc$
is a subset $\scrm$ of the class $\{\hbox{objects in }\scrc\}$,
together with a function $M\mapsto d_M:\scrm\to\{0,1,2,\ldots\}$.

\vskip.1in
Elements of $\scrm$ are called the {\bf models} of $(\scrm,d_\cdot)$.
For each $M\in\scrm$, $d_M$ is called the {\bf dimension} of $M$.
We will adopt the notation:

For all integers $n\ge0$,
$\scrm_n:=\{M\in\scrm\,|\,d_M=n\}$.

\vskip.1in\noindent
{\it Example.}
For example, consider $\scrc=\{\hbox{topological spaces}\}$
with $\scrm=\{\Delta^0,\Delta^1,\ldots\}$, the set of simplices,
and with $d_\cdot$ defined by $d_{\Delta^n}=n$.
Then $(\scrm,d_\cdot)$ is a model system on $\scrc$.
In this case, for all integers $n\ge0$, we have
$\scrm_n=\{\Delta^n\}$.

\vskip.1in
For any category $\scrc$, for any functor
$T_\bullet:\scrc\to\{\hbox{chain complexes}\}$,
for any $n\in\Z$, we will define
$T_n:\scrc\to\{\hbox{additive Abelian groups}\}$
by: $T_n(C)=(T_\bullet(C))_n$.

Recall that a chain complex $\scrc_\bullet$ is said to
be {\bf nonnegative} if, for all integers $n<0$,
we have $C_n=0$.

\vskip.1in\noindent
{\it Definition.}
Let $\scrc$ be a category.
Let $(\scrm,d_\cdot)$ be a model system on $\scrc$.
Let
$$T_\bullet:\scrc\to\{\hbox{nonnegative chain complexes}\}$$
be a functor.
Let $\displaystyle{\scrc_0:={\mathop\bigcup_{M\in\scrm}}\,T_{d_M}(M)}$.
A {\bf $T_\bullet$ model set} on $(\scrm,d_\cdot)$
is a function
$$M\mapsto e_M:\scrm\to\scrc_0$$
such that, for all $M\in\scrm$,
we have $e_M\in T_{d_M}(M)$.

\vskip.1in\noindent
{\it Example.}
For example, consider $\scrc=\{\hbox{topological spaces}\}$
with $\scrm=\{\Delta^0,\Delta^1,\ldots\}$, the set of simplices,
and with $d_\cdot$ defined by $d_{\Delta^n}=n$.
Consider $T_\bullet=S_\bullet$, the singular chain functor,
which we recall is defined by $S_n(X)=\Z[C(\Delta^n,X)]$.
For all integers $n\ge0$, let
$$e_{\Delta^n}:=\id_{\Delta^n}\in C(\Delta^n,\Delta^n)
\subseteq\Z[C(\Delta^n,\Delta^n)]=S_n(\Delta^n).$$
Then $e_\cdot$ is a $T_\bullet$ model set on $(\scrm,d_\cdot)$.

\vskip.1in\noindent
{\it Definition.}
Let $\scrc$ be a category.
Let $(\scrm,d_\cdot)$ be a model system on $\scrc$.
For all integers $n\ge0$, let $\scrm_n:=\{M\in\scrm\,|\,d_M=n\}$.
Let
$$T_\bullet:\scrc\to\{\hbox{nonnegative chain complexes}\}$$
be a functor.
Let $e_\cdot$ be a $T_\bullet$ model set on $(\scrm,d_\cdot)$.
We say that $e_\cdot$ is a {\bf basis} for $T_\bullet$
on $(\scrm,d_\cdot)$ if, for all $X\in\scrc$, for all integers $n\ge0$,
we have that
$$\{(Tf)(e_M)\,|\,M\in\scrm_n,f\in\Hom(M,X)\}$$
is a basis for $T_n(X)$.

\vskip.1in\noindent
{\it Definition.}
Let $\scrc$ be a category.
Let $(\scrm,d_\cdot)$ be a model system on $\scrc$.
Let
$$T_\bullet:\scrc\to\{\hbox{nonnegative chain complexes}\}$$
be a functor.
We say that $T_\bullet$ is {\bf free} on $(\scrm,d_\cdot)$
if there exists a basis for $T_\bullet$ on $(\scrm,d_\cdot)$.

\vskip.1in
Note that, in the preceding example, $e_\cdot$ is a basis
for $T_\bullet$ on $(\scrm,d_\cdot)$, so we see that
$T_\bullet$ is free on $(\scrm,d_\cdot)$.

\vskip.1in\noindent
{\it Definition.}
Let $\scrc$ be a category.
Let $\scrm$ be a subset of the class $\{\hbox{objects in }\scrc\}$.
Let
$$U_\bullet:\scrc\to\{\hbox{nonnegative chain complexes}\}$$
be a functor.
We say that $U_\bullet$ is {\bf$\scrm$-acyclic} if,
for all $M\in\scrm$, for all integers $n>0$,
we have $H_n(U_\bullet M)=0$.

\vskip.1in\noindent
{\it Example.}
Consider $\scrc=\{\hbox{topological spaces}\}$
with $\scrm=\{\Delta^0,\Delta^1,\ldots\}$, the set of simplices.
Let $d_\cdot$ be defined by $d_{\Delta^n}=n$.
Let $I:=[0,1]$.
Let $U_\bullet:\scrc\to\{\hbox{nonnegative chain complexes}\}$
be defined by $U_\bullet(X)=S_\bullet(X\times I)$.
Then $U_\bullet$ is $\scrm$-acyclic.

\vskip.1in\noindent
{\it Definition.}
Let $\scrc$, $\scrd$ and $\scre$ be categories.
Let $\scra:\scrc\to\scrd$, $\scrb:\scrc\to\scrd$
and $\scrf:\scrd\to\scre$ be functors.
Let $\tau:\scra\to\scrb$ be a natural transformation.
Then we denote by $\scrf\tau$
the natural transformation from
$\scrf\scra:\scrc\to\scre$ to $\scrf\scrb:\scrc\to\scre$
defined by $(\scrf\tau)_C=\scrf(\tau_C)$.

\vskip.1in
We may therefore write $\scrf\tau_C$ without fear of ambiguity.

\vskip.1in\noindent
{\it Definition.}
Let $\scrc$ be a category.
Let
$$T_\bullet,U_\bullet:\scrc\to\{\hbox{nonnegative chain complexes}\}$$
be functors.
Let $\tau,\mu:T_\bullet\to U_\bullet$
be natural transformations.
A natural transformation $h:T_\bullet\to U_{\bullet+1}$
is said to be a {\bf natural chain homotopy} from $\tau$ to $\mu$,
if, for all $X\in\scrc$, we have that $h_X:T_\bullet X\to U_{\bullet+1}X$
is a chain homotopy from 
$\tau_X:T_\bullet X\to U_\bullet X$ to 
$\mu_X:T_\bullet X\to U_\bullet X$.
We say that $\tau$ and $\mu$ are {\bf naturally chain homotopic}
if there exists a natural chain homotopy from $\tau$ to $\mu$.

\vskip.1in
Note that if $\tau$ and $\mu$ are naturally chain homotopic,
then $H_\bullet\tau=H_\bullet\mu$.

\vskip.1in\noindent
{\it Definition.}
Let $\scrc$ be a category.
Let
$$T_\bullet,U_\bullet:\scrc\to\{\hbox{nonnegative chain complexes}\}$$
be functors.
Let $\iota_T:T_\bullet\to T_\bullet$ be the identity natural transformation.
Let $\iota_U:U_\bullet\to U_\bullet$ be the identity natural transformation.
We say that $T_\bullet$ and $U_\bullet$ are
{\bf naturally chain homotopy equivalent}
if there are natural transformations
$\tau:T_\bullet\to U_\bullet$
and $\tau':U_\bullet\to T_\bullet$
such that
\itemitem{(1)} $\tau'\tau$ is naturally chain homotopic to $\iota_T$; and
\itemitem{(2)} $\tau\tau'$ is naturally chain homotopic to $\iota_U$.

\vskip.1in
If $T_\bullet$ and $U_\bullet$ are naturally chain homotopy equivalent,
then $H_\bullet T_\bullet$ is equivalent to $H_\bullet U_\bullet$.
This, in turn implies, for all $X\in\scrc$,
that $(H_\bullet T_\bullet)(X)\cong(H_\bullet U_\bullet)(X)$.

\vskip.1in\noindent
{\it Acyclic Models Theorem.}
Let $\scrc$ be a category.
Let $(\scrm,d_\cdot)$ be a model system on $\scrc$.
Let
$$T_\bullet,U_\bullet:\scrc\to\{\hbox{nonnegative chain complexes}\}$$
be functors.
Assume that $T_\bullet$ is free on $(\scrm,d_\cdot)$
and that $U_\bullet$ is acyclic on $\scrm$.
Then:
\itemitem{(1)}
For any natural transformation $\phi:H_0T_\bullet\to H_0U_\bullet$,
there exists a natural transformation $\tau:T_\bullet\to U_\bullet$
such that $H_0\tau=\phi$; and
\itemitem{(2)}
Let $\tau,\mu:T_\bullet\to U_\bullet$
be natural transformations and assume that $H_0\tau=H_0\mu$.
Then $\tau$ and $\mu$ are naturally chain homotopic.

\vskip.1in\noindent
{\it Corollary.}
Let $\scrc$ be a category.
Let $(\scrm,d_\cdot)$ be a model system on $\scrc$.
Let
$$T_\bullet,U_\bullet:\scrc\to\{\hbox{nonnegative chain complexes}\}$$
be functors.
Assume that $T_\bullet$ is free on $(\scrm,d_\cdot)$
and that $U_\bullet$ is acyclic on $\scrm$.
Let $\tau,\mu:T_\bullet\to U_\bullet$
be natural transformations and assume that $H_0\tau=H_0\mu$.
Then $H_\bullet\tau=H_\bullet\mu$.

\vskip.1in
Consequently, for all $X\in\scrc$,
we have $H_\bullet(\tau_X)=H_\bullet(\mu_X)$.

\vskip.1in\noindent
{\it Corollary.}
Let $\scrc$ be a category.
Let $(\scrm,d_\cdot)$ be a model system on $\scrc$.
Let
$$T_\bullet,U_\bullet:\scrc\to\{\hbox{nonnegative chain complexes}\}$$
be functors.
Assume that $T_\bullet$ and $U_\bullet$ are free on $(\scrm,d_\cdot)$.
Assume that $T_\bullet$ and $U_\bullet$ are acyclic on $\scrm$.
Assume that $H_0T_\bullet$ is equivalent to $H_0U_\bullet$.
Then $T_\bullet$ and $U_\bullet$ are naturally chain
homotopy equivalent.

\vskip.1in
Consequently $H_\bullet T_\bullet$
is equivalent to $H_\bullet U_\bullet$.
Consequently, for all $X\in\scrc$,
we have $(H_\bullet T_\bullet)(X)\cong(H_\bullet U_\bullet)(X)$.

\vskip.1in\noindent
{\it Proof:}
Let $\iota_T:T_\bullet\to T_\bullet$ and
$\iota_U:U_\bullet\to U_\bullet$ denote the identity natural
transformations.
Let $\iota_T^H:H_0T_\bullet\to H_0T_\bullet$ and
$\iota_U^H:H_0U_\bullet\to H_0U_\bullet$ denote the identity
natural transformations.

Choose $\phi:H_0T_\bullet\to H_0U_\bullet$
and $\phi':H_0U_\bullet\to H_0T_\bullet$
such that $\phi'\phi=\iota_T^H$ and such that $\phi\phi'=\iota_U^H$.
By the existence part of the Acyclic Models Theorem,
choose $\tau:T_\bullet\to U_\bullet$
and $\tau':U_\bullet\to T_\bullet$
such that $H_0\tau=\phi$ and such that $H_0\mu=\phi'$.

We have $H_0(\tau'\tau)=\phi'\phi=\iota_T^H=H_0\iota_T$
and $H_0(\tau\tau')=\phi\phi'=\iota_U^H=H_0\iota_U$.
So, by the uniqueness part of the Acyclic Models Theorem,
we see that
$\tau'\tau$ is naturally chain homotopic to $\iota_T$
and that 
$\tau\tau'$ is naturally chain homotopic to $\iota_U$.
{\bf QED}

\vskip.1in\noindent
{\it Application.}
For example, consider $\scrc=\{\hbox{topological spaces}\}$
with $\scrm=\{\Delta^0,\Delta^1,\ldots\}$, the set of simplices,
and with $d_\cdot$ defined by $d_{\Delta^n}=n$.
Let $I:=[0,1]$.

Let $T_\bullet:=S_\bullet$ be the singular chain functor,
which is defined by $S_n(X)=\Z[C(\Delta^n,X)]$.
Let $U_\bullet:\scrc\to\{\hbox{nonnegative chain complexes}\}$
be defined by $U_\bullet(X)=S_\bullet(X\times I)$.

Define $\tau:T_\bullet\to U_\bullet$ by
$\tau_X=(\cdot,0)_*:S_\bullet X\to S_\bullet(X\times I)$.
Define $\mu:T_\bullet\to U_\bullet$ by
$\mu_X=(\cdot,1)_*:S_\bullet X\to S_\bullet(X\times I)$.
By Exercise 18A below, $H_0\tau=H_0\mu$, so,
by the first corollary to the Acyclic Models Theorem,
we see that $H_\bullet\tau=H_\bullet\mu$.

\vskip.1in\noindent
{\bf EXERCISE 18A:}
Show, in the notation of the previous application,
that $H_0\tau=H_0\mu$.

\vskip.1in\noindent
{\it Corollary.}
Let $X$ and $Y$ be topological spaces
and let $f,g:X\to Y$ be homotopic maps.
Then $f_*,g_*:H_\bullet X\to H_\bullet Y$ are equal.

\vskip.1in\noindent
{\it Proof:}
By the previous application, we see that
$H_\bullet\tau_X=H_\bullet\mu_X$.
Since
$$H_\bullet\tau_X=(\cdot,0)_*:H_\bullet X\to H_\bullet(X\times I)$$
and
$$H_\bullet\mu_X=(\cdot,1)_*:H_\bullet X\to H_\bullet(X\times I)$$
we conclude that
$(\cdot,0)_*:H_\bullet X\to H_\bullet(X\times I)$
and 
$(\cdot,1)_*:H_\bullet X\to H_\bullet(X\times I)$
are equal.

Let $h:X\times I\to Y$ be a homotopy from $f$ to $g$.
Then $h\circ(\cdot,0)=f$ and $h\circ(\cdot,1)=g$.
Then
$$h_*\circ(\cdot,0)_*=f_*:H_\bullet X\to H_\bullet Y$$
and
$$h_*\circ(\cdot,1)_*=g_*:H_\bullet X\to H_\bullet Y.$$
Then
$f_*=h_*\circ(\cdot,0)_*=h_*\circ(\cdot,1)_*=g_*$.
{\bf QED}

\vskip.1in
This justifies our point of view that the Acyclic Models
Theorem is ``just'' a generalization of the basic fact
that two homotopic maps induce the same map on homology.
Since we know this basic fact, we consider the
Acyclic Models Theorem to be proved.
For a full proof, see
Vick's book ``Homology'' or Rotman's book on algebraic topology.

Our next application of Acyclic Models is the Eilenberg-Zilber
Theorem, which we recall states:

\vskip.1in\noindent
{\it Eilenberg-Zilber Theorem.}
Let $X$ and $Y$ be topological spaces.
Then
$H_\bullet(X\times Y)$ is naturally isomorphic
to $H_\bullet((S_\bullet X)\otimes(S_\bullet Y))$.

\vskip.1in\noindent
{\it Proof:}
Let $\scrc$ be the category of ordered pairs of topological spaces.
Let
$$\scrm:=\{(\Delta^k,\Delta^l)\,|\,k,l\in\{0,1,2,\ldots\}\}.$$
Define $d_\cdot:\scrm\to\{0,1,2\ldots\}$
by $d_{(\Delta^k,\Delta^l)}=k+l$.

Recall that 
$S_\bullet:\{\hbox{topological spaces}\}\to
\{\hbox{nonnegative chain complexes}\}$
denotes the singular chain complex functor.
Define
$T_\bullet,U_\bullet:\scrc\to\{\hbox{nonnegative chain complexes}\}$
by $T_\bullet(X,Y)=S_\bullet(X\times Y)$
and $U_\bullet(X,Y)=(S_\bullet(X))\otimes(S_\bullet(Y))$.
It suffices to show that $T_\bullet$ and $U_\bullet$
are naturally chain homotopy equivalent.

\vskip.1in\noindent
{\bf EXERCISE 18B:}
Show that
$H_0T_\bullet,H_0U_\bullet:\scrc\to\{\hbox{additive Abelian groups}\}$
are equivalent.

\vskip.1in\noindent
{\bf EXERCISE 18C:}
Show that $T_\bullet$ and $U_\bullet$ are free on $(\scrm,d_\cdot)$.

\vskip.1in\noindent
{\bf EXERCISE 18D:}
Show that $T_\bullet$ and $U_\bullet$ are acyclic on $\scrm$.

\vskip.1in
Then, by the second corollary following the statement
of the Acyclic Models Theorem, we conclude that
$T_\bullet$ and $U_\bullet$ are naturally chain
homotopy equivalent.
{\bf QED}

\vskip.1in
We now begin a new topic: The ring structure in cohomology.

Recall that a {\bf ring} is an additive Abelian group $R$
together with a $\Z$-bilinear ``multiplication map''
$(a,b)\mapsto ab:R\times R\to R$
such that
\itemitem{(1)} $\forall a,b,c\in R$, $a(b+c)=ab+ac$,
$(a+b)c=ac+bc$ and $a(bc)=(ab)c$; and
\itemitem{(2)} there exists $1\in R$ such that, for all $a\in R$,
we have $1a=a1=a$.

\noindent
The ring $R$ is {\bf commutative} if, for all $a,b\in R$,
we have $ab=ba$.

\vskip.1in\noindent
{\it Definition.}
A {\bf pre-graded ring} is a graded group $R_\bullet$
together with a ``multiplication map''
$\mu:R_\bullet\otimes R_\bullet\to R_\bullet$.
By convention, for all $p,q\in\Z$,
for all $a\in R_p$, $b\in R_q$ we denote
$\mu(a\otimes b)$ by $ab$.
A pre-graded ring $R_\bullet$ is said to be
a {\bf graded ring} if
\itemitem{(1)} for all $p,q,r\in\Z$,
for all $a\in R_p$, $b\in R_q$, $c\in R_r$,
we have $a(b+c)=ab+ac$, $(a+b)c=ac+bc$ and $a(bc)=(ab)c$; and
\itemitem{(2)} there exists $1\in R_0$ such that, for all $p\in\Z$,
for all $a\in R_p$, we have $1a=a1=a$.

\noindent
The graded ring $R_\bullet$ is said to be {\bf graded commutative}
if, for all $p,q\in\Z$, for all $a\in R_p$, for all $b\in R_q$,
we have $ab=(-1)^{pq}ba$.

\vskip.1in
A graded commutative graded ring will simply be called
a {\bf graded commutative ring}.
Note that there is a forgetful functor
$$\scrf:\{\hbox{graded commutative rings}\}\to\{\hbox{graded groups}\},$$
in which one simply forgets the multiplication map.

Let $R$ be a commutative ring whose underlying additive Abelian group
is denoted~$A$.
(That is, $A$ is equal to: $R$, after forgetting about multiplication.)
Our next goal is to construct a functor
$$H^\bullet(\cdot;R):\{\hbox{topological spaces}\}\to
\{\hbox{graded commutative rings}\}$$
such that $\scrf\circ(H^\bullet(\cdot;R))=H^\bullet(\cdot;A):
\{\hbox{topological spaces}\}\to\{\hbox{graded groups}\}$.

We will also indicate the existence of topological spaces
$X$ and $Y$ such that, with $R=\Z$, we have
$$H^\bullet(X;R)\cong H^\bullet(Y;R),\qquad\hbox{but}\qquad
H^\bullet(X;A)\not\cong H^\bullet(Y;A).$$
This shows that, for the case $R=\Z$,
the functor $H^\bullet(\cdot;R)$ is better at distinguishing topological
spaces than is the functor $H^\bullet(\cdot;A)$.
This is true for other rings $R$, as well.

Finally, we note that we could set up our definitions so that a
slightly more sophisticated version of $H(\cdot;R)$ takes values in the
category of ``graded commutative $R$-algebras''.  We do not wish to
define this term, since the definitions are already somewhat
complicated.
However, we do comment that there is a forgetful functor
$$\{\hbox{graded commutative $R$-algebras}\}\to
\{\hbox{graded commutative rings}\}$$
and our $H(\cdot;R)$ is simply obtained by composing this more
sophisticated $H(\cdot;R)$ with this forgetful functor.  Note that it
might be that this more sophisticated $H(\cdot;R)$ is better at
distinguishing topological spaces that our $H(\cdot;R)$ is.

Note that, given a topological space $X$, the meaning of
$H^\bullet(X):=H^\bullet(X;\Z)$ is now ambiguous, depending on whether
we think of $\Z$ as a ring (in which case $H^\bullet(X)$ is a graded
commutative ring), or as a group (in which case $H^\bullet(X)$ is a
graded group).
Generally, the reader must simply figure this out from context.
The same comment applies with $\Z$ replaced by $\Z/2$
or with many other commutative rings,
since the underlying additive Abelian group
is usually written with the exact same notation as the ring itself.

With those general comments behind us, we fix a topological space $X$
and a commutative ring $R$,
and proceed to define the commutative graded ring $H(X;R)$.
The (contravariant)
functoriality of $H(\bullet;R)$ will be left as an unassigned exercise for
the reader.

Fix $p,q\in\Z$.
Define $f:\Delta^p\to\Delta^{p+q}$ by
$f(t_0,\ldots,t_p)=(t_0,\ldots,t_p,0,\ldots,0)$
and define $b_p:\Delta^q\to\Delta^{p+q}$ by
$b(t_0,\ldots,t_q)=(0,\ldots,0,t_0,\ldots,t_q)$.
For all $\sigma\in C(\Delta^{p+q},X)$,
we define $\sigma_f:=\sigma\circ f\in C(\Delta^p,X)$
and $\sigma_b:=\sigma\circ b\in C(\Delta^q,X)$.
For all $\alpha\in S^p(X;R)$, for all $\beta\in S^q(X;R)$,
define $\alpha\cup\beta\in S^{p+q}(X;R)$ to be the
$\Z$-linear extension to $S_{p+q}(X)$ of
the map
$C(\Delta^{p+q},X)\quad\to\quad R$
given by
$$\sigma\mapsto(\alpha(\sigma_f))(\beta(\sigma_b)).$$
Note that, in this formula, we use the multiplication in $R$.
That is, $R$ must be a ring, or this would not make sense.
Note also that
$$(\alpha,\beta)\qquad\mapsto\qquad\alpha\cup\beta\qquad:\qquad
(S^p(X;R))\times(S^q(X;R))\qquad\to\qquad S^{p+q}(X;R)$$
is $R$-bilinear.

We leave it as an unassigned exercise to show,
for all $p,q\in\Z$, for all $\alpha\in S^p(X;R)$, $\beta\in S^q(X;R)$,
we have
$d(\alpha\cup\beta)=[(d\alpha)\cup\beta]+[(-1)^p][\alpha\cup(d\beta)]$.
Using this, it is an easy (unassigned) exercise to show,
for all $p,q\in\Z$,
for all $\alpha\in Z^p(X;R)$, $\beta\in Z^q(X;R)$,
that $\alpha\cup\beta\in Z^{p+q}(X;R)$.

Fix $p,q\in\Z$ and let $\alpha'\in H^p(X;R)$ and let $\beta'\in H^q(X;R)$.
We define $\alpha'\cup\beta'\in H^{p+q}(X;R)$, as follows:
Let
$$\eqalign{
a&:Z^p(X;R)\to H^p(X;R),\cr
b&:Z^q(X;R)\to H^q(X;R),\cr
c&:Z^{p+q}(X;R)\to H^{p+q}(X;R)}$$
be the canonical maps.
Choose $\alpha\in Z^p(X;R)$, $\beta\in Z^q(X;R)$
such that $a(\alpha)=\alpha'$ and $b(\beta)=\beta'$.
We define $\alpha'\cup\beta':=c(\alpha\cup\beta)$.
We leave it as an unassigned exercise show that this
map is well-defined.
({\it Hint:} Use the above-mentioned $R$-bilinearity
together with
$d(\alpha\cup\beta)=[(d\alpha)\cup\beta]+[(-1)^p][\alpha\cup(d\beta)]$.)

We leave it as an unassigned exercise to show,
for any topological space $X$, for any commutative ring $R$,
that $(H^\bullet(X;R),\cup)$ is a commutative graded ring.

\vskip.1in\noindent
{\it Example.}
$H^\bullet(\R P^n;\Z/2)$ is isomorphic
to the commutative graded ring
$\displaystyle{{(\Z/2)[x]}\over{(x^{n+1})}}$

\vskip.1in
A map $f:S^3\to S^3$ will be said to be {\bf odd}
if, for all $x\in S^3$, we have $f(-x)=-(f(x))$.

\vskip.1in\noindent
{\it Lemma.}
Let $\psi:S^3\to S^3$ be odd.
Then $\psi$ is not homotopic to a constant map.

\vskip.1in\noindent
{\it Proof:}
Let $p:S^3\to\R P^3$ be the canonical map.
Because $\psi:S^3\to S^3$ is odd, it follows
that there is a continuous map $\chi:\R P^3\to\R P^3$
such that $\chi\circ p=p\circ\psi$.

\vskip.1in\noindent
{\bf EXERCISE 19B:}
Let $s\in\R P^3$ and let $t:=\chi(s)$.
Show that
$$\chi_*\qquad:\qquad\pi_1(\R P^3,s)\qquad\to\qquad\pi_1(\R P^3,t)$$
is nontrivial.

\vskip.1in\noindent
{\bf EXERCISE 19C:}
Show that $p_*:H_3(S^3)\to H_3(\R P^3)$ is nontrivial.
(Note: You may assume, without proof, the fact that
the antipodal map $A:S^n\to S^n$
induces a map $A_*:H_n(S^n)\to H_n(S^n)$
such that, for all $z\in H_n(S^n)$,
we have $A_*(z)=(-1)^{n+1}z$.)

\vskip.1in
Since $H_3(S^3)\cong\Z\cong H_3(\R P^3)$, it follows,
from Exercise 19C, that the homomorphism
$p_*:H_3(S^3)\to H_3(\R P^3)$ is injective.

Recall that $\pi_1(\R P^3,s)\cong\Z/2\cong\pi_1(\R P^3,t)$.
Then, by Exercise 19B and the Hurewicz Theorem,
we see that $\chi_*:H_1(\R P^3)\to H_1(\R P^3)$
is nontivial.
Then, as $H_1(\R P^3)\cong\Z/2$,
we conclude that $\chi_*:H_1(\R P^3)\to H_1(\R P^3)$
is the identity map.

We have $\Ext(H^0(\R P^3),\Z/2)=\{0\}$.
Then, by the Universal Coefficients Theorem,
$H^1(\R P^3;\Z/2)\cong\Hom(H_1(\R P^3),\Z/2)\cong\Z/2$.
Also, by naturality in the Universal Coefficients Theorem, the map
$\chi^*:H^1(\R P^3;\Z/2)\to H^1(\R P^3;\Z/2)$
is obtained by applying the functor $\Hom(\cdot,\Z/2)$
to the identity map
$$\chi_*:H_1(\R P^3)\to H_1(\R P^3).$$
Then $\chi^*:H^1(\R P^3;\Z/2)\to H^1(\R P^3;\Z/2)$
is the identity map.

Let $u$ be the nontrivial element of $H^1(\R P^3;\Z/2)$.
Then $\chi^*(u)=u$.
Let
$$v:=u\cup u\cup u\in H^3(\R P^3;\Z/2).$$
Then, by the graded ring structure on $H^\bullet(\R P^3;\Z/2)$,
described in the preceding Example, we see that $v\ne0$.
Moreover, as
$\chi^*:H^\bullet(\R P^3;\Z/2)\to H^\bullet(\R P^3;\Z/2)$
is a homomorphism of graded commutative rings,
and as $\chi^*(u)=u$,
we see that $\chi^*(v)=v$.
Then
$$\chi^*:H^3(\R P^3;\Z/2)\to H^3(\R P^3;\Z/2)$$
is nontrivial.

We have $H^2(\R P^3)=\{0\}$, so $\Ext(H^2(\R P^3),\Z/2)=\{0\}$.
Thus, another application of naturality in the Universal Coefficients
shows that
$$\chi^*:H^3(\R P^3;\Z/2)\to H^3(\R P^3;\Z/2)$$
is obtained by applying $\Hom(\cdot,\Z/2)$ to
$$\chi_*:H_3(\R P^3)\to H_3(\R P^3).$$
Then $\chi_*:H_3(\R P^3)\to H_3(\R P^3)$ is nontrivial.
So, as $H_3(\R P^3)\cong\Z$, we conclude that
$\chi_*:H_3(\R P^3)\to H_3(\R P^3)$ is injective.

Since $p_*:H_3(S^3)\to H_3(\R P^3)$ and
$\chi_*:H_3(\R P^3)\to H_3(\R P^3)$ are both injective,
it follows that $(\chi\circ p)_*:H_3(S^3)\to H_3(\R P^3)$
is injective.
Then, as $\chi\circ p=p\circ\psi$,
we conclude that $p_*\circ\psi_*:H_3(S^3)\to H_3(\R P^3)$
is injective, so
$\psi_*:H_3(S^3)\to H_3(S^3)$
is injective, and therefore nonzero.
Then $\psi:S^3\to S^3$ cannot be homotopic to a constant map.
{\bf QED}

\vskip.1in
We now use the above example and lemma to prove the

\vskip.1in\noindent
{\it Ham Sandwich Theorem.}
Let $\lambda$ denote Lebesgue measure in $\R^3$.
Let $A$, $B$ and $C$ be measurable subsets of $\R^3$,
all of which have finite Lebesgue measure.
Then there exists a half-space $H\subseteq\R^3$
such that
$$\lambda(A\cap H)=\lambda(A\backslash H),\qquad
\lambda(B\cap H)=\lambda(B\backslash H)\qquad\hbox{and}\qquad
\lambda(C\cap H)=\lambda(C\backslash H).$$

\vskip.1in
NOTE: This result implies that, given a ham sandwich with a piece
of wheat bread, a piece of ham and a piece of rye bread,
one can, with a single stroke of the knife, bisect each of
the three parts of the sandwich.

\vskip.1in\noindent
{\it Proof:}
Define $\rho:\{\hbox{subsets of }\R^3\}\to\R^3$ by
$$\rho(H)=
((\,\lambda(A\cap H))-(\lambda(A\backslash H))\,,\,
(\lambda(B\cap H))-(\lambda(B\backslash H))\,,\,
(\lambda(C\cap H))-(\lambda(C\backslash H))\,).$$
We wish to show, for some half-space $H$ of $\R^3$,
that $\rho(H)=0$.

Define $i:\R^3\to\R^4$ by $i(w)=(w,1)$.
Let $\cdot$ denote the dot product in $\R^4$.
Define $\chi:S^3\to\{\hbox{subsets of }\R^3\}$
by $\chi(v)=\{w\in\R^3\,|\,v\cdot(i(w))\ge0\}$.
Let $\phi:=\rho\circ\chi:S^3\to\R^3$.
Then $\phi$ is continuous.
We wish to show, for some $v\in S^3$, that $\phi(v)=0$.

\vskip.1in\noindent
{\bf EXERCISE 19A:}
Show, for all $v\in S^3$, that $\phi(-v)=-(\phi(v))$.

\vskip.1in
By Exercise 19A, $\phi:S^3\to\R^3$ is odd.
It is an unassigned exercise to show that 
$\phi:S^3\to\R^3$ is continuous.
The remainder of the proof shows that the image of any odd continuous
function $S^3\to\R^3$ must contain the origin $0:=(0,0,0)$.
Assume, for all $v\in S^3$, that $\phi(v)\ne0$.
We aim for a contradiction.

Define $\psi:S^3\to S^3$ by
$$\psi(v)=
\left(\quad{{\phi(v)}\over{\|\phi(v)\|}}\quad,\quad0\quad\right).$$
Then $\psi$ is continuous.
Also, for all $v\in S^3$,
we have $\psi(-v)=-(\psi(v))$,
{\it i.e.}, $\psi$ is odd.
Moreover,
$$\psi(S^3)\qquad\subseteq\qquad S^2\times\{0\}\qquad\subsetneq\qquad S^3,$$
so $\psi:S^3\to S^3$ is not surjective.
It follows that $\psi:S^3\to S^3$ is homotopic to a constant map.

However the preceding lemma asserts that
any odd continuous
function $S^3\to S^3$ cannot be homotopic to a constant map.
{\bf QED}

\vskip.1in
Our next ``application'' of algebraic topology is the

\vskip.1in\noindent
{\it Brouwer Fixed-Point Theorem.}
Let $n\ge1$ be an integer and let $D$ denote the
closed unit ball in $\R^n$.
Let $f:D\to D$ be continuous.
Then there exists $x\in D$ such that $f(x)=x$.

\vskip.1in\noindent
{\it Proof:}
The case $n=1$ is an exercise in calculus, so we assume
that $n\ge2$.
Suppose the result is false. We aim for a contradiction.

For all $x\in X$, let $L_x:=\{(1-t)(f(x))+tx\,|\,t\in\R\}$
be the line through $f(x)$~and~$x$.
Then $R_x:=\{(1-t)(f(x))+tx\,|\,t>0\}$
is that connected component of $L_x\backslash\{f(x)\}$
containing $x$.
We leave it as an unassigned exercise in geometry to show,
for all $a,b\in D$, that, if $L$ denotes the line through
$a$ and $b$, then the connected component $R$ of $L\backslash\{a\}$
which contains $b$ has the property that $R\cap S^{n-1}$ contains
exactly one point.
For all $x\in X$, let $g(x)$ be
the unique element of $R_x\cap S^{n-1}$.
Then $g:D\to S^{n-1}$ is continuous and, for all $x\in S^{n-1}$,
we have $g(x)=x$.

Let $i:S^{n-1}\to D$ be the inclusion map.
Let $\id:S^{n-1}\to S^{n-1}$ be the identity map.
Then $g\circ i=\id$.
Then the composition of
$i_*:H_{n-1}(S^{n-1})\to H_{n-1}(D)$
with $g_*:H_{n-1}(D)\to H_{n-1}(S^{n-1})$
is the identity map $\id_*:H_{n-1}(S^{n-1})\to H_{n-1}(S^{n-1})$.
On the other hand, $D$ is contractible, so $H_{n-1}(D)=\{0\}$,
so this composite must be trivial.
However, $H_{n-1}(S^{n-1})\cong\Z$,
so the map
$\id_*:H_{n-1}(S^{n-1})\to H_{n-1}(S^{n-1})$
cannot be both the identity map and trival at the same time,
so we have a contradiction.
{\bf QED}

\vskip.1in
When $R$ is a ring, we have seen that $H^\bullet(X;R)$
has the structure of a graded commutative ring.
We now explain how,
if $R$ is a ring and $M$ is an $R$-module,
then $H^\bullet(X;M)$ has the structure of
a graded $(H^\bullet(X;R))$-module.
To do so, we must, for all $p,q\in\Z$,
define a multiplication map
$$H^p(X;R)\times H^q(X;M)\to H^{p+q}(X;M).$$

Let $X$ be a topological space, let $R$ be a ring
and let $M$ be an $R$-module.
Let $p,q\in\Z$.
Define $f:\Delta^p\to\Delta^{p+q}$ by
$f(t_0,\ldots,t_p)=(t_0,\ldots,t_p,0,\ldots,0)$
and define $b_p:\Delta^q\to\Delta^{p+q}$ by
$b(t_0,\ldots,t_q)=(0,\ldots,0,t_0,\ldots,t_q)$.
For all $\sigma\in C(\Delta^{p+q},X)$,
we define $\sigma_f:=\sigma\circ f\in C(\Delta^p,X)$
and $\sigma_b:=\sigma\circ b\in C(\Delta^q,X)$.

Given $\alpha\in S^p(X;R)$ and $\beta\in S^q(X;M)$,
define $\alpha\cup\beta:S_{p+q}(X)\to R$ by:
for all $\sigma\in C(\Delta^{p+q},X)$,
$(\alpha\cup\beta)(\sigma)=(\alpha(\sigma_f))(\beta(\sigma_b))
\in RM\subseteq M$.
It is an unassigned exercise to show that this map
$$(\alpha,\beta)\mapsto\alpha\cup\beta:
S^p(X;R)\times S^q(X;M)\to S^{p+q}(X;M)$$
restricts to a map
$$(\alpha,\beta)\mapsto\alpha\cup\beta:
Z^p(X;R)\times Z^q(X;M)\to Z^{p+q}(X;M)$$
and then factors to a map
$$(\alpha,\beta)\mapsto\alpha\cup\beta:
H^p(X;R)\times H^q(X;M)\to H^{p+q}(X;M).$$
Here, $\alpha\cup\beta$ is called the {\bf cup product}
of $\alpha$ and $\beta$.
In the above, two important cases are:
\itemitem{(1)} $R$ any ring and $M=R$; and
\itemitem{(2)} $R=\Z$ and $M$ any additive Abelian group.

\noindent
This concludes our discussion of ``cup products''
Next, we define the ``cap'' product.

Again, let $X$ be a topological space, let $R$ be a ring
and let $M$ be an $R$-module.
Let $p,q\in\Z$.
We will define a bilinear map
$$(a,\beta)\mapsto a\cap\beta:
H_{p+q}(X;R)\times H^q(X;M)\to H_p(X;M).$$

As before,
define $f:\Delta^p\to\Delta^{p+q}$ by
$f(t_0,\ldots,t_p)=(t_0,\ldots,t_p,0,\ldots,0)$
and define $b_p:\Delta^q\to\Delta^{p+q}$ by
$b(t_0,\ldots,t_q)=(0,\ldots,0,t_0,\ldots,t_q)$.
As before,
for all $\sigma\in C(\Delta^{p+q},X)$,
we define $\sigma_f:=\sigma\circ f\in C(\Delta^p,X)$
and $\sigma_b:=\sigma\circ b\in C(\Delta^q,X)$.

For all $\beta\in S^q(X;M)$,
define $\Phi_\beta:S_{p+q}(X;R)\to S_p(X;M)$
to be the $R$-linear extension of
$$\sigma\mapsto
(\beta(\sigma_b))\sigma_f:
C(\Delta^{p+q},X)\to S_p(X;M).$$
Then, for all $a\in S_{p+q}(X;R)$, for all $\beta\in S^q(X;M)$,
define $a\cap\beta=\Phi_\beta(a)$.
It is an unassigned exercise to show that this map
$$(a,\beta)\mapsto a\cap\beta:
S_{p+q}(X;R)\times S^q(X;M)\to S_p(X;M)$$
restricts to a map
$$(a,\beta)\mapsto a\cap\beta:
Z_{p+q}(X;R)\times Z^q(X;M)\to Z_p(X;M)$$
and then factors to a map
$$(a,\beta)\mapsto a\cap\beta:
H^p(X;R)\times H^q(X;M)\to H_p(X;M).$$
Here, $a\cap\beta$ is called the {\bf cap product}
of $a$ and $\beta$.

This concludes our discussion of cap products.
We now define the Kronecker pairing.

Let $X$ be a topological space and let $q\in\Z$.
Recall that $S^q(X)=\Hom(S_q(X),\Z)$.
Define $\Phi_0:S^q(X)\times S_q(X)\to\Z$ by
$\Phi_0(f,s)=f(s)$.
Let $\Phi_1:Z^q(X)\times Z_q(X)\to\Z$
be the restriction of $\Phi_0$ to $Z^q(X)\times Z_q(X)$.
Let $\alpha:Z^q(X)\to H^q(X)$
and $\beta:Z_q(X)\to H_q(X)$ be the canonical maps.
Define $\Phi:H^q(X)\times H_q(X)\to\Z$ by
$\Phi(\alpha(a),\beta(b))=\Phi_1(a,b)$;
we leave it as an unassigned exercise to show that this
is well-defined.
This map $\Phi:H^q(X)\times H_q(X)\to\Z$
is called the {\bf Kronecker pairing} determined by $X$.

We leave it as an unassigned exercise to verify:
If $p,q,r\in\Z$, if $\alpha\in H^p(X)$, if $\beta\in H^q(X)$
if $c\in H_r(X)$ and if $p+q=r$,
then $\Phi(\alpha\cup\beta,c)=\Phi(\alpha,c\cap\beta)$.
That is, ``cupping with $\beta$ is adjoint to capping with $\beta$,
with respect to the Kronecker pairing.''

We now start on orientation.

\vskip.1in\noindent
{\it Definition.}
Let $M$ be a connected $d$-dimensional topological manifold
and let $S$ be a subset of $M$.
We say that $M$ is {\bf direct orientable} along $S$
if $H_d(M,M\backslash S)$ is isomorphic to the
additive group $\Z$.
A {\bf direct orientation} of $M$ along $S$
is an isomorphism $H_d(M,M\backslash S)\to\Z$.

\vskip.1in\noindent
{\bf EXERCISE 20A:}
Let $d\ge1$ be an integer.
Let $M$ be a connected $d$-dimensional topological manifold
and let $m\in M$.
Show that $M$ is direct orientable along $\{m\}$.

\vskip.1in\noindent
{\bf EXERCISE 20B:}
Let $d\ge1$ be an integer.
Let $M$ be a connected $d$-dimensional topological manifold
and let $m\in M$.
Show that there is a neighborhood $V$ of $m$ in $M$
such that $M$ is direct orientable along $V$.

\vskip.1in
Note that, if $M$ is a topological manifold
and if $m\in M$, then there exist exactly two
direct orientations of $M$ along $\{m\}$.

\vskip.1in\noindent
{\it Definition.}
Let $M$ be a topological manifold.
For each $m\in M$, let $\Sigma_m$ be the set
of direct orientations of $M$ along $\{m\}$.
A {\bf preorientation} of $M$
is a map
$\displaystyle{\tau:M\to{\mathop\bigcup_{m\in M}}\,\Sigma_m}$
such that, for all $m\in M$, we have $\tau(m)\in\Sigma_m$.

\vskip.1in\noindent
{\it Definition.}
Let $M$ be a topological manifold
and let $\tau$ be a preorientation of $M$.
Let $d:=\dim(M)$ and let $V\subseteq M$.
For all $v\in V$,
let $\phi_v:H_d(M,M\backslash V)\to H_d(M,M\backslash\{v\})$
be the map induced by the inclusion
$M\backslash V\subseteq M\backslash\{v\}$.
We say that $\tau$ is {\bf coherent} along $V$ if
there is a direct orientation $\mu:H_d(M,M\backslash V)\to\Z$
of $M$ along $V$ such that, for all $v\in V$, we have
$(\tau(v))\circ\phi_v=\mu$.

\vskip.1in\noindent
{\it Definition.}
Let $M$ be a topological manifold
and let $\tau$ be a preorientation of $M$.
We say that $\tau$ is an {\bf orientation} of $M$
if, for all $m\in M$, there is an open neighborhood $V$ of $m$ in $M$
such that $\tau$ is coherent along $V$.

\vskip.1in
We will assume the next result without proof.

\vskip.1in\noindent
{\it Theorem.}
Let $M$ be a manifold and let $A$ be an Abelian group.
Let $d:=\dim(M)$.
Then, for all integers $k>d$, we have $H_k(M;A)=0$.
Moreover, if $M$ is noncompact, then we have $H_d(M;A)=0$.

\vskip.1in\noindent
{\bf EXERCISE 20C:}
Let $M$ be a $d$-dimensional topological manifold
and let $m\in M$.
Assume that $H_d(M)\ne\{0\}$.
Let $\phi:H_d(M)\to H_d(M,M\backslash\{m\})$
be the map induced by the inclusion
$\emptyset\subseteq M\backslash\{m\}$.
Show that $\phi:H_d(M)\to H_d(M,M\backslash\{m\})$
is an isomorphism.
({\it Hint:} Using long exact sequences and the preceding theorem,
show that $\phi:H_d(M)\to H_d(M,M\backslash\{m\})$ is injective.
Then, using Exercise 20A, show that $H_d(M)$ is isomorphic to $\Z$.
Note that, if $\psi:\Z\to\Z$ is not surjective,
then, for some finite cyclic group $A$,
we have $\psi\otimes\id_A:\Z\otimes A\to\Z\otimes A$ is not injective.
It therefore suffices to show, for all finite cyclic groups $A$, that
$\phi\otimes\id_A:(H_d(M))\otimes A\to(H_d(M,M\backslash\{m\}))\otimes A$
is injective.
For any additive Abelian group $A$,
let $\phi_A:H_d(M;A)\to H_d(M,M\backslash\{m\};A)$
be the map induced by the inclusion
$\emptyset\subseteq M\backslash\{m\}$.
Using long exact sequences and the preceding theorem,
show that $\phi_A:H_d(M;A)\to H_d(M,M\backslash\{m\};A)$ is injective.
Conclude, by naturality in the Universal Coefficients Theorem, that
$\phi\otimes\id_A:(H_d(M))\otimes A\to(H_d(M,M\backslash\{m\}))\otimes A$
is injective.)

\vskip.1in
We will assume the next result without proof.

\vskip.1in\noindent
{\it Theorem.}
Let $M$ be a compact, connected topological manifold.
Then any orientation on $M$ is coherent along $M$.

\vskip.1in\noindent
{\it Corollary.}
Let $M$ be a compact, connected, $d$-dimensional topological manifold.
Then the following are equivalent:
\itemitem{(A)} $M$ is orientable;
\itemitem{(B)} $H_d(M)\ne\{0\}$;
\itemitem{(C)} $H_d(M)$ is isomorphic to $\Z$; and
\itemitem{(D)} $M$ is direct orientable along $M$.

\vskip.1in\noindent
{\it Proof:}
By Exercise 20A, we see, for all $m\in M$,
that $H_d(M,M\backslash\{m\})\cong\Z$.
For all $m\in M$, let
$\phi_m:H_d(M)\to H_d(M,M\backslash\{m\})$
be the map induced by the inclusion
$\emptyset\subseteq M\backslash\{m\}$.

The equivalence of (B) and (C) follows from Exercise 20C.
The equivalence of (C) and (D) follows
from the definition of ``direct orientable''.

If there is a orientation on $M$, then,
by the theorem, it is coherent on $M$,
and it follows from the definition of ``coherent''
that, for all $m\in M$, the map
$$\phi_m:H_d(M)\to H_d(M,M\backslash\{m\})$$
is an isomorphism.
Thus (A) implies (C).

It remains to show that (C) implies (A).
Let $\rho:H_d(M)\to\Z$ be an isomorphism.
By Exercise 20C, we see, for all $m\in M$,
that $\phi_m:H_d(M)\to H_d(M,M\backslash\{m\})$ is an isomorphism.
Define an orientation $\tau$ by: for all $m\in M$,
$\tau(m):=\rho\circ\phi_m^{-1}:H_d(M,M\backslash\{m\})\to\Z$.
Then $\tau$ is coherent along $M$,
and so $\tau$ is an orientation.
{\bf QED}

\vskip.1in\noindent
{\it Definition.}
Let $M$ be a compact, connected, orientable, $d$-dimensional manifold.
A {\bf fundamental class} on $M$ is a generator of $H_d(M)$.

\vskip.1in
We will assume the next result without proof.

\vskip.1in\noindent
{\it Poincare Duality.}
Let $M$ be a compact, connected, orientable, $d$-dimensional
topological manifold.
Let $A$ be an additive Abelian group.
Let $z$ be a fundamental class on $M$.
Let $q\in\Z$.
Then the map
$\omega\mapsto z\cap\omega:H^q(M;A)\to H_{d-q}(M;A)$
is an isomorphism.

\vskip.1in
The basic intuition behind Poincare Duality is as follows.
Suppose $M$ is a compact, connected, orientable, $d$-dimensional
topological manifold.
Let $q\in\Z$.
Let $a\in H_q(M)$ and let $b\in H_{d-q}(M)$.
At some intuitive level, because $a$ and $b$ have
complementary dimensions, it should be true that
there is an intersection of $a$ with $b$ that is finite
in size.
To compute the ``size'' $n$ of this intersection,
proceed as follows:
\itemitem{(1)} Choose almost any $q$-cycle $a_0$ in $a$
and almost any $(d-q)$-cycle $b_0$ in $b$; then
\itemitem{(2)}take the union $A\subseteq M$ of the
images of the parametric simplices
in $a_0$; then
\itemitem{(3)} take the union $B\subseteq M$ of the parametric simplices
in $b_0$; then
\itemitem{(4)} develop some careful bookkeeping system to tell the
number $n_p$ of times any point of $p\in A\cap B$ is covered by a simplex
in $a_0$ and one in $b_0$; and then
\itemitem{(5)} let $n:=\sum n_p$.

\noindent
Of course, even if we could do the careful bookkeeping required,
we would need to show that $n$ is independent of the choice of $a_0$
and $b_0$, so, to accomplish all this is a daunting program.

On the other hand, with Poincare Duality and the Kronecker pairing,
we can obtain this number $n$ more easily:
Let $\Phi:H^\bullet(M)\times H_\bullet(M)\to\Z$ be the Kronecker pairing.
Let $D:H^\bullet(M)\to H_{d-\bullet}(M)$ be the isomorphism
defined by $D(\omega)=z\cap\omega$.
Then we define the {\bf intersection number} of $a$ and $b$
(with respect to $z$)
to be $\Phi(D^{-1}(a),b)$.
We will show below that this is not commutative in $a$ and $b$,
although its absolute value is.
That is, we have $\Phi(D^{-1}(a),b)=\pm[\Phi(D^{-1}(b),a)]$.
The absolute value of this number may be thought of, intuitively,
as the ``size'' of the intersection of $a$ with $b$.

For $p,q\in\Z$, given $a\in H_p(M)$ and $b\in H_q(M)$
we define
$$a\sqcap b:=D((D^{-1}a)\cup(D^{-1}b))\in H_{p+q-d}(M),$$
and thus give a sense of what it means to ``intersect'' a pair
of homology classes.
This kind of intersection theory is perhaps the primary motivation
for Poincare Duality.
Note that, because $H^\bullet(M)$ is a graded commutative
ring, it follows that
$$a\sqcap b\qquad:=\qquad(-1)^{(d-p)(d-q)}\,\,(b\sqcap a).$$

Let $z$ is a fundamental class on a $d$-dimensional
compact, orientable manifold $M$.
Define an isomorphism $\Psi:H_0(M)\to\Z$
by $\Psi(c)=\Phi(D^{-1}(c),z)$.
Let $p,q\in\Z$ and assume that $p+q=d$.
Let $a\in H_p(M)$ and $b\in H_q(M)$.
Then the intersection number of $a$ and $b$ is
equal to $\Psi(a\sqcap b)$, because
$\Phi(D^{-1}a,b)=
\Phi(D^{-1}a,z\cap(D^{-1}b))=
\Phi((D^{-1}a)\cup(D^{-1}b),z)=
\Phi(D^{-1}(a\sqcap b),z)=\Psi(a\sqcap b)$.
Note that, as
$a\sqcap b:=(-1)^{(d-p)(d-q)}\,(b\sqcap a)$,
this implies that
$$\Phi(D^{-1}a,b)\qquad=\qquad(-1)^{(d-p)(d-q)}\,\,[\Phi(D^{-1}b,a)].$$
That is, the intersection number only commutes up to sign.

We now go carefully through an example.

Let $M:=\R^2/\Z^2$, the $2$-torus.
Then $d:=\dim(M)=2$.
Let $p:=1$ and $q:=1$.
Let $f:\R^2\to M$ be the canonical map.
Let $u:=(0,0)$, $v:=(0,1)$, $w:=(1,1)$ and $x:=(1,0)$.

Recall, for any $r,s,t\in\R^2$, that $[rst]$ denotes
the affine parametric $2$-simplex $\Delta^2\to\R^2$
which maps $e_0^2$ to $r$, $e_1^2$ to $s$ and $e_2^2$ to $t$.
Similary $[rs]:\Delta^1\to\R^2$ is the unique affine map
sending $e_0^1$ to $r$ and $e_1^1$ to $s$.

Let $\hatz:=f_*([uvw]-[uxw])\in Z_2(M)$.
Let $z$ denote the image of $\hatz$ under the
canonical map $Z_2(M)\to H_2(M)$.
Then $z$ is a fundamental class of $M$.

Let $k:=(u+v)/2$, $l:=(v+w)/2$, $m:=(w+x)/2$
and $n:=(u+x)/2$.
Let $\hata:=f_*[nl]\in Z_1(M)$ and $\hatb:=f_*[km]\in Z_1(M)$.

We now posit the existence of $\hatalpha\in Z^1(M)$
such that, intuitively, for any parametric $1$-simplex $\sigma$,
$\hatalpha(\sigma)$ counts the number of times that
$\sigma$ crosses $\hata$ from left to right
(after lifting from $M$ into $[0,1]\times[0,1]$).
All we will use is that $\hatalpha(\hatb)=1$,
that $\hatalpha(f_*[vw])=1$
and that $\hatalpha(f_*[xw])=0$.
We ask our reader to accept the existence of such an element
$\hatalpha\in Z^1(M)$.
Let $\alpha$ denote the image of $\hatalpha$
under the canonical map $Z^1(M)\to H^1(M)$.

We now see, from the definition of cap product, that
$$\hatz\cap\hatalpha\qquad=\qquad
(\hatalpha(f_*[uvw]_b))\,\,(f_*[uvw]_f)\,\,-\,\,
(\hatalpha(f_*[uxw]_b))\,\,(f_*[uxw]_f).$$
From the definition of front and back faces,
we have $[uvw]_b=[vw]$, $[uvw]_f=[uv]$,
$[uxw]_b=[xw]$ and $[uxw]_f=[ux]$.
Substituting, and using the assumed properties of $\hatalpha$,
we get $\hatz\cap\hatalpha=f_*[uv]$.
Since $f_*[uv]$ is homologous to $\hata$,
if we take the image of the equation
$\hatz\cap\hatalpha=f_*[uv]$
under the canonical map $Z_1(M)\to H_1(M)$,
we get $z\cap\alpha=a$.
That is, $D(\alpha)=a$.
Then the intersection number of $a$ and $b$
is given by
$\Phi(D^{-1}a,b)=\Phi(\alpha,b)
=\hatalpha(\hatb)=1$.
If one pictures $a$ and $b$ by picturing $\hata$ and $\hatb$,
then this is a reasonable result, since $\hata$ and $\hatb$
do intersect in just one point.

\vskip.1in\noindent
{\it Definition.}
Let $X$ be a topological space
and let $p\in\Z$.
Let $\sigma\in\Z_p(X)$.
Choose an integer $j\ge0$,
choose $c_1,\ldots,c_j\in\Z\backslash\{0\}$
and choose $\sigma_1,\ldots,\sigma_j\in C(\Delta^p,X)$
such that $\sigma=c_1\sigma_1+\cdots+c_j\sigma_j$.
Then we define
$\im(\sigma):=(\sigma_1(\Delta^p))\cup\cdots\cup(\sigma_j(\Delta^p))$.

\vskip.1in
The next result, which we state without proof,
is based on the intuition that there is a connection
between intersection number and actual intersection.

\vskip.1in\noindent
{\it Theorem.}
Let $M$ be a compact, connected, oriented manifold.
Let $d:=\dim(M)$ and let $k\in\Z$.
Let $a_0\in Z_k(M)$ and $b_0\in Z_{d-k}(M)$.
Let $a$ and $b$, be, respectively, the images
of $a_0$ and $b_0$ under the canonical maps
$Z_k(M)\to H_k(M)$ and $Z_{d-k}(M)\to H_{d-k}(M)$.
Fix a fundamental class $z$ on $M$.
Assume that the intersection number (with respect to $z$)
of $a$ and $b$ is $\ne0$.
Then $(\im(a_0))\cap(\im(b_0))\ne\emptyset$.

\vskip.1in\noindent
{\it Corollary.}
Let $f:S^2\to S^2$ be homotopic to the identity map $S^2\to S^2$.
Then there exists $p_0\in S^2$ such that $f(p_0)=p_0$.

\vskip.1in
{\it Note 1:} The proof given below works, not just for $S^2$,
but for any compact, orientable
topological manifold whose Euler number is nonzero.

{\it Note 2:} One may use this result to argue that if $M$
is a smooth compact, orientable manifold with nonzero Euler number
then, for any smooth vector field $V$ on $M$,
there exists $m_0\in M$ such that $V_{m_0}=0$.
(This requires ideas from differential topology, which is
covered {\it next}, so we are actually getting ahead of ourselves here.)
Here's why: Let $\phi_t$ denote the time $t$ flow of $V$.
For all integers $j\ge0$, let $M_j:=\{m\in M\,|\,\phi_{1/2^j}(m)=m\}$.
The preceding corollary (together with Note 1 above)
shows, for all integers $n\ge0$, that $M_n\ne\emptyset$.
Moreover, we have $M_0\supseteq M_1\supseteq M_2\supseteq\cdots$.
Then, by the finite intersection property, there exists
$m_0\in\cap_{j=1}^\infty\,M_j$.
Then, for all $i,j\in\Z$, if $j\ge0$, then $\phi_{i/2^j}(m_0)=m_0$.
Since $\{i/2^j\,|\,i,j\in\Z,i\ge0\}$ is dense in $\R$,
we conclude, for all $t\in\R$, that $\phi_t(m_0)=m_0$.
Then $V_{m_0}=0$.

We now proceed to the proof of the corollary.

\vskip.1in\noindent
{\it Proof:}
Let $M:=S^2\times S^2$.

Let $z$ be a fundamental class on $S^2$.
Choose $z_0\in Z_2(S^2)$ such that
the image of $z_0$ under the canonical map
$Z_2(S^2)\to H_2(S^2)$ is a fundamental class $z$ of $S^2$.

Let $I:S^2\to M$ be defined by $I(p)=(p,p)$
and let $F:S^2\to M$ be defined by $F(p)=(p,f(p))$
As $f$ is homotopic to the identity,
it follows that $I$ and $F$ are homotopic.
We wish to show that 
$(I(S^2))\cap(F(S^2))\ne\emptyset$.

Let $a_0:=I_*(z_0)\in Z_2(M)$
and let $b_0:=F_*(z_0)\in Z_2(M)$.
Let $a:=I_*(z)\in H_2(M)$ and let $b:=F_*(z)\in H_2(M)$.

Since $z$ is a fundamental class on $S^2$,
it follows that $a\cup a$ is a fundamental class on~$M$.

Since $\im(a_0)\subseteq I(S^2)$
and $\im(b_0)\subseteq F(S^2)$,
it suffices to show $(\im(a_0))\cap(\im(b_0))\ne\emptyset$.
Then, by the preceding theorem,
it suffices to show that the intersection number of $a$~and~$b$
(with respect to $a\cup a$)
is nonzero.

However, as $I$ and $F$ are homotopic,
it follows that $I_*:H_2(S^2)\to H_2(M)$
and $F_*:H_2(S^2)\to H_2(M)$ are equal.
Then $a=I_*(z)=F_*(z)=b$.

Then the intersection number of $a$ and $b$
is equal to that of $a$ and $a$.
By the following theorem, this number is
$$\sum_i\,(-1)^i[\dim(H_i(S^2;\R))]=1-0+1-0+0-0+0-\cdots=2.\qquad{\bf QED}$$

Given a compact, connected, oriented topological manifold $X$,
note that $M:=X\times X$ is also compact, connected and oriented.
Let $d:=\dim(X)$.
Let $I:X\to M$ be the diagonal map $I(x)=(x,x)$.
Let $z$ be a fundamental class on $X$ and let $a:=I_*(z)\in H_d(M)$.
Then $a\cup a$ is a fundamental class on $M$.
The preceding proof indicates that we will often be interested
in calculating the intersection number of $a$ with itself.

This is a well-known problem, but to give the answer, we need
to first define the ``Euler number'' of a topological space.
This, in turn, requires that we define Betti numbers.

\vskip.1in\noindent
{\it Definition.}
Let $X$ be a topological space and let $i\in\Z$.
Then the {\bf$i$th Betti number} of~$X$
is defined as $\beta_i(X):=\dim(H_i(X;\R))$.

\vskip.1in
For any Abelian group $A$, we have $\Tor(A,\R)=0$, so,
by the Universal Coefficients Theorem,
$H_i(X;\R)\cong(H_i(X))\otimes\R$.
Note that, if $H_i(X)=\Z^k\oplus F$
and if $F$ is finite,
then $(H_i(X))\otimes\R=\R^k$,
which implies that $\dim(H_i(X;\R))=k$.
Thus, if $H_i(X)$ is finitely generated,
we have $\beta_i(X)=\rank(H_i(X))$.

For any mapping $i\mapsto a_i:\Z\to\R$,
if $\{i\in\Z\,|\,a_i\ne0\}$ is finite,
then we define $\sum'\,a_i:=\sum\,(-1)^ia_i$.
This is called the {\bf alternating sum} of the $a_i$.

We'll say that a topologial space
{\bf admits an Euler number} if
$\{i\in\Z\,|\,\beta_i(X)\ne0\}$
is finite and if, for all $i\in\Z$, $\beta_i(X)$ is finite.
In this case, we define $\chi(X):=\sum'\,\beta_i(X)$.
This is called the {\bf Euler number} of $X$;
it is the alternating sum of the Betti numbers of $X$.

\vskip.1in
We omit the proof of the next result.

\vskip.1in\noindent
{\it Theorem.}
Let $X$ be a compact, connected, oriented topological manifold $X$.
and let $M:=X\times X$.
Let $d:=\dim(X)$.
Let $I:X\to M$ be the diagonal map $I(x)=(x,x)$.
Let $z$ be a fundamental class on $X$ and let $a:=I_*(z)\in H_d(M)$.
Then the intersection number of $a$ and $a$
(with respect to the fundamental class $a\cup a\in H_{2d}(M)$)
is equal to $\chi(X)$.

\vskip.1in
Some remarks on the calculations of Euler numbers:
Let $G_\bullet$ be a graded group.
We'll say that $G_\bullet$ {\bf admits an Euler number} if
$\{i\in\Z\,|\,G_i\otimes\R\ne\{0\}\}$ is finite
and if, for all $i\in\Z$, $G_i\otimes\R$ is finite-dimensional.
In this case,
we define the {\bf Euler number} of $G_\bullet$
to be $\chi(G_\bullet):=\sum'\dim(G_i\otimes\R)$.

Note that if $X$ is a topological space admitting an Euler number,
then $H_\bullet(X)$ admits an Euler number and
then $\chi(X)=\chi(H_\bullet(X))$.

Let $(C_\bullet,\partial)$ be a chain complex.
We say that $(C_\bullet,\partial)$ {\bf admits an Euler number}
if its underlying graded group $C_\bullet$ does,
and, in this case, we define its {\bf Euler number}
to be $\chi(C_\bullet,\partial)=\chi(C_\bullet)$.
Then we have the following elementary lemma.

\vskip.1in\noindent
{\it Lemma.}
Let $C_\bullet$ be a chain complex
admitting an Euler number.
Then $H_\bullet(C_\bullet)$ admits an Euler number and
$\chi(C_\bullet)=\chi(H_\bullet(C_\bullet))$.

\vskip.1in\noindent
{\it Proof:}
Let $H_\bullet:=H_\bullet(C_\bullet)$.
We wish to show that $\chi(C_\bullet)=\chi(H_\bullet)$.

Let $Z_\bullet:=Z_\bullet(C_\bullet)$.
Let $B_\bullet:=B_\bullet(C_\bullet)$.
For all $i\in\Z$,
define $c_i:=\dim(C_i\otimes\R)$,
$h_i:=\dim(H_i\otimes\R)$,
$z_i:=\dim(Z_i\otimes\R)$,
$b_i:=\dim(B_i\otimes\R)$.

For $i\in\Z$,
since $H_i=Z_i/B_i$, we get $h_i=z_i-b_i$.
For $i\in\Z$,
since there is a surjection $C_i\to B_{i-1}$
with kernel $Z_i$,
we have $C_i/Z_i\cong B_{i-1}$,
and so $c_i=z_i+b_{i-1}$.

Then $\chi(C_\bullet)=\sum'\,c_i
=(\sum'\,z_i)+(\sum'\,b_{i-1})=
(\sum'\,z_i)-(\sum'\,b_i)=\sum'\,h_i=\chi(H_\bullet)$.
{\bf QED}

\vskip.1in
Now assume that $X$ is the realization of a finite two-dimensional
simplicial complex.
(For example, you might consider carefully the case where $X$ is the
surface of a tetrahedron, which is homeomorphic to $S^2$.)
Let $v$ be the number of vertices in the simplicial complex,
let $e$ be the number of edges
and let $f$ be the number of faces.
The simplicial complex gives rise to a skeletal filtration on $X$,
and let $C_\bullet$ denote the resulting cellular chain complex.

The wonderful thing about Euler numbers is that, to find
$\chi(C_\bullet)$, we need only know the ranks of the terms
of $C_\bullet$.
The boundary maps (which are harder to analyze) can be ignored.
In the case under discussion,
the simplical complex is two-dimensional,
so, for all integers $i\notin\{0,1,2\}$,
we have that $C_i=\{0\}$.
Moreover, we have $\rank(C_0)=v$,
$\rank(C_1)=e$ and $\rank(C_2)=f$.
Then $\chi(C_\bullet)=v-e+f$.

Via spectral sequences, we verified that
$H_\bullet(X)\cong H_\bullet(C_\bullet)$.
Then, by the preceding lemma,
we have $\chi(H_\bullet(X))=\chi(C_\bullet)$.
Then $\chi(X)=\chi(H_\bullet(X))=\chi(C_\bullet)=v-e+f$.

In particular, when $X$ is the surface of a tetrahedron,
we have $\chi(X)=4-6+4=2$.
So since $X$ is homeomorphic to $S^2$,
we see that $\chi(S^2)=2$.
Alternatively, viewing $\chi(S^2)$ as the alternating
sum of the Betti numbers of $S^2$,
we get $\chi(S^2)=1-0+1-0+0-0+\cdots=2$.

\vskip.1in
Next topic: There exist two topological spaces $X$ and $Y$ such that
$H_\bullet(X)\cong H_\bullet(Y)$ but such that the graded rings
$H^\bullet(X)$ and $H^\bullet(Y)$ are not isomorphic.
(Note that, by Universal Coefficients, the graded groups
$H^\bullet(X)$ and $H^\bullet(Y)$ are isomorphic.)
Specifically, let $X:=\C P^2$ and let $Y:=S^2\vee S^4$.
We leave it as an unassigned exercise to show, for all $k\in\{0,2,4\}$, that
$H_k(X)\cong\Z\cong H_k(Y)$.
We leave it as an unassigned exercise to show, for all integers
$k\notin\{0,2,4\}$, that
$H_k(X)\cong\{0\}\cong H_k(Y)$.
Recall that, by the graded ring structure on $H^\bullet(X)$,
we have $[H^2(X)][H^2(X)]\ne\{0\}$.
However, if $x,y\in H^2(Y)$, then we can argue that $x\cup y=0$,
as follows:
Let $\phi:H^2(Y)\to H^2(S^2)$ and $\psi:H^4(Y)\to H^4(S^2)$
be the maps induced by the inclusion
$S^2\hookrightarrow Y$.
Let $\phi':H^2(Y)\to H^2(S^4)$ and $\psi':H^4(Y)\to H^4(S^4)$
be the maps induced by the inclusion
$S^4\hookrightarrow Y$.
Then $\psi'(x\cup y)=[\phi'(x)]\cup[\phi'(y)]$,
by functoriality of $H^\bullet$ from topological spaces to graded rings.
We have $\psi(x\cup y)\in H^4(S^2)=\{0\}$.
Moreover, $\phi'(x),\phi'(y)\in H^2(S^4)=\{0\}$.
Then $\psi(x\cup y)=0$ and $\psi'(x\cup y)=[\phi'(x)]\cup[\phi'(y)]=0$.
By Mayer-Vietoris, the map
$\psi\oplus\psi':H^4(Y)\to[H^4(S^2)]\oplus[H^4(S^4)]$
is an isomorphism, and so $x\cup y=0$.

Note that most of the algebraic topological tools we have described
have trouble distinguishing between homotopy equivalent topological
spaces. An amazing exception is the dimension of a topological manifold,
which shows, for example that $S^1$ and $S^1\times\R$ are not
homeomorphic. This follows from the Invariance of Domain
theorem, which implies that two manifolds of different dimension
cannot be homeomorphic. Recall that this result is proved
by algebraic topological techniques, something truly remarkable.
Generally, however, when one is presented with to homotopy
equivalent topological manifolds of the same dimension,
it is hard to know if they are homeomorphic.

In this context, the following result is quite important:

\vskip.1in\noindent
{\it Weak Version of the Mostow Rigidity Theorem.}
Let $M$ and $N$ be compact hyperbolic manifolds.
If $M$ and $N$ are homotopy equivalent, then
$M$ is homeomorphic to $M$.

\vskip.1in
We do not define ``hyperbolic'' here, but note that the class
of hyperbolic manifolds is very broad and
is of interest to many manthematicians.
You may want to talk to Prof.~Al Marden if you'd like to hear
more about this theorem.

Also, if the dimension of $M$ or $N$ is $\ge3$, then
homotopy equivalent does not just imply homeomorphic but
actually implies isometric (which means isomorphic in the
category of hyperbolic manifolds).
Finally, there are many other strengthenings of this theorem
that have been noted over the years, and there is a related
subject, ``superrigidity'', which is quite an active
area of current research.

\vskip.1in
Our next topic is the {\it degree} of a map.

\vskip.1in\noindent
{\it Definition.}
Let $M$ be a compact, orientable, $d$-dimensional topological manifold.
Let $f:M\to M$ be continuous.
The {\bf degree} of $f$, denoted $\deg(f)$, is the unique $n\in\Z$
such that, for all $x\in H_dM$, we have $f_*(x)=nx$.

\vskip.1in\noindent
{\bf EXERCISE 21A:}
Let $p:=(0,1)\in S^1$.
Define $R:S^1\to S^1$ by $R(x,y)=(-x,y)$,
so that $R(p)=p$.
Let $R_*:\pi_1(S^1,p)\to\pi_1(S^1,p)$ be the induced map
on fundamental groups.
Let $\gamma\in\pi_1(S^1,p)$.
Show that $R_*(\gamma)=\gamma^{-1}$.

\vskip.1in
Then, by the Hurewicz Theorem, it follows that $R_*:H_1S^1\to H_1S^1$
is multiplication by $-1$, i.e., that $R:S^1\to S^1$ has degree $-1$.
Similarly, one can argue:

\vskip.1in\noindent
{\it Example.}
The map $(cos(t),\sin(t))\mapsto(\cos(nt),\sin(nt)):S^1\to S^1$
has degree $n$.

\vskip.1in
We collect, without proof, some basic (easily proved) properties
of degree:

\vskip.1in\noindent
{\it Lemma.}
Let $M$ be a compact, orientable topological manifold.
Then:
\itemitem{(1)} If $f,g:M\to M$ are homotopic maps, then $\deg(f)=\deg(g)$.
\itemitem{(2)} If $\id:M\to M$ is the identity map, then $\deg(\id)=1$.
\itemitem{(3)} If $f:M\to M$ is a constant map, then $\deg(f)=0$.
\itemitem{(4)} If $f,g:M\to M$, then $\deg(f\circ g)=[\deg(f)][\deg(g)]$.

\vskip.1in\noindent
{\it Theorem.}
Let $d\ge1$ be an integer.
Then
\itemitem{(1)} for any integer $n\ge1$, there exists $f:S^d\to S^d$
such that $\deg(f)=n$; and
\itemitem{(2)} for any $f,g:S^d\to S^d$, we have:
$f$ is homotopic to $g$ iff $\deg(f)=\deg(g)$.

\vskip.1in
That is, the degree map gives a bijection between the set of integers
and the set of homotopy classes of maps $S^d\to S^d$.

In what follows (using a process called ``suspension''),
we will show, for any integer $d\ge2$ and any integer $n$,
that if there is a continuous map $S^{d-1}\to S^{d-1}$
of degree $n$, then
there is a continuous map $S^d\to S^d$ of degree $n$.
By the preceding Example, for any integer~$n$,
there is a continuous map $S^1\to S^1$ of degree $n$.
Thus, by induction, Conclusion~(1) of the preceding theorem is true.
The ``only if'' part of (2) follows from (1) of the preceding lemma.
The ``if'' part of (2) is hard and is, in fact, beyond the
scope of this course.

We now define the suspension of a map.
Let $J:=[-1,1]$.
The {\bf suspension functor}
$\Sigma:\{\hbox{topological spaces}\}\to\{\hbox{topological spaces}\}$
is defined by letting $\Sigma X$ be the quotient of $X\times J$
by the smallest equivalence relation $R$ on $X$
satisfying both
\itemitem{(a)} $X\times\{0\}$ is an equivalence class of $R$; and
\itemitem{(b)} $X\times\{1\}$ is an equivalence class of $R$.

\noindent
That is, we form $X\times J$, then collapse $X\times\{0\}$ to
a point, and then collapse $X\times\{1\}$ to a point. The result
is $\Sigma X$.
For spheres, note that $\Sigma S^d$ is homeomorphic to $S^{d+1}$,
as follows: Define $\phi:S^d\times J\to S^{d+1}$ by:
$\phi(x,t)=(x\sqrt{1-t^2},t)$ and note that $\phi$
factors to a homeomorphism $\psi:\Sigma S^d\to S^{d+1}$.
For any continuous map $f:S^{d-1}\to S^{d-1}$,
we define $\Sigma_1f:S^d\to S^d$ by
$\Sigma_1f:=\psi\circ(\Sigma f)\circ\psi^{-1}$.
For all $x\in S^{d-1}$, for all $t\in J$, we have
($\Sigma_1f)(x\sqrt{1-t^2},t)=((f(x))\sqrt{1-t^2},t)$.
Sometimes $\Sigma_1f$ is called the {\bf suspension} of $f$.

\vskip.1in\noindent
{\it Theorem.}
Let $d\ge1$ be an integer and
let $f:S^d\to S^d$ be continuous.
Then we have: $\deg(\Sigma_1f)=\deg(f)$.

\vskip.1in\noindent
{\it Proof:}
For concretness, we will prove this for the case $d=1$.
The proof for other dimensions is completely analogous.

Let $k:=\deg(f)$.
Then $f_*:H_1S^1\to H_1S^1$ is
multiplication by $k$.

Let $F:=\Sigma_1f$
and let $\Lambda:=F_*:H_2S^2\to H_2S^2$.
We wish to show that $\Lambda$ is multiplication by $k$.

Let $E$ be the equator of $S^2$.
Let
$$S^2_+:=\{(x,y,z)\in S^2\,|\,z\ge0\},\qquad
S^2_-:=\{(x,y,z)\in S^2\,|\,z\le0\}.$$
Let $F^0:=F|E:E\to E$,
let $F^+:=F|S^2_+:S^2_+\to S^2_+$
and let $F^-:=F|S^2_-:S^2_-\to S^2_-$.
Note that $F^0:E\to E$ is isomorphic
(in the arrow category of the category of topological spaces)
to $f:S^1\to S^1$.
It follows that $F^0_*:H_1E\to H_1E$
is isomorphic
(in the arrow category of the category of additive Abelian groups)
to $f_*:H_1S^1\to H_1S^1$.
Thus $F^0_*:H_1E\to H_1E$ is multiplication by $k$.


Let
$$\tildeS^2:=(S^2,E),\qquad
\tildeS^2_+:=(S^2_+,E),\qquad
\tildeS^2_-:=(S^2_-,E).$$
Let $\tildeF:=(F,F^0):\tildeS^2\to\tildeS^2$.
Let
$$\tildeF^+:=(F^+,F^0):\tildeS^2_+\to\tildeS^2_+,\qquad
\tildeF^-:=(F^-,F^0):\tildeS^2_-\to\tildeS^2_+.$$

By functoriality of long exact sequences,
we see that $\tildeF^+_*:H_2\tildeS^2_+\to H_2\tildeS^2_+$
is isomorphic (in the arrow category of the
category of additive Abelian groups)
to $F^0_*:H_1E\to H_1E$ 
Thus $\tildeF^+_*:H_2\tildeS^2_+\to H_2\tildeS^2_+$
is multiplication by $k$.
Similarly,
$\tildeF^-_*:H_2\tildeS^2_-\to H_2\tildeS^2_-$
is multiplication by $k$.

Let $A:=H_2\tildeS^2_+\oplus H_2\tildeS^2_-$
and $\Phi:=\tildeF^+_*\oplus\tildeF^-_*:A\to A$.
Then $\Phi$ is multiplication by~$k$.
Let $B:=H_2\tildeS^2$
and $\Psi:=\tildeF_*:B\to B$.
Let $\beta^+:\tildeS^2_+\to\tildeS^2$
and $\beta^-:\tildeS^2_-\to\tildeS^2$ be the inclusion maps.
Let $\alpha:=\beta^+_*\oplus\beta^-_*:A\to B$.

We leave it as an unassigned exercise to show
that $\alpha\circ\Phi=\Psi\circ\alpha$.
({\it Hint:} Note that $\Phi=(\tildeF^+\coprod\tildeF^-)_*$
and $\alpha=(\beta^+\coprod\beta^-)_*$, where
$\coprod$ denotes disjoint union of topological pairs.
Argue that there is a commutative diagram in the
category of topological pairs whose image under $H_2$
yields $\alpha\circ\Phi=\Psi\circ\alpha$.)

We also leave it as an unassigned exercise to show
that $\alpha$ is surjective.
({\it Hint:} In fact, $S^2$ is obtained from $E$
by attaching two cells, namely $S^2_+$ and $S^2_-$.
It therefore follows from our general theory about
the cellular chain comples that $H_2(S^2,E)\cong\Z^2$.
More specifically, we have that the inclusion maps
$(S^2_+,E)\subseteq(S^2,E)$ and $(S^2_-,E)\subseteq(S^2,E)$ induce
an isomorphism
$(H_2(S^2_+,E))\oplus(H_2(S^2_-,E))\to H_2(S^2,E)$.
This is exactly the statement that $\alpha$ is an
isomorphism, and is therefore, in particular, surjective.)

Since $\Phi$ is multiplication by $k$,
since $\alpha\circ\Phi=\Psi\circ\alpha$
and since $\alpha$ is surjective,
it follows that $\Psi:B\to B$ is multiplication by $k$.

Let $\gamma:H_2S^2\to H_2\tildeS^2$
be induced by the inclusion $(S^2,\emptyset)\subseteq(S^2,E)=\tildeS^2$.
By the long exact sequence of $(S^2,E)$,
we see that $\gamma$ is injective.
Recall that $\Lambda:=F_*:H_2S^2\to H_2S^2$.
By functoriality of long exact sequences,
we have $\Psi\circ\gamma=\gamma\circ\Lambda$.
So, since $\Psi$ is multiplication by $k$,
it follows that $\Lambda$ is multiplication by $k$, as desired.
{\bf QED}

\vskip.1in\noindent
{\it Corollary.}
Let $d\ge1$ be an integer.
Then any reflection of $S^d$ has degree $-1$.

\vskip.1in\noindent
{\it Proof:}
We leave it as an unassigned exercise to show that
any two reflections of $S^d$ are conjugate by
a rotation. (In fact, just take any rotation that
moves the fixed hyperplane of the first reflection to the
fixed hyperplane of the second.)
Consequently, in the arrow category of the category of
topological spaces, any two reflections of $S^d$
are isomorphic.
Thus if one of them has degree $-1$, they all do.

By Exercise 21A and the Hurewicz Theorem,
some reflection of $S^1$ has degree $-1$.

We now proceed by induction.
Now assume that there is some reflection $R$ of $S^{d-1}$
which has degree $-1$.
We wish to show that some reflection of $S^d$ has degree $-1$.

We leave it as an unassigned exercise to verify that
$\Sigma_1R:S^d\to S^d$ is a reflection.
By the preceding theorem, $\deg(\Sigma_1R)=\deg(R)=-1$.
{\bf QED}

\vskip.1in\noindent
{\it Corollary.}
Let $d\ge1$ be an integer.
Let $A:S^d\to S^d$ be the antipodal map defined by $A(p)=-p$.
Then $\deg(A)=(-1)^{d+1}$.

\vskip.1in\noindent
{\it Proof:}
For all integers $i\in[0,d]$,
let $R_i:S^d\to S^d$ be the reflection defined by
$$R_i(x_0,\ldots,x_d)=(x_0,\ldots,x_{i-1},-x_i,x_{i+1},\ldots,x_d);$$
by the preceding corollary, $\deg(R_i)=-1$.

Since $A=R_0\circ\cdots\circ R_d$,
$\deg(A)=[\deg(R_0)][\deg(R_1)]\cdots[\deg(R_d)]=(-1)^{d+1}$.
{\bf QED}

\vskip.1in
For integers $d\ge1$, for $x,y\in\R^d$,
we write $x\perp y$ to mean $x\cdot y=0$,
where $\cdot$ denote the Euclidean dot product.

\vskip.1in\noindent
{\it Lemma.}
Let $n\ge1$ be an integer and let $f:S^n\to S^n$ be continuous.
Assume, for all $x\in S^n$, that $f(x)\ne-x$.
Then $f$ is homotopic to the identity map.

\vskip.1in\noindent
{\it Proof:}
For all $x\in S^n$, let $x^\perp:=\{y\in\R^{n+1}\,|\,x\perp y\}$.
For all $x\in S^n$, let $\phi_x:x^\perp\to S^n\backslash\{-x\}$
be stereographic projection, given by
$\phi_x(y)=[x+y]/[\|x+y\|]$.
For all $t\in[0,1]$,
let $f_t:S^n\to S^n$ be defined by
$f_t(x)=\phi_x(t\cdot\phi_x^{-1}(f(x)))$.
Then $(t,x)\mapsto f_t(x):I\times S^n\to S^n$ is a homotopy
from the identity to $f$.
{\bf QED}

\vskip.1in
We now give a proof that you can't comb the hairs on a hedgehog:

\vskip.1in\noindent
{\it Theorem.}
Let $f:S^2\to\R^2$ be continuous and assume, for all $x\in S^2$,
that $f(x)\perp x$.
Then there exists $x_0\in S^2$ such that $f(x_0)=0$.

\vskip.1in\noindent
{\it Proof:}
Assume, to the contrary, that, for all $x\in S^2$,
we have $f(x)\ne0$.
Define $g:S^2\to\R^2$ by $g(x)=[f(x)]/[\|f(x)\|]$.
Let $I:S^2\to S^2$ be the identity map,
defined by $I(x)=x$.
Let $A:S^2\to S^2$ be the antipodal map,
defined by $A(x)=-x$.
Let $h:=A\circ g:S^2\to S^2$.
Then, for all $x\in S^2$, we have
$g(x)\notin\{x,-x\}$, which implies that
$g(x)\ne-x$ and $h(x)\ne-x$.
Then, by the preceding lemma, both $g$ and $h$ are
homotopic to $I$.
Then $A\circ h$ is homotopic to $A\circ I$.
That is, $g$ is homotopic to $A$.
On the other hand $g$ is homotopic to $I$.
Then $A$ is homotopic to $I$.
However, the degree of $A$ is $-1$,
whereas the degree of $I$ is $1$, a contradiction.
{\bf QED}

\vskip.1in
Recall that the lemma preceding the Ham Sandwich Theorem
asserts

\vskip.1in\noindent
{\it Lemma.}
Let $\psi:S^3\to S^3$ be odd.
Then $\psi$ is not homotopic to a constant map.

\vskip.1in
The proof given is valid all odd dimensional spheres, {\it i.e.},

\vskip.1in\noindent
{\it Lemma.}
Let $n>0$ be an odd integer.
Let $\psi:S^n\to S^n$ be odd.
Then $\psi$ is not homotopic to a constant map.

\vskip.1in
We now show that, by using suspension, we may extend the preceding
lemma to {\it all} spheres, {\it i.e.},

\vskip.1in\noindent
{\it Lemma.}
Let $n>0$ be an integer.
Let $\psi:S^n\to S^n$ be odd.
Then $\psi$ is not homotopic to a constant map.

\vskip.1in
NOTE: Since, for any $x\in S^n$, $S^n\backslash\{x\}$
is contractible, it is a corollary of the above
lemma that an odd map $S^n\to S^n$ must be surjective.

\vskip.1in\noindent
{\it Proof:}
If $n$ is odd, then we are done by the preceding lemma,
so assume that $n$ is even, and that $\psi$ is homotopic
to a constant map.
We aim for a contradiction.

We leave it as an unassigned exercise to show that,
because $\psi:S^n\to S^n$ is odd, it follows that
$\Sigma_1\psi:S^{n+1}\to S^{n+1}$ is odd.
We leave it as an unassigned exercise to show that,
because $\psi:S^n\to S^n$ is homotopic to thd a constant map,
it follows that $\Sigma_1\psi:S^{n+1}\to S^{n+1}$ is
homotopic to thd a constant map.

This then contradicts the preceding lemma
(with $n$ replaced by $n+1$ and with $\psi$ replaced by $\Sigma_1\psi$).
{\bf QED}


\vskip.1in\noindent
{\it Temperature Pressure Theorem.}
Let $f:S^2\to\R^2$ be continuous.
Then there exists $p\in S^2$ such that
$f(p)=f(-p)$.

\vskip.1in
NOTE: This result implies that, at any moment in time, there is a point
$p$ on the Earth (which we assume to be a perfect sphere, centered
at the origin in $\R^3$) such that:
\itemitem{(1)} the temperature at $p$ is the same as at $-p$; and
\itemitem{(2)} the pressure at $p$ is the same as at $-p$.

\vskip.1in\noindent
{\it Proof:}
Define $g:S^2\to\R^2$ by $g(p)=(f(p))-(f(-p))$.
Then $g$ is odd.
Assume for all $p\in S^2$,
that $g(p)\ne0$.
We aim for a contradiction.

Let $h:S^2\to S^1$ be defined by $h(p)=[g(p)]/[\|g(p)\|]$.
Then $h$ is odd.
Let $i:S^1\to S^2$ be the ``equator inclusion map'', given
by $i(x)=(x,0)$.
Then the image of $i\circ h:S^2\to S^2$ is contained in the
equator, so $i\circ h:S^2\to S^2$ is not surjective.

For any point $x\in S^2$, recall that $S^2\backslash\{x\}$ is
contractible.
Thus any map $S^2\to S^2$ which is not surjective
is homotopic to a constant map.
Consequently, $i\circ h:S^2\to S^2$ is homotopic to a constant map.

However, $h$ is odd and $i$ is even, so $i\circ h$ is odd,
so we have a contradiction to the preceding lemma.
{\bf QED}

\vskip.1in
Finally, we discuss the classification of surfaces.

\vskip.1in\noindent
{\it Definition.}
Let $d\ge1$ be an integer and let $M$ be a topological $d$-manifold.
For all $m\in M$, a map $f:\R^d\to M$ will be called
{\bf an $m$-centered chart} if
\itemitem{(1)} $f$ is open, continuous and injective; and
\itemitem{(2)} $f(0)=m$.

\vskip.1in\noindent
{\it Definition.}
Let $d\ge1$ be an integer.
Let $M$ and $N$ be connected topological $d$-manifolds.
Let $m\in M$ and $n\in N$.
Let $f:\R^d\to M$ be an $m$-centered chart and
let $g:\R^d\to M$ be an $n$-centered chart.
Define $h:(f(\R^d))\backslash\{m\}\to(g(\R^d))\backslash\{n\}$
by $h(f(x))=g(x/\|x\|^2)$.
Let $R$ be the smallest equivalence relation on $M\coprod N$
whose graph contains the graph of~$h$.
Define $M\#_{f,g}N:=(M\coprod N)/R$.

\vskip.1in\noindent
{\bf EXERCISE 21B:}
Show that $M\#_{f,g}N$ is a connected topological $d$-manifold.

\vskip.1in\noindent
{\it Fact.}
Let $d\ge1$ be an integer.
Let $M$ and $N$ be connected topological $d$-manifolds.
Let $m,m'\in M$ and $n,n'\in N$.
Let $f:\R^d\to M$ be an $m$-centered chart and
let $g:\R^d\to M$ be an $n$-centered chart.
Let $f':\R^d\to M$ be an $m'$-centered chart and
let $g':\R^d\to M$ be an $n'$-centered chart.
Then $M\#_{f,g}N$ is homeomorphic to $M\#_{f',g'}N$.

\vskip.1in
Because of the preceding Fact,
we often simply write $M\#N$, which is well-defined up to homeomorphism.
We call $M\#N$ {\bf the connected sum} of $M$ and $N$.
Note, {\it e.g.}, that $\T^2\#\T^2$ is the genus two orientable surface.

We leave it as an unassigned exercise to show that
$\#$ is associative and commutative.
We leave it as an unassigned exercise to show,
for any connected topological $d$-manifold~$M$, that
$S^d\#M$ is homeomorphic to $M$.

We will write $X\cong Y$ to mean that the topological spaces
$X$ and $Y$ are homeomorphic.
A {\bf surface} is a $2$-manifold.
The following is difficult, and we present it without proof:

\vskip.1in\noindent
{\it Theorem.} For any compact connected surface $X$,
either $X\cong S^2$, or there exist an integer $n\ge1$
and $X_1,\ldots,X_n\in\{\T^2,\R P^2\}$
such that $X\cong X_1\#\cdots\#X_n$.

\vskip.1in
To summarize the preceding theorem, we say that the $\#$-closure of
$\T^2$ and $\R P^2$ is the collection of all compact connected
surfaces.
One naturally wonders when two such expressions $X_1\#\cdots\#X_n$
are homeomorphic, and we next devise a scheme to answer this.

Let $S$ be a finite set and let $w$ be an element of the free
group $\langle S\rangle$ generated by $S$.
We obtain a topological space $T(w)$ as follows:
Let $P$ be a regular polygon in $\R^2$
with as many edges as there are factors in $w$.
(E.g., if $w=aba^{-1}b^{-1}$, then $P$ should have $4$ edges.)
Pick one vertex as a starting vertex
and label each edge, traveling clockwise, with
one factor in $w$.
Now let $X$ be the precompact connected component of $\R^2\backslash P$
and let $T(w)$ be the topological space obtained by
identifying the edges of $X\cup P$ according to their labelings.
(Note that $T(aba^{-1}b^{-1})$ is homemorphic to
the $2$-torus $\T^2:=S^1\times S^1$.)

We have $T(aa)\cong\R P^2\cong T(bb)$,
so $\R P^2\#\R P^2\cong T(aabb)$.
If we cut the $aabb$ polygon along a line going from

\centerline{the vertex at the middle of the path of $aa$}

\noindent
to

\centerline{the vertex at the middle of the path of $bb$,}

\noindent
then label the new edges along the cut both as $c$,
and then glue the resulting two triangles
($acb$ and $abc^{-1}$) along $b$,
we find that $T(aabb)\cong T(aca^{-1}c)$.

Algebraically, this amounts to
\itemitem{(1)} picking a two letter subword $w'$
(specifically $ab$) in the word $aabb$; then
\itemitem{(2)} replacing that subword $w'$ with
a new letter, say $c$, yielding a new word $w_0$
(in our case $w_0=acb$); then
\itemitem{(3)} picking one of the letters, say $b$,
in that subword $w'$; then
\itemitem{(4)} solving $c=w'$ for $b$, getting
an equation of the form $b=w''$ (in our case, we solve
$c=ab$ for $b$ and get $b=a^{-1}c$); and then
\itemitem{(5)} replacing every $b$ in $w_0$ with $w''$
(in our case we get, in the end, $aca^{-1}c$).

Since $K:=T(aca^{-1}c)$ is the Klein bottle, we have shown
that the connected sum of two projective planes is the Klein bottle $K$.
Remark: The projective plane $\R P^2$ is sometimes called a {\bf cross cap},
so one says that the Klein bottle is a connected sum of two cross caps.

\vskip.1in\noindent
{\it Lemma.}
We have $\R P^2\#K\cong\R P^2\#\T^2$.

\vskip.1in\noindent
{\it Proof:}
Let $X:=\R P^2\#K$.
Since $\R P^2\cong T(aa)$ and $K\cong T(bcbc^{-1})$,
we get $X\cong T(aabcbc^{-1})$.
Replacing $ab$ with $d$ and $a$ with $db^{-1}$,
we obtain $X\cong T(db^{-1}dcbc^{-1})$.
Replacing $dc$ by $e$ and $c$ by $d^{-1}e$,
we obtain $X\cong T(db^{-1}ebe^{-1}d)$.

A cyclic permutation of $db^{-1}ebe^{-1}d$ is $ddb^{-1}ebe^{-1}$,
so we get $X\cong T(ddb^{-1}ebe^{-1})$.
Replacing $b$ by $f^{-1}$, we get
$X\cong T(ddfef^{-1}e^{-1})$.
Then, as
$$\R P^2\cong T(dd)\qquad\hbox{and}\qquad\T^2\cong T(fef^{-1}e^{-1}),$$
we conclude that $X\cong\R P^2\#\T^2$, as desired.
{\bf QED}

\vskip.1in
Let $n$ be a positive integer
and let $X_1,\ldots,X_n\in\{\T^2,\R P^2\}$.
Let $X:=X_1\#\cdots\#X_n$.
Since $K\cong\R P^2\#\R P^2$, the preceding lemma
yields $T\#\R P^2\cong\R P^2\#\R P^2\#\R P^2$.
Thus, if $\R P^2\in\{X_1,\ldots,X_n\}$,
then $X$ is homeomorpic to a connected sum
of projective planes (because any $\T^2$ appearing in the
connected sum $X_1\#\cdots\#X_n$ can be replaced by $\R P^2\#\R P^2$).
Either $\R P^2\notin\{X_1,\ldots,X_n\}$,
in which case $X$ is a connected sum of $2$-tori,
or else $\R P^2\in\{X_1,\ldots,X_n\}$,
in which case $X$ is a connected sum of projective planes.

This observation, combined with the preceding theorem,
shows that every compact connected surface is homeomorphic to one of
\itemitem{(1)} $S^2$; or
\itemitem{(2)} a connected sum of $2$-tori; or
\itemitem{(3)} a connected sum of projective planes.

\noindent
We leave it as an unassigned exercise for the reader to use algebraic
topological techniques developed in this course
to show that, of the surfaces described in (1), (2) and (3) above,
no two are homeomorphic.
This then gives a complete classification of compact connected surfaces.

\vskip.1in\noindent
{\it Question:}
Can one do something similar for compact connected $3$-manifolds?

\vskip.1in
This is an extremely difficult question and has been worked on by
many, many people.
An attempt at answering this question is the
{\bf Thurston Geometrization Conjecture}, which is too complicated
even to state here, but which would similarly break any compact
connected $3$-manifold up into pieces each of which belongs to a
well-studied list.

\vskip.1in
{\bf Last updated: Tuesday 8 March at 12noon}
\remember that this set of notes is complete -- we now move to notes3

\end
