Notes for Math 8301, Manifolds and Topology, Fall 2004 The Fundamental Group and Covering Spaces

Definition. B_{δ}(x) = (x − δ, x + δ) ⊆ ℝ. Definition. A function f : ℝ → ℝ is continuous at x if, for all ϵ > 0, there exists δ > 0 such that f(B_{δ}(x)) ⊆ B_{ϵ}(f(x)), i.e., B_{δ}(x) ⊆ f^{−1}(B_{ϵ}(f(x))), Note: For ϕ : A → B and B_{0} ⊆ B, we define ϕ^{−1}(B_{0}) = {a ∈ A | ϕ(a) ∈ B_{0}}. Definition. A function f : ℝ → ℝ is continuous if, for all x ∈ ℝ, we have: f is continuous at x. Question: How do we define continuity for f : ℝ^{m} → ℝ^{n}? How about f : S^{n} → ℝ^{m}, where S^{n} := {x ∈ ℝ^{n+1} | d(x, 0) = 1}? How about f : S^{n} → S^{m}? Let’s try to formulate a general definition of continuity that will include all these definitions as special cases. Definition. An N-space consists of

(1) a set X and

(2) a function x ↦ N_{x} : X → {subsets of {subsets of X}}

such that (A) for all x ∈ X, we have: N_{x} ≠ ∅; and

(B) for all x ∈ X, for all N ∈ N_{x}, we have: x ∈ N.

Example. X = ℝ, N_{x} = {B_{δ}(x) | δ > 0}. Definition. For all x ∈ X, an “N-neighborhood” of x is an element of N_{x}. Definition. Let X and Y be N-spaces and let f : X → Y be a function. For a point x ∈ X, we say that f : X → Y is N-continuous at x if, for any N-neighborhood V of f(x), there exists an N-neighborhood U of x such that f(U) ⊆ V, i.e., U ⊆ f^{−1}(V). We say that f is N-continuous if f is N-continuous at every point x ∈ X. Definition. A category consists of

(1) a class C of objects;

(2) for all C, C′ ∈ C, a set Hom(C, C′) of morphisms or arrows from C to C′;

(3) for all C, C′, C′′ ∈ C, a composition function Hom(C, C′) × Hom(C′, C′′) → Hom(C, C′′); and

(4) for all C ∈ C, an identity arrow id_{C} ∈ Hom(C, C) such that

(A) for all C, C′ ∈ C, for all f ∈ Hom(C, C′), we have f ∘ id_{C} = id_{C′} ∘ f = f; (B) for all C, C′, C′′, C′′′ ∈ C, for all f ∈ Hom(C, C′), for all g ∈ Hom(C′, C′′), for all

h ∈ Hom(C′′, C′′′), we have (f ∘ g) ∘ h = f ∘ (g ∘ h);

and probably other properties. EXERCISE 1A: Look up, and write out the defintion of a category.

Example. N S := {N-spaces}. That is, N S is the category of N-spaces, together with N-continuous maps, together with the usual composition of maps, together with the usual identity maps.

EXERCISE 1B:

(1) For all X ∈ N S, show that x ↦ x : X → X is N-continuous.

(2) For all X, Y, Z ∈ N S, for all N-continuous f : X → Y, for all N-continuous g : Y → Z, show that g ∘ f : X → Z is N-continuous.

Let {Sets} denote the category of sets (and functions). For any X ∈ N S, let X_{set} be the underlying set of X. For X, Y ∈ N S, for f ∈ Hom(X, Y), let f_{set} : X_{set} → Y_{set} be the underlying set function of f. Then X ↦ X_{set} : N S → {Sets} is an example of a functor. (Really, we should say X ↦ X_{set} together with f ↦ f_{set} forms a functor.) Note that, for X, Y, Z ∈ N S, for f ∈ Hom(X, Y), for g ∈ Hom(Y, Z), we have (g ∘ f)_{set} = g_{set} ∘ f_{set}. This is one of the basic properties of a functor.

EXERCISE 1C: Look up, and write down the definition of a functor from one category to another. Be sure to note that some functors are covariant, while others are contravariant.

Example. Let {Rings} denote the category of (not necessarily commutative) rings with unit, together with unit-preserving ring homomorphisms. Let {Groups} denote the category of groups, with group homomorphisms. Then (R, ·, +) ↦ (R, +) is a functor from {Rings} to {Groups} (and even takes values in the subcategory of Abelian groups). Then (R, ·, +) ↦ ({units in R}, ·) is another functor from {Rings} to {Groups}.

Definition. Let C be any category. Then there is an identity functor Id_{C} : C → C defined by Id_{C}(C) = C. (Note: This defines the functor on objects, and it is left to you you to guess what the functor does on arrows.) (Note: Well, okay, I’ll tell you this time – it leaves each arrow fixed, i.e., for all f : C → C′ in C, we define Id_{C}(f) = f. In general, though, it is an expectation that you should be able to figure this kind of thing out, but always feel free to ask if it’s unclear.)

Fix, until END OF DISCUSSION #1, some X ∈ N S. Let’s say we’ve identified some points of X that are “happy”.

Let 0 ∈ X. If we say “points close to x are happy” or “points sufficiently close to x are happy” or “all points sufficiently close to x are happy”, we mean that there is an N-neighborhood U of x such that, for all u ∈ U, we have: u is happy. Now suppose that y is sufficiently close to x that it is guaranteed to be happy. That is, suppose that y ∈ U. Question: Does it automatically follow that all points sufficiently close to y are guaranteed to be happy by being in U? In other words, can we conclude that there is a neighborhoos V of y such that V ⊆ U.

Answer: No. For example, let X_{set} = ℝ, and for all x ∈ X, let N_{x} := {[x−δ, x+δ] | δ > 0}. Then X = (X_{set}, x ↦ N_{x}) is an N-space.

Suppose that [−1, 1] is exactly the set of happy points. Let x := 0, y := −1. Then U = [−1, 1] is an N-neighborhood of x and y ∈ U, but there is no N-neighborhood V of y such that V ⊆ U.

So even though y is happy, we have no “room to maneuver” because the kinds of N-neigborhoods in X are, somehow, very unforgiving, in the sense that, if you have a point that is in an N-neighborhood, and you then perturb it ever so slightly, you may find yourself outside the N-neighborhood.

Definition. We say that an N-space X is forgiving if, for all x ∈ X, for any Nneighborhood U of x, for any y ∈ U, there is an N-neighborhood V of y such that V ⊆ U.

END OF DISCUSSION #1

Fix, until END OF DISCUSSION #2, some X ∈ N S. Let’s say we’ve identified some points of X that are “happy” and some point that are “dull”. So X breaks up into happy dull points, unhappy dull points, happy sharp points and unhappy sharp points.

Let x ∈ X and assume that all points sufficiently close to x are happy. That is, there is an N-neighborhood U of x such that every element of U is happy. Assume, furthermore that all points sufficiently close to x are dull. That is, there is an N-neighborhood V of x such that every element of V is happy. Does it automatically follow that there is a neighborhood W of x such that W ⊆ U ∪ V, which guarantees that all points sufficiently close to x are both happy and dull?

The answer is no. It’s not an assigned problem, but you should think about constructing a counterexample.

Definition. An N-space X is intersection-stable if, for all x ∈ X, for any N-neighborhoods U and V of x, there is an N-neighborhood W of x such that W ⊆ U ∩ V.

END OF DISCUSSION #2

Definition. An N-space is said to be topological if it is both forgiving and intersectionstable.

Definition. Let T N S denote the category of topological N-spaces (with N-continuous maps).

Definition. Let X be a set. A topology on X is a subset O of {subsets of X} such that (A) ∅, X ∈ O; (B) for all S ⊆ O, we have ∪S ∈ O; and (C) for all finite F ⊆ O, we have ∩F ∈ O.

Definition. A topological space consists of (1) a set X; and

(2) a topology on X.

Definition. An open subset of X is an element of O. A subset of X is said to be open or open in X if it is an open subset of X.

Definition. Let X and Y be topological spaces and let f : X → Y be a function defined on their underlying sets. We say that f is continuous if, for any open V ⊆ Y, we have that f^{−1}(V) is open in X.

What motivates the definition of topological space and of this definition of continuity?

Definition. Let T S denote the category of topological spaces, with continuous maps.

Definition. Define a functor F : T N S → T S by

F((X, x ↦ N_{x})) = (X, {∪S | S ⊆ ∪_{x∈X}N_{x}}).

Here, I leave it to you to guess what F does on arrows. Note that {open sets in FX} is exactly the closure of {N-neighborhoods of points of X} under union.

Define a functor G : T S → T N S by

G((Y, O)) = (Y, y ↦ {U ∈ O | y ∈ U}).

EXERCISE 1D: Show, for all X ∈ T N S, that FX ∈ T S. (You must show that the open subsets of FX are closed under finite intersection and arbitrary union, and that both ∅ and X are open in FX.)

EXERCISE 1E: Show, for all Y ∈ T S, that GY ∈ T N S.

EXERCISE 1F: Show, for all Y ∈ T S, that FGY = Y. (You must show that a subset of Y is open in FGY iff it is open in Y.)

The preceding exercise asks you to prove that FG = Id_{T S}.

We now show that GFX is not necessarily equal to X. That is, we show that FG ≠ Id_{T N S}:

Example. Let X := (ℝ, x ↦ {(x − δ, x + δ) | δ > 0}). Then the set of open subsets of FX is exactly the closure of {(x − δ, x + δ) | x ∈ ℝ, δ > 0} under union. In particular, (−1, 1) ∪ (2, 4) is open in FX. Then (−1, 1) ∪ (2, 4) is an N-neighborhood of 0 in GFX. However it is not an N-neighborhood of 0 in X.

Therefore GFX ≠ X.

Definition. An arrow (or morphism) f : C → C′ in a category C is an isomorphism if there exists an arrow g : C′ → C such that f ∘ g = id_{C′} and such that g ∘ f = id_{C}.

That is, an isomorphism is simply an invertible arrow.

Definition. In the category T S, an isomorphism will be called a homeomorphism. In the category T N S, an isomorphism will be called an N-homeomorphism.

Definition. Let A and B be functors from a category C to a category D. A natural transformation τ : A → B associates to each object C ∈ C an arrow τ_{C} : AC → BC, provided that this association has the property that: for any arrow f : C → C′ in C, we have (Bf) ∘ τ_{C} = τ_{C′} ∘ (Af).

Note that, for their to be a natural transformation from A to B, the domains of A and B must agree. Moreover, their targets must agree as well.

Let A and B be functors from a category C to a category D. It may be helpful to picture AC as a subcategory inside of D. In some sense, A is a “parametric” subcategory of D, where the objects are parametrized by objects in C. Similarly, BC is a parametric subcategory of D. With this intution, a transformation τ : A → B may be thought of as a parametric family of arrows in D, one for each object in C. For each C ∈ C, the corresponding arrow runs from AC to BC.

Moreover, for each arrow in C, we now get a diagram in D (with four objects and four arrows). If these diagrams all commute, then we say that the transformation is “natural”.

Example. Let V be the category of finite dimensional vector spaces, and linear transformations. Let I : V → V be the (covariant) identity functor. Let D : V → V be the (contravariant) functor defined by D(V) = V^{∗} := Hom(V, ℝ). Let DD : V → V be the (covariant) functor DD := D∘D obtained by composing D with itself. Then, for all V ∈ V, we have DD(V) = V^{∗∗}. For all V ∈ V, let τ_{V} : V → V^{∗∗} be defined by (τ_{V} v)(l) = l(v). (Note: The map τ_{V} v : V^{∗} → ℝ is sometimes called the “evaluational at v”.) Then τ : I → DD is a natural tranformation, as Exercise 2B will verify. By contrast, the next exercise shows that there is no natural transformation from I to D, even though, for all V ∈ V, the vector spaces V and V^{∗} are isomorphic, since they have the same dimension.

EXERCISE 2A: Let V := ℝ^{2} and let f : V → V^{∗} be an isomorphism. For any linear map g : V → V, let g^{∗} : V^{∗} → V^{∗} be defined by g^{∗}(l) = l ∘ g. Show that there exists an isomorphism g : V → V such that f ≠ g^{∗} ∘ f ∘ g.

EXERCISE 2B: As in the example above, for any real vector space V, let τ_{V} : V → V^{∗∗} be defined by (τ_{V} v)(l) = l(v). Let W and X be real vector spaces and let g : W → X be a linear transformation. Define g^{∗} : X^{∗} → W^{∗} by g^{∗}(l) = l ∘ g and define g^{∗∗} : W^{∗∗} → X^{∗∗} by g^{∗∗}(l) = l ∘ g^{∗}. Show that g^{∗∗} ∘ τ_{W} = τ_{X} ∘ g.

Definition. Let C be a category. Then the arrow category of C, Arr(C), is a category whose objects are the arrows in C, and for which, for all pairs of arrows f : C_{1} → C_{2}, f′ : C′_{1} → C′_{2} in C, we define Hom(f, f′) to be the collection of all those

(g_{1}, g_{2}) ∈ Hom(C_{1}, C′_{1}) × Hom(C_{2}, ×C′_{2})

such that g_{2} ∘ f = f′ ∘ g_{1}.

We leave it as an exercise to guess what the definitions of compositions and identity arrows in Arr(C).

Definition. Given a category C, we define two functors Dom_{C}, Tar_{C} : Arr(C) → C by: for all f : D → T in C, we set Dom_{C}(f) = D and Tar_{C}(f) = T.

With this terminology in place, if A, B : C → D are two functors, then a natural transformation μ : A → B is equivalent to a functor M : C → Arr(D) satisfying both Dom_{D} ∘ M = A and Tar_{D} ∘ M = B.

Now recall the functors F : T N S → T S and G : T S → T N S defined above. Recall that FG is the identity functor on T S, but that GF is not the identity functor on T N S. We will make clear next a sense in which FG is “close” to the identity on T N S.

Definition. Let C and D be categories and let A : C → D be a functor. Define a natural transformation ι^{A} : A → A by ι_{C}^{A} := id_{AC} : AC → AC. This natural transformation is called the identity on A.

Definition. Let C and D be categories, let P, Q, R : C → D be functors and let α : P → Q and β : Q → R be natural transformations. We define a natural transformation βα : P → R by (βα)_{C} = β_{C} ∘ α_{C}.

Definition. We say that two functors A, B : C → D are equivalent (and write A ∼ B) if there exist natural transformations τ : A → B and μ : B → A such that μτ = ι^{A} and τμ = ι^{B}.

Intuitively, if two functors are equivalent, then they have the same “information” in them. If I’m an expert on one of the functors and you’re an expert on the other, then we’ll probably spend our days quibbling over notation, but will be well aware that we both know the same stuff.

EXERCISE 2C: Let A, B : C → D be functors. Let τ : A → B be a natural transformation. Suppose, for all C ∈ C, that τ_{C} : AC → BC is an isomorphism. Show that there is a unique natural transformation μ : B → A such that μτ = ι^{A} and τμ = ι^{B}.

EXERCISE 2D: Let A, B : C → D be functors. Let τ : A → B and μ : B → A be natural transformations such that μτ = ι^{A} and τμ = ι^{B}. Show, for all C ∈ C, that τ^{C} : AC → BC is an isomorphism.

EXERCISE 2E: Recall the functors F : T N S → T S and G : T S → T N S defined above. Recall that FG = Id_{T S} but that GF ≠ Id_{T N S}. Prove that GF ∼ Id_{T N S}.

Definition. A category C is said to be equivalent to a category D (write C ∼ D) if there are functors P : C → D and Q : D → C such that PQ ∼ Id_{D} and QP ∼ Id_{C}.

Intuitively, if two categories are equivalent, then they have the same “information” in them. If I’m an expert on one of the categories and you’re an expert on the other, then we’ll probably spend our days quibbling over notation, but will be well aware that we both know the same stuff.

The preceding exercise asks you to demonstrate that T N S is equivalent to T S. Which is better, T N S or T S? In some sense, neither, since they’re equivalent.

On the other hand, with a little thought, you will readily see that F : T N S → T S is not “one-to-one”. To almost any topological space Y, there are several topological Nspaces whose images under F are all equal to Y. For example, (ℝ, x ↦ {(x − δ, x + δ)}) and (ℝ, x ↦ {(x − 2δ, x + δ)}) have the same image under F.

So, in some sense, T N S is redundant, with more copies of each space than one really needs. From this perspective, T S is better than T N S.

Finally, a few broad words about mathematics. Typically a working mathematician develops expertise in a category, e.g., the category T S of topological spaces. A fundamental question in topology is to determine, given two topological spaces, whether they are homeomorphic, but, more generally, whatever your “category of choice” may be, you’ll want to be able to form many examples of objects in it, and, in many cases to show that two such examples are not isomorphic in the category.

If one can form a functor F that goes from your category of choice C to some other category D, and if tools are available for distinguishing objects in D, then, given C, C′ ∈ C, if you are able to show that FC and FC′ are not isomorphic in D, then you can immediately conclude that C and C′ are not isomorphic in C. (See the next exercise.)

For example, there is a functor called “first homology”, and denoted H_{1} which runs from topological spaces to Abelian groups. Given two topological spaces, if one is able to show that their first homology groups are not isomorphic, then the spaces are not homeomorphic.

This kind of argument will be central to much of this course!

EXERCISE 2F: Show that functors carry isomorphisms to isomorphisms. That is, show that, if F : C → D is a functor, if f : C → C′ is an isomorphism in C, then Ff : FC → FC′ is an isomorphism in D.

In the next definition, we make the convention that ∪∅ = ∅.

Definition. Let X be a set and let S ⊆ {subsets of X}. The union closure of S is {∪A | A ⊆ S}. The finite intersection closure in X of S is {X} ∪ {∩A | A ⊆ S, 0 < |A| < ∞}.

Definition. Let X be a set and let O be a topology on X. Let B ⊆ {subsets of X}. We say that B is a basis for O if the union closure of B is O. We say that B is a subbasis for O if the finite intersection closure in X of B is a basis for O.

Note that any basis for a topology is a subbasis.

EXERCISE 2G: Let X be a set and let S ⊆ {subsets of X}. Show that S is a subbasis of a topology on X. (That is, show that the union closure of the finite intersection closure in X of S is a topology on X.)

EXERCISE 2H: Let X be a set and let B ⊆ {subsets of X}. Show that B is a basis of a topology on X if and only if, both of the following conditions hold:

(1) ∪B = X; and

(2) for all B, B′ ∈ B, for all x ∈ B∩B′, there exists B′′ ∈ B such that x ∈ B′′ ⊆ B∩B′.

Definition. Let O and O′ be topologies on a set X. We say that O is finer (or stronger) than O′ if O ⊇ O′. We say that O is coarser (or weaker) than O′ if O ⊆ O′.

Remark. Let X and I be sets and, for all i ∈ I, let O_{i} be a topology on X. Then ⋂_{i∈I} O_{i} is

a topology on X.

The following is a restatement of the preceding remark.

Remark. Let Ξ ⊆ {topologies on X}. Then ∩Ξ is a topology on X.

The preceding two remarks are asserting that, given a collection of topologies on a set, there exists a unique topology which is finest among:

those topologies coarser than every topology in the collection.

Question: How about the other way around? That is, given a collection of topologies on a set, does there there exist a unique topology which is coarsest among:

those topologies finer than every topology in the collection.

Answer: We decided in class that the answer is yes. This can probably be argued using Zorn’s lemma, but more directly, one can simply take the union of the topologies in the collection and then take the intersection of all the topologies containing that union. By the preceding two remarks, that intersection is again a topology. Yet another way to say the same thing is that you take the union of the topologies in the collection, then take the finite intersection closure of that union, then take the union closure of that.

Definition. Let X be a set and let

S ⊆ {subsets of X}.

Then the topology generated by S is the coarsest topology among those topologies containing S.

Note that, if B is a basis or subbasis of a topology O on a set X, then O is the topology generated by B.

Definition. Let X and Y be sets, let f : X → Y be a function and let

S ⊆ {subsets of Y }.

Then we define f^{∗}S := {f^{−1}(S) | S ∈ S}.

EXERCISE 2I: Let X and Y be sets, let f : X → Y be a function and let O be a topology on Y. Show that f^{∗}O is a topology on X.

Remark. Let X and Y be sets, let f : X → Y be a function and let O be a topology on Y. Then f^{∗}O is the coarsest topology making f continuous.

Definition. Let Y be a topological space and let X ⊆ Y. Then the inherited topology on X (or relative topology on X or subspace topology on X) is:

{X ∩ U | U is open in Y }.

Remark. Let Y be a topological space and let X ⊆ Y. Let i : X → Y be the inclusion, defined by i(x) = x. Let O := {open subsets of Y }. Then i^{∗}O is the inherited topology on X. It is the coarsest topology making i continuous.

Definition. Let I be a set and, for all i ∈ I, let X_{i} be a topological space. Let X := ∏_{i∈I} X_{i}.

For all i ∈ I, let p_{i} : X → X_{i} be the ith coordinate projection map. For all i ∈ I, let O_{i} be the set of all open subsets of X_{i}. Then the product topology on X is the topology generated by ⋃_{i∈I} p_{i}^{∗}O_{i}.

Note that the product topology is the coarsest among those that make all the projection maps p_{i} continuous.

In general, whenever a product of topological spaces is formed, we will give it the product topology, unless otherwise specified. In particular, when X and Y are topological spaces, X × Y is a topological space as well.

Definition. Let X be a topological space, let Y be a set and let f : X → Y be a function. The quotient topology on Y is the finest making f continuous. (See Exercise 3A below.)

Now let’s start making some topological spaces!

Definition. Let n ≥ 0 be an integer. The standard topology on ℝ^{n} is the topology generated by {B_{r}(x) | x ∈ ℝ^{n}, r > 0}.

Definition. Let n ≥ 0 be an integer. We define S^{n} := {x ∈ ℝ^{n+1} | d(x, 0) = 1}. The standard topology on S^{n} is the topology inherited from ℝ^{n+1}.

Definition. If X is a topological space, and if x ∈ X, then an open neighborhood of x in X is an open subset U of X such that x ∈ U. If X is a topological space, and if S ⊆ X, then an open neighborhood of S in X is an open subset U of X such that S ⊆ U.

Definition. To say that a topological space X is Hausdorff means: for all x, x′ ∈ X, if x ≠ x′, then there are open neighborhoods U of x and U′ of x′ in X such that U ∩ U′ = ∅.

Definition. Let S be a subset of a topological space X. We say that S is closed in X if X\S is open in X.

Note that {closed subsets of X} is closed under finite union and arbitrary intersection.

Definition. Let X be a topological space and assume, for all x ∈ X, that {x} is closed in X. We say that X is regular if the following condition holds: For all x_{0} ∈ X, for any closed subset C_{0} of X, if x_{0} ≠ C_{0}, then there exist open nieghborhoods U and V of x_{0} and C_{0} in X such that U ∩ V = ∅. We say that X is normal if the following condition holds: For any closed subsets C, C′ of X, if C ∩ C′ = ∅, then there exist open nieghborhoods U and U′ of C and C′ in X such that U ∩ U′ = ∅.

Definition. Let X be a topological space and let S ⊆ X. The closure of S in X is the intersection of all closed subsets of X that contain S, and is denoted \overline{S} or Cl_{X}(S). The interior of S in X is the union of all open subsets of X contained in S, and is denoted S^{∘} or Int_{X}(S). The boundary of S in X is \overline{S}\S^{∘}, and is denoted ∂S or Bd_{X}(S).

Since an intersection of closed sets is closed and a union of open sets is open, it follows that the closure of S is the smallest closed set containing S, while the interior of S is the largest open set contained in S.

Let S be a subset of a topological space X. We say that S is dense in X if Cl_{X}(S) = X.

Definition. We say that a topological space is discrete if all of its subsets are open. We say that a topological space X is indiscrete if its only open sets are ∅ and X.

Definition. Let X be a Hausdorff topological space. We say X is compact if, for any S ⊆ {open subsets of X}, we have:

(∗) if ∪S = X, then there is a finite subset F of S such that ∪F = X.

That is, a Hausdorff topological space is said to be compact if every open cover has a finite subcover, in the following terminology: Definition. Let X be a topological space. An open cover of X is a subset U of {open subsets of X} such that ∪U = X. Given two covers U, V of X, we say that U is a subcover of V if U ⊆ V. Question: Is S^{2} homeomorphic to ℝ^{2}? Fact. S^{2} is compact, whereas ℝ^{2} is not. Answer to preceding question: No. Definition. Let X be a Hausdorff topological space. Then X is said to be locally compact if, for all x ∈ X, there exists an open neighborhood U of x in X such that \overline{U} is compact. Example. ℝ is locally compact, and any finite product of locally compact topological spaces is again locally compact. Consequently, for all integers n ≥ 1, ℝ^{n} is locally compact. However infinite products of locally compact topological spaces are generally not locally compact, e.g., ∏_{i=1}^{∞} ℝ is not locally compact.

Next Question: Is ℝ homeomorphic to ℝ^{2}? EXERCISE 2J: For all p, q ∈ ℝ^{2}, show that ℝ^{2}\{p} is homeomorphic to ℝ^{2}\{q}. EXERCISE 2K: Let X and Y be topological spaces and let x ∈ X. Assume that X is homeomorphic to Y. Show that there exists y ∈ Y such that X\{x} is homeomorphic to Y \{y}. Definition. Let S be a subset of a topological space X. We say that S is clopen in X if S is both closed an open in X. Definition. We say that a topological space X is connected if it has no clopen sets other than ∅ and X. A topological space is disconnected if it is not connected. Fact. ℝ\{0} is disconnected, while ℝ^{2}\{(0, 0)} is connected. Remark. ℝ is not homeomorphic to ℝ^{2}. Proof: Assume, for a contradiction that ℝ and ℝ^{2} are homeomorphic. By Exercise 2K, choose y ∈ ℝ^{2}\{(0, 0)} such that ℝ\{0} is homeomorphic to ℝ^{2}\{y}. By Exercise 2J, ℝ^{2}\{y} is homeomorphic to ℝ^{2}\{(0, 0)}. Then ℝ\{0} is homeomorphic to ℝ^{2}\{(0, 0)}. However, by the fact above, ℝ\{(0} is disconnected, while ℝ^{2}\{(0, 0)} is connected. QED

In the above proof, we used, without comment, the fact that if two topological spaces are homeomorphic and if one is connected, than the other is as well. That is, “connectedness is a homeomorphism invariant”. In fact, for a property of topological spaces to be useful, it is typically a homeomorphism invariant, and we will leave it as an implicit exercise to verify that the properties we may introduce below are, in fact, homeomorphism invariants. Typically, such arguments are straightforward.

Next Question: Is ℝ^{2} homeomorphic to ℝ^{3}?

Note that both ℝ^{2}\{(0, 0)} and ℝ^{3}\{(0, 0, 0)} are connected. Nevertheless, some variant of the argument given above will work to show that the answer to the above question is “no”.

Definition. A pointed topological space consists of (1) a topological space X; and

(2) a point x ∈ X.

We call x the basepoint of the pointed topological space (X, x).

Definition. The category of pointed topological spaces (with basepoint preserving continuous maps) is denoted PT S. The forgetful functor (X, x) ↦ XPT S → T S is denoted by FB. An isomorphism in PT S is called a basepoint-preserving homeomorphism.

The next Fact is nontrivial, but we will assume it for now, and prove it in the next few lectures.

Fact. There is a functor π_{1} : PT S → {groups} such that:

(1) for all x ∈ X := ℝ^{2}\{(0, 0)}, we have that π_{1}(X, x) is isomorphic to the additive group of integers; and

(2) for all y ∈ Y := ℝ^{3}\{(0, 0, 0)}, we have that π_{1}(Y, y) is isomorphic to the trivial group.

Remark. For all topological spaces X, Y, for all x ∈ X, if X is homeomorphic to Y then there exists y ∈ Y such that (X, x) is basepoint-preserving homeomorphic to (Y, y).

Remark. Let X := ℝ^{2}\{(0, 0)} and Y := ℝ^{3}\{(0, 0, 0)}. Then X and Y are not homeomorphic.

Proof: Say for a contradiction that X is homeomorphic to Y. Choose x ∈ X. By the second to last Remark, choose y ∈ Y such that (X, x) is basepoint-preserving homeomorphic to (Y, y). Let π_{1} : PT S → {groups} be the functor described in the preceding Fact.

Then, by Exercise 2F, π_{1}(X, x) is isomorpic to π_{1}(Y, y), in the category of groups. We say that π_{1}(X, x) is the fundamental group of X (with respect to x). Then, by the preceding fact, the additive group of integers is isomorphic to the trivial group, contradiction. QED

Corollary. ℝ^{2} is not homeomorphic to ℝ^{3}.

Given the preceding Remark, the preceding Corollary is argued exactly as in the proof that ℝ and ℝ^{2} are not homeomorphic.

Our goal now becomes to define and study the functor π_{1} : PT S → {groups} of the preceding Fact.

Definition. A space X is locally compact if every point of X has an open neighborhood whose closure in X is compact.

Definition. Let X and Y be topological spaces and assume that X locally compact. Recall that Hom(X, Y) is the set of continuous maps from X to Y. Let K denote the set of all compact subsets of X. Let O denote the set of all open subsets of Y. For all K ∈ K, for all U ∈ O, let let W(K, U) := {f ∈ Hom(X, Y) | f(K) ⊆ U}. The topology on Hom(X, Y) generated by the subbasis {W(K, U) | K ∈ K, U ∈ O} is called the compact-open topology on Hom(X, Y). We let C(X, Y) denote the topological space whose underlying set is Hom(X, Y) and whose topology is the compact-open topology.

One has the general feeling that a space is locally compact if it is “almost” finite dimensional. In particular, if I is a set and, for all i ∈ I, we have a Hausdorff topological space X_{i}, then ∏_{i∈I} X_{i} is locally compact iff {i ∈ I | X_{i} is noncompact} is finite. The

topological space C(ℝ, ℝ) does not feel anywhere near finite dimensional, so one would guess that it is not locally compact, and that is, in fact true:

Remark. C(ℝ, ℝ) is not locally compact.

Sketch of proof: Define f : ℝ → ℝ by f(x) = 0. Then f ∈ C(ℝ, ℝ). Let U be an open neighborhood of f in C(ℝ, ℝ). Assume, for a contradiction, that \overline{U} is compact.

Choose ϵ > 0 and a compact subset K ⊆ ℝ such that

U_{0} := {g ∈ C(ℝ, ℝ) | |g| < ϵ on K} ⊆ U.

(It is an unassigned exercise to prove existence of such an ϵ and K.) For all integers j ≥ 1, let g_{j} : ℝ → ℝ be defined by g_{j}(x) = [ϵ/2][sin(jx)]; then g_{j} ∈ U_{0}, so g_{j} ∈ U. We leave it as an unassigned exercise to show that g_{j} has no convergent subsequence in C(ℝ, ℝ). After all, to what could these functions possibly converge? (Recall that, in any compact Hausdorff topological space, any sequence has a convergent subsequence.) QED

Fact. Let X and Y be locally compact topological spaces. For all f ∈ C(X, C(Y, Z)), define α_{f} ∈ C(X × Y, Z) by α_{f}(x, y) = (f(x))(y). Then f ↦ α_{f} : C(X, C(Y, Z)) → C(X × Y, Z) is a homeomorphism.

If X is a locally compact topological space and if (Y, d) is a metric space, then the d-uniform on compacta topology on Hom(X, Y) is the topology generated by the basis {f ∈ Hom(X, Y) | ∀k ∈ K, d(f(k), f_{0}(k)) < ϵ}, where f_{0} ranges over Hom(X, Y), where K ranges over compact subsets of X and where ϵ ranges over the positive real numbers. It is a fact that, for the topology on Y generated by the basis

{open balls with respect to the metric d},

the compact-open topology on Hom(X, Y) agrees with the d-uniform on compacta topology on Hom(X, Y).

Definition. Let X be a topological space and let I := [0, 1]. Let x, x′ ∈ X. A path in X from x to x′ is an element γ ∈ C(I, X) such that both γ(0) = x and γ(1) = x′. A loop in X at x is a path from x to x.

Definition. Let X be a topological space and let I := [0, 1]. Let x_{0}, x_{1} ∈ X and let

P_{x_{0}}^{x_{1}} := {γ ∈ C(I, X) | γ(0) = x_{0}, γ(1) = x_{1}}

be the topological space of paths in X from x_{0} to x_{1}. (Give P_{x_{0}}^{x_{1}} the inherited topology from C(I, X).) Let γ_{0}, γ_{1} ∈ P_{x_{0}}^{x_{1}}. An endpoint fixed homotopy from γ_{0} to γ_{1} is a path in P_{x_{0}}^{x_{1}} from γ_{0} to γ_{1}. We say that γ_{0} and γ_{1} are endpoint fixed homotopic if there is an endpoint fixed homotopy from γ_{0} to γ_{1}.

Following the preceding Fact, we may equivalently define an endpoint fixed homotopy from γ_{0} to γ_{1} to be a continuous H : I × I → X such that

(1) for all s ∈ I, we have H(s, 0) = x_{0};

(2) for all s ∈ I, we have H(s, 1) = x_{1};

(3) for all t ∈ I, we have H(0, t) = γ_{0}(t); and

(4) for all t ∈ I, we have H(1, t) = γ_{1}(t). We have some point set topology exercises:

EXERCISE 3A: Let X be a topological space, let Y be a set and let f : X → Y be a function. Let O := {U ⊆ Y | f^{−1}(U) is open in X}. Show that O is the finest topology on Y making f continuous.

EXERCISE 3B: Let X be a topological space, let S ⊆ X and let x_{0} ∈ X. Show that x_{0} ∈ ∂S iff the following condition holds: For all open neighborhoods U of x_{0} in X, we have U ∩ S ≠ ∅ ≠ U\S.

EXERCISE 3C: Let X be a topological space and let S ⊆ X. Show that X\(S^{∘}) = \overline{X\S}.

EXERCISE 3D: Let X be a topological space and let U be an open subset of X. Show that (∂U)^{∘} = ∅. That is, show that Int_{X}(Bd_{X}(U)) = ∅.

Definition. Let (X, x) be a pointed topological space. Define an equivalence relation ∼ on P_{x}^{x}(X) by: For all γ, γ′ ∈ P_{x}^{x}(X), γ ∼ γ′ means that γ and γ′ are endpoint fixed homotopic, i.e., that P_{γ}^{γ′}(P_{x}^{x}(X)) ≠ ∅. We define π_{1}(X, x) := (P_{x}^{x}(X))/ ∼.

Note that this defines a functor π_{1} : {pointed topological spaces} → {sets}. Eventually we’ll redefine π_{1} as a functor π_{1} : {pointed topological spaces} → {groups}, and the current π_{1} will be the compostion with that redefined π_{1}, followed by the forgetful functor {sets} → {groups}. For the moment, however, this definition will suffice for our purposes.

Recall that, in any topological space Z, if we define an equivalence relation ≅ on Z by z ≅ z′ iff P_{z}^{z′}(Z) ≠ ∅, then the equivalence classes of ≅ are called path components of Z. With that terminology, π_{1}(X, x) is simply the set of path components in P_{x}^{x}(X).

EXERCISE 3E: Given an arrow f : (X, x) → (Y, y) in the category

{pointed topological spaces}, define an arrow π_{1}(f) : π_{1}(X, x) → π_{1}(Y, y) in {sets}.

As is typical with covariant functors, we use f_{∗} as an abbreviation for π_{1}(f). (For contravariant functors, f^{∗} is typical as an abbreviation for the result of applying the functor to an arrow f.)

Recall that we aim to prove that ℝ^{2} and ℝ^{3} are not homeomorphic. We already observed that it suffices to show that X := ℝ^{2}\{(0, 0)} and Y := ℝ^{3}\{(0, 0, 0)} are not homeomorphic. It now suffices to show, for all x ∈ X and all y ∈ Y, that |π_{1}(X, x)| ≥ 2, while |π_{1}(Y, y)| = 1.

Definition. We say that a topological space X is path connected if X has only one path component, i.e., if the following condition holds: for all x, x′ ∈ X, we have P_{x}^{x′}(X) ≠ ∅.

We will argue, in Exercise 3G, that if X is path connected, then, for any x, x′ ∈ X, we have that π_{1}(X, x) and π_{1}(X, x′) are bijective, i.e., isomorphic in the category {sets}. Eventually this will be refined to saying that π_{1}(X, x) and π_{1}(X, x′) are isomorphic in the category {groups}.

Definition. Let x, x′, x′′ ∈ X ∈ T S. For all γ ∈ P_{x}^{x′}(X), for all γ′ ∈ P_{x′}^{x′′}(X), we define γ∥γ′ ∈ P_{x}^{x′′}(X) by

(γ∥γ′)(t) = \{ \begin{matrix}γ(2t), & if t ∈ [0, 1/2]; \\ γ′(2t − 1), & if t ∈ [1/2, 1]. \end{matrix}

The path γ∥γ′ is called the concatenation of γ with γ′.

Definition. Let x, x′ ∈ X ∈ T S. For all γ ∈ P_{x}^{x′}(X), we let [γ] denote the endpoint fixed homotopy class of γ, i.e., we define

[γ] := {δ ∈ P_{x}^{x′}(X) | P_{γ}^{δ}(P_{x}^{x′}(X)) ≠ ∅}.

EXERCISE 3F: Let a, b, c, d ∈ X ∈ T S. Let γ ∈ P_{a}^{b}(X), δ ∈ P_{b}^{c}(X) and ϵ ∈ P_{c}^{d}(X). Show that [γ∥(δ∥ϵ)] = [(γ∥δ)∥ϵ].

Exercise 3F asserts that concatenation is “associative up to homotopy”. Note that it is not associative, i.e., note that it frequently happens that γ∥(δ∥ϵ) ≠ (γ∥δ)∥ϵ.

Definition. Let a, b ∈ X ∈ T S. For all γ ∈ P_{a}^{b}(X), we define \overleftarrow{γ} ∈ P_{b}^{a}(X) by (\overleftarrow{γ})(t) = γ(1 − t).

EXERCISE 3G: Let a, b ∈ X ∈ T S and let γ ∈ P_{a}^{b}(X). Show that

[δ] ↦ [\overleftarrow{γ}∥δ∥γ] : π_{1}(X, a) → π_{1}(X, b)

is a well-defined bijection.

Part of the the preceding exercise is to show that, if [δ] = [δ′], then [\overleftarrow{γ}∥δ∥γ] = [\overleftarrow{γ}∥δ′∥γ]. This is what is mean by proving “well-definedness”.

One summarizes Exercise 3G by saying that “π_{1} is essentially independent of the basepoint”. This will even be true after redefining π_{1} in a group theoretic way.

We can define a category of “pathed topological spaces”, each object of which consists of a topological space X, together with a path [0, 1] → X. Let

A, B : {pathed topological spaces} → {sets}

be defined by

A(X, γ) = π_{1}(X, γ(0)), B(X, γ) = π_{1}(X, γ(1)).

Then the argument of Exercise 3G shows that A and B are equivalent functors, and this is the key point of that exercise. So, for x_{0}, x_{1} ∈ X, the bijection π_{1}(X, x_{0}) → π_{1}(X, x_{1}) is “natural up to choosing a path connecting the basepoints x_{0} and x_{1}”.

Note that, since X := ℝ^{2}\{(0, 0)} and Y := ℝ^{3}\{(0, 0, 0)} are both path-connected, it now suffices to show that there exist x ∈ X and y ∈ Y such that |π_{1}(X, x)| ≥ 2 and such that |π_{1}(Y, y)| = 1. (Before, we needed to show this for all x ∈ X, y ∈ Y, so the problem is now formally easier.)

Definition. Let I := [0, 1]. Let X, Y ∈ T S and let f, g : X → Y be continuous. A homotopy from f to g is a continuous map H : I × X → Y such that H(0, ·) = f and H(1, ·) = g. We say that f is homotopic to g, if there exists a homotopy from f to g.

EXERCISE 3H: Let X, Y ∈ T S, let f, g : X → Y be continuous and let x ∈ X. Assume that f is homotopic to g. As usual, let f_{∗} := π_{1}(f) : π_{1}(X, x) → π_{1}(Y, f(x)) and let g_{∗} := π_{1}(g) : π_{1}(X, x) → π_{1}(Y, g(x)). Prove that there is a bijection

b : π_{1}(Y, f(x)) → π_{1}(Y, g(x))

such that b ∘ f_{∗} = g_{∗}. (Hint: If H is a homotopy from f to g, then H(·, x) is a path from f(x) to g(x). The needed bijection then comes from Exercise 3G.)

One summarizes Exercise 3H by the buzzphrase: “Homotopic maps induce the same map on π_{1}.” However, to be technically precise one needs to remember that the two maps on π_{1}, namely f_{∗} and g_{∗}, yield different basepoints on Y, namely f(x) and g(x). A precise interpretation of this buzzphrase must take this into account.

Definition. Let X, Y ∈ T S. We say that X is homotopy equivalent to Y (or that X and Y have the same homotopy type) if there are continuous maps f : X → Y and g : Y → X such that both of the following conditions hold:

(1) g ∘ f : X → X is homotopic to id_{X} : X → X; and

(2) f ∘ g : Y → Y is homotopic to id_{Y} : Y → Y.

EXERCISE 3I: Let X, Y ∈ T S and assume that both X and Y are path-connected. Assume that X is homotopy equivalent to Y. Show, for any x ∈ X and any y ∈ Y, that π_{1}(X, x) is bijective to π_{1}(Y, y), i.e., that π_{1}(X, x) is isomorphic to π_{1}(Y, y) in the category {sets}.

Once again, the preceding exercise will strengthen to saying that π_{1}(X, x) and π_{1}(Y, y) are isomorphic in the category {groups}, once we’ve redefined the functor π_{1}.

EXERCISE 3J: Let n ≥ 2 be an integer. Show that S^{n−1} := {x ∈ ℝ^{n} | d(0, x) = 1} is homotopy equivalent to Q := ℝ^{n}\{0}. (Hint: Use the maps p ↦ p : S^{n−1} → Q and q ↦ q/(d(0, q)) : Q → S^{n−1}.

Given the last two exercises, our goal is now to show that there exist a ∈ S^{1} and b ∈ S^{2} such that |π_{1}(S^{1}, a)| ≥ 2 and such that |π_{1}(S^{2}, b)| = 1.

Definition. Let X be a topological space and x ∈ X. We say that X is x-avoidable if, for all a, b ∈ X\{x}, for all γ ∈ P_{a}^{b}(X), there exists γ′ ∈ P_{a}^{b}(X\{x}) such that γ is endpoint fixed homotopic to γ′.

EXERCISE 4A: Let n ≥ 2 be an integer. Let U be an open, convex, nonempty subset of ℝ^{n}. Let X be a topological space and assume that X is homeomorphic to U. Show, for all x ∈ X, that X is x-avoidable. (Hint: Start by showing, for all a, b ∈ U, for all γ, γ′ ∈ P_{a}^{b}(U), that γ is endpoint fixed homotopic to γ′.)

Lemma. Let X be a topological space and let x ∈ X. Assume that {x} is a closed subset of X. Let U be an open neighborhood of x in X. Assume that U is x-avoidable. Then X is x-avoidable.

Proof: Let a, b ∈ X\{x} and let γ ∈ P_{a}^{b}(X). We wish to show, for some γ ∈ P_{a}^{b}(X\{x}), that γ is endpoint fixed homotopic to γ′.

Let V := X\{x}. Then {U, V } is an open cover of X. Let δ > 0 be a Lebesgue number for the open cover {γ^{−1}(U), γ^{−1}(V)} of [0, 1]. Choose s_{0}, …, s_{n} ∈ [0, 1] such that 0 = s_{0} < s_{1} < ⋯ < s_{n−1} < s_{n} = 1 and such that, for all integers i ∈ [1, n], we have s_{i} −s_{i−1} < δ/2. Then, by definition of Lebesgue number, for all integers i ∈ [1, n], we have either γ([s_{i−1}, s_{i}]) ⊆ U or γ([s_{i−1}, s_{i}]) ⊆ V. For all integers i ∈ [1, n], let a_{i} := γ(s_{i−1}) and b_{i} := γ(s_{i}), and define γ_{i} ∈ P_{a_{i}}^{b_{i}}(X) by γ_{i}(t) = γ(s_{i−1}(1−t)+s_{i}t); then γ_{i} ∈ P_{a_{i}}^{b_{i}}(U) or γ_{i} ∈ P_{a_{i}}^{b_{i}}(V). Moreover, γ is endpoint fixed homotopic to γ_{1}∥ ⋯ ∥γ_{n}. (You may parenthesize this however you find convient.) For all integers i ∈ [1, n], choose γ′_{i} ∈ P_{a_{i}}^{b_{i}}(X\{x}) such that γ_{i} is endpoint fixed homotopic to γ′_{i}. (Note that if γ_{i} ∈ P_{a_{i}}^{b_{i}}(V), then, as V = X\{x}, we may simply set γ′_{i} := γ_{i}. On the other hand, if γ_{i} ∈ P_{a_{i}}^{b_{i}}(U), then we use that U is x-avoidable.) Let γ′ := γ′_{1}∥ ⋯ ∥γ′_{n}. (Again, parenthesize as you wish.) Then γ is endpoint fixed homotopic to γ′ and γ′ ∈ P_{a}^{b}(X\{x}). QED

EXERCISE 4B: Show, for all p ∈ S^{2}, that S^{2}\{p} is homeomorphic to ℝ^{2}.

Remark. For all q ∈ S^{2}, S^{2} is q-avoidable.

Proof: Fix p ∈ S^{2}\{q}. Note that {p} is closed in S^{2}. Let U := S^{2}\{p}. By Exercise 4A and Exercise 4B, we see that U is q-avoidable. Then, by the preceding lemma, we conclude that S^{2} is q-avoidable, as well. QED

Corollary. Let q, y ∈ S^{2} and assume that q ≠ y. Then the inclusion map (S^{2}\{q}, y) → (S^{2}, y) induces a surjection π_{1}(S^{2}\{q}, y) → π_{1}(S^{2}, y).

We can finally prove:

Fact. For all y ∈ S^{2}, we have |π_{1}(S^{2}, y)| = 1.

Proof: We have π_{1}(S^{2}, y) ≠ ∅ (why?), so |π_{1}(S^{2}, y)| ≥ 1. It suffices to show that |π_{1}(S^{2}, y)| ≤ 1.

Fix q ∈ S^{2}\{y}. By the preceding corollary, we see that |π_{1}(S^{2}, y)| ≤ |π_{1}(S^{2}\{q}, y)| so it suffices to show that |π_{1}(S^{2}\{q}, y)| = 1.

By Exercise 4B, it suffices to prove, for all z ∈ ℝ^{2}, that |π_{1}(ℝ^{2}, z)| = 1. Let Z := {z}. Then |π_{1}(Z, z)| = 1, so, by Exercise 3I, it suffices to show that ℝ^{2} is homotopy equivalent to Z.

Let i : Z → ℝ^{2} be the inclusion map and define j : ℝ^{2} → Z by j(x) = z. Then i ∘ j is the identity on Z. Moreover, by convexity of ℝ^{2}, we see that any two maps ℝ^{2} → ℝ^{2} are homotopic, so j ∘ i is homotopic to the identity on ℝ^{2}. QED

We now aim to show, for some x ∈ S^{1}, that |π_{1}(X, x)| ≥ 2.

Let x := (1, 0) ∈ S^{1} ⊆ ℝ^{2}. Define γ, γ′ ∈ P_{x}^{x}(S^{1}) by γ(t) = x and γ′(t) = (cos(2πt), sin(2πt)). We wish to show that γ is not endpoint fixed homotopic to γ′.

Definition. Let X, Y and A be topological spaces. Let π : X → Y be continuous. Let f : A → Y be continuous. A (full) π-lift of f is a continuous map f̂: A → X such that π ∘ f̂ = f. Let A_{0} ⊆ A. A (partial) π-lift of f on A_{0} is a continuous map f̂_{0} : A_{0} → X such that π ∘ f̂_{0} = f|A_{0}.

Definition. Let X, Y and A be topological spaces. Let π : X → Y be continuous. Let A_{0} ⊆ A. We say that π has unique (A_{0}, A) lifting if, for all continuous f : A → Y, for any partial π-lift f̂_{0} of f on A_{0}, there is a unique full π-lift f̂ of f such that f̂|A_{0} = f̂_{0}.

One may think of f̂_{0} as “initial data”, and the problem is to find a “solution” f̂ of the “lifting problem” given by (A, f, π), but the solution must fit the given initial data.

Definition. Let X, Y and A be topological spaces. Let π : X → Y be continuous. Let A_{0} ⊆ A. We say that π has local unique (A_{0}, A) lifting if, for all y ∈ Y, there exists an open neighborhood V of y in Y such that π|[π^{−1}(V)] : π^{−1}(V) → V has unique (A_{0}, A) lifting.

A topological pair is a pair (A_{0}, A) such that A is a topological space and A_{0} is a subset of A. These are the objects of a category. (Here, an arrow (A_{0}, A) → (B_{0}, B) is a continuous map f : A → B such that f(A_{0}) ⊆ B_{0}.) If (A_{0}, A) and (B_{0}, B) are isomorphic in the category of topological pairs (i.e., if there is a homeomorphism A → B which carries A_{0} to B_{0}), then

(1) unique (A_{0}, A) lifting is equivalent to unique (B_{0}, B) lifting; and

(2) local unique (A_{0}, A) lifting is equivalent to local unique (B_{0}, B) lifting.

Unassigned exercise: Show that ({0}, [0, 1]) is isomorphic, in the category of topological pairs, to ({5}, [5, 5.3]).

Definition. Let X, X′ and Y be topological spaces and let π : X → Y and π′ : X′ → Y be continuous. We will say that π and π′ are Y -homeomorphic if there is a homeomorphism h : X → X′ such that π′ ∘ h = π.

Fix a topological space Y. A topological space over Y consists of a topological space X and a continuous map X → Y. Then topological spaces over Y form the objects of a category. (What are the arrows?) Then π and π′ are Y -homeomorphic iff we have: (X, π) and (X′, π) are isomorphic in the category of topological spaces over Y.

Remark. Let A, X, X′ and Y be topological spaces and let π : X → Y and π′ : X′ → Y be Y -homeomorphic continuous maps. Let A_{0} ⊆ A. Then

(1) π has unique (A_{0}, A) lifting iff π′ has unique (A_{0}, A) lifting; and

(2) π has locally unique (A_{0}, A) lifting iff π′ has locally unique (A_{0}, A) lifting.

Definition. Let X, X′, Y and Y′ be topological spaces and let π : X → Y and π′ : X′ → Y′ be continuous. We say that π and π′ are homeomorphic if there are homeomorphisms α : X → X′ and β : Y → Y′ such that π′ ∘ α = β ∘ π.

Note that π and π′ are homeomorphic iff they are isomorphic objects in the category Arr(T S).

The next remark generalizes the preceding one.

Remark. Let A, X, X′, Y and Y′ be topological spaces and let π : X → Y and π′ : X′ → Y′ be homeomorphic continuous maps. Let A_{0} ⊆ A. Then

(1) π has unique (A_{0}, A) lifting iff π′ has unique (A_{0}, A) lifting; and

(2) π has locally unique (A_{0}, A) lifting iff π′ has locally unique (A_{0}, A) lifting.

Definition. Let X and Y be topological spaces and let π : X → Y be continuous. We say that π is a trivial covering map if there exists a discrete topological space D such that π is Y -homeomorphic to (y, d) ↦ y : Y × D → Y.

Unassigned exercise: Show that π : X → Y is a trivial covering map iff there exist topological spaces D and Y′ such that D is discrete and such that π is homeomorphic to (y, d) ↦ y : Y′ × D → Y′.

Suggestion for the next two exercises: Find a general result about lifting for trivial covering maps that will imply both of the next two exercises, and you can do both at once.

EXERCISE 4C: Let X and Y be topological spaces. Let π : X → Y be a trivial covering map. Show that π has unique ({0}, [0, 1]) lifting.

EXERCISE 4D: Let X and Y be topological spaces. Let π : X → Y be a trivial covering map. Let I := [0, 1] and let L := (I × {0}) ∪ ({0} × I). Show that π has unique (L, I^{2}) lifting.

Unassigned exercise: Show that any trivial covering map has unique ((1, 0), S^{1}) lifting.

Definition. Let X and Y be topological spaces. Let π : X → Y be continuous. We say that π is a covering map if, for all y ∈ Y, there exists an open neighborhood V of y in Y such that π|[π^{−1}(V)] : π^{−1}(V) → V is a trivial covering map.

EXERCISE 4E: Show that t ↦ (cos(2πt), sin(2πt)) : ℝ → S^{1} is a covering map.

Remark. Let X and Y be topological spaces. Let π : X → Y be a covering map. Let I := [0, 1] and let L := (I × {0}) ∪ ({0} × I). Then π has both local unique ({0}, I) lifting and local unique (L, I^{2}) lifting.

Proof: This follows immediately from Exercise 4C and Exercise 4D. QED

Theorem. Let I := [0, 1]. Then local unique ({0}, I) lifting implies unique ({0}, I) lifting.

This is an example of a “local to global” result, which puts it among a very important class of theorems in mathematics.

Proof: We will prove existence, and leave uniqueness as an unassigned exercise.

Let π : X → Y have local unique ({0}, I) lifting. Let f : I → Y be continuous, and let x_{0} ∈ π^{−1}(f(0)). We wish to show that there is a continuous map f̂: I → X such that both π ∘ f̂ = f and f̂(0) = x_{0}.

Fix an open cover V of Y such that, for all V ∈ V, we have that π|[π^{−1}(V)] : π^{−1}(V) → V of any element of the open cover has unique ({0}, I) lifting.

Let δ be a Lebesgue number for the f-pullback {f^{−1}(V) | V ∈ V} of V to I. Choose a positive integer n such that 1/n < δ/2. For all integers i ∈ [1, n], let a_{i} := (i − 1)/n, let b_{i} := i/n, let I_{i} := [a_{i}, b_{i}], and let f_{i} := f|I_{i}.

Let i ∈ [1, n] be an integer. Note that, in the category of topological pairs, defined above, the topological pair ({0}, I) is isomorphic to the topological pair ({a_{i}}, [a_{i}, b_{i}]), so that unique ({0}, I) lifting is equivalent to unique ({a_{i}}, [a_{i}, b_{i}]) lifting. By definition of Lebesgue number, there is some V ∈ V such that f(I_{i}) ⊆ V. Consequently, for all x ∈ π^{−1}(f(a_{i})), there exists a unique ϕ̂ : [a_{i}, b_{i}] → X such that both π ∘ ϕ̂ = f_{i} and ϕ(a_{i}) = x.

Using this, choose f̂_{1} : [a_{1}, b_{1}] → X such that both π ∘ f̂_{1} = f_{1} and f̂_{1}(a_{1}) = x_{0}. Then let x_{1} := f̂_{1}(b_{1}).

Then choose f̂_{2} : [a_{2}, b_{2}] → X such that both π ∘ f̂_{2} = f_{2} and f̂_{2}(a_{2}) = x_{1}. Then let x_{2} := f̂_{1}(b_{2}).

Next, choose f̂_{3} : [a_{3}, b_{3}] → X such that both π ∘ f̂_{3} = f_{3} and f̂_{3}(a_{3}) = x_{2}. Then let x_{3} := f̂_{1}(b_{3}).

Continuing on, we obtain f̂_{1}, …, f̂_{n}. Finally, define f̂ : I → X by: for all integers i ∈ [1, n], f̂|I_{i} = f̂_{i}. QED

Theorem. Let I := [0, 1] and let L := (I × {0}) ∪ ({0} × I). Then local unique (L, I^{2}) lifting implies unique (L, I^{2}) lifting.

This is another local to global result. It is proved in a similar way to the theorem preceding it, but instead of dividing I into small subintervals, we divide I^{2} into small squares, and then, to begin, solve the lifting problem on the bottom row of squares, left to right. Then we solve the lifting problem on the next row of squares up, again moving from left to right. Continuing, we get a solution on all squares, and then piece the results together to get a lift on all of I^{2}.

Question: Let A be a topological space and let A_{0} be a subset of A. Is it automatically true that local unique (A_{0}, A) lifting implies unique (A_{0}, A) lifting?

Proposition. Let X and Y be topological spaces. Let π : X → Y be a covering map. Let I := [0, 1], let I^{2} := I × I and let L := (I × {0}) ∪ ({0} × I). Then π has both unique ({0}, I) lifting and unique (L, I^{2}) lifting.

Proof: This follows from the last two theorems, together with the last remark. QED

EXERCISE 4F: Let I := [0, 1]. Let X and Y be topological spaces and let π : X → Y be a covering map. Let γ : I → Y be constant and let γ̂ : I → X be a π-lift of γ. Show that γ̂ is also constant. (Hint: Use the fact that, by the preceding proposition, π has unique ({0}, [0, 1]) lifting.)

Theorem. Let I := [0, 1]. Let X and Y be topological spaces. Let π : X → Y be a covering map. Let γ, γ′ : I → Y be continuous and assume that γ and γ′ are endpoint fixed homotopic. Let γ̂, γ′ : I → X be π-lifts of γ, γ′, respectively. Assume that γ̂(0) = γ′(0). Then γ̂(1) = γ′(1), and γ̂ is endpoint fixed homotopic to γ′.

Proof: Let I^{2} := I × I and L := (I × {0}) ∪ ({0} × I). By the preceding proposition, π has unique ({0}, I) lifting and also has unique (L, I^{2}) lifting. Let x_{0} := γ̂(0) = γ′(0). Since γ and γ′ are endpoint fixed homotopic, let h : I^{2} → Y be a continuous map such that:

(1) h(0, ·) = γ;

(2) h(1, ·) = γ′;

(3) h(·, 0) is constant; and

(4) h(·, 1) is constant. We have π(x_{0}) = π(γ̂(0)) = γ(0) = h(0, 0). Then, by (3), for all s ∈ I, we have h(s, 0) = π(x_{0}). By unique (L, I^{2}) lifting, choose a π-lift ĥ : I^{2} → X of h : I^{2} → Y such that

(5) ĥ(0, ·) = γ̂; and

(6) for all s ∈ I, we have ĥ(s, 0) = x_{0}.

By (6), ĥ(1, 0) = x_{0}. By (2), ĥ(1, ·) is a π-lift of γ′. So, since γ′ is another π-lifts of γ′, and since ĥ(1, 0) = x_{0} = γ′(0), it follows, from unique ({0}, I) lifting, that ĥ(1, ·) = γ′. In particular, we have ĥ(1, 1) = γ′(1).

By (4), ĥ(·, 1) is a π-lift of a constant, so, by Exercise 4F, we conclude that ĥ(·, 1) is constant. In particular, we have ĥ(0, 1) = ĥ(1, 1).

By (5), ĥ(0, 1) = γ̂(1). Then γ̂(1) = ĥ(0, 1) = ĥ(1, 1) = γ′(1). Moreover, ĥ is an endpoint fixed homotopy from γ̂ to γ′ in X. QED

Alternate Proof: Let I^{2} := I × I and L := (I × {0}) ∪ ({0} × I). Let U := L ∪ ({1} × I). By the preceding proposition, π has unique ({0}, I) lifting and also has unique (L, I^{2}) lifting. It is an unassigned exercise to show that (L, I^{2}) is isomorphic in the category of topological pairs to (U, I^{2}). Then π also has unique (U, I^{2}) lifting.

Let x_{0} := γ̂(0) = γ′(0). Since γ and γ′ are endpoint fixed homotopic, let h : I^{2} → Y be a continuous map such that:

(1) h(0, ·) = γ;

(2) h(1, ·) = γ′;

(3) h(·, 0) is constant; and

(4) h(·, 1) is constant.

We have π(x_{0}) = π(γ̂(0)) = γ(0) = h(0, 0). Then, by (3), for all s ∈ I, we have h(s, 0) = π(x_{0}). By unique (U, I^{2}) lifting, choose a π-lift ĥ : I^{2} → X of h : I^{2} → Y such that

(5) ĥ(0, ·) = γ̂;

(6) ĥ(1, ·) = γ′; and
(7) for all s ∈ I, we have ĥ(s, 0) = x_{0}.

By (4), ĥ(·, 1) is a π-lift of a constant, so, by Exercise 4F, we conclude that ĥ(·, 1) is constant. In particular, we have ĥ(0, 1) = ĥ(1, 1). Then, using (5) and (6), we get γ̂(1) = ĥ(0, 1) = ĥ(1, 1) = γ′(1). QED

Recall that we asked the following question:

Question: Let A be a topological space and let A_{0} be a subset of A. Is it automatically true that local unique (A_{0}, A) lifting implies unique (A_{0}, A) lifting?

We can now answer that question in the negative:

EXERCISE 5A: Define π_{0} : ℝ → S^{1} by π_{0}(t) = (cos(2πt), sin(2πt)). Let x := (1, 0) ∈ S^{1}. Show that π_{0} has local unique ({x}, S^{1}) lifting, but does not have unique ({x}, S^{1}) lifting.

EXERCISE 5B: Let X be a topological space and let x, x′, x′′ ∈ X. Let γ, γ′ ∈ P_{x}^{x′}(X) and let δ, δ′ ∈ P_{x′}^{x′′}(X). Assume that γ is endpoint fixed homotopic to γ′. Assume that δ is endpoint fixed homotopic to δ′. Show that γ∥δ is endpoint fixed homotompic to γ′∥δ′.

Recall that, for all γ ∈ P_{x}^{x}(X), we have defined [γ] to be the endpoint fixed homotopy class of γ, i.e., the set of all elements of P_{x}^{x}(X) which are endpoint fixed homotopic to γ. Recall that π_{1}(X, x) := {[γ] | γ ∈ P_{x}^{x}(X)} is the set of all endpoint fixed homotopy classes of elements of P_{x}^{x}(X).

Definition. Let X be a topological space and let x ∈ X. For all [γ], [δ] ∈ π_{1}(X, x), we define [γ][δ] := [γ∥δ].

Exercise 5B tells us that this multiplication on π_{1}(X, x) is well-defined, i.e., that if [γ] = [γ′] and if [δ] = [δ′], then [γ∥δ] = [γ′∥δ′].

EXERCISE 5C: Let X be a topological space and let x ∈ X. Let Γ := π_{1}(X, x). Show that under the multiplication defined above, Γ is a group. That is, show all of the following:

(1) For all a, b, c ∈ Γ, we have (ab)c = a(bc).

(2) There exists a unique e ∈ Γ such that, for all a ∈ Γ, we have ea = ae = a.

(3) For all a ∈ Γ, there exists b ∈ Γ such that ab = ba = e.

In Exercise 3E, given an arrow f : (X, x) → (Y, y) in {pointed topological spaces}, you were asked to define f_{∗} = π_{1}(f) : π_{1}(X, x) → π_{1}(Y, y). The definition remains the same: (π_{1}(f))([γ]) = [f ∘ γ]. It is now an unassigned exercise to show that this map π_{1}(f) is a homomorphism of groups, i.e., is an arrow in the category of groups.

Recall that we defined the functor π_{1} : {pointed topological spaces} → {sets}. We have now redefined it as a functor π_{1} : {pointed topological spaces} → {groups}. From now on, if we want to think of π_{1} as a functor into {sets}, then we will write πset_{1}.

Define functors α, ω : {pathed topological spaces} → {pointed topological spaces} by α(X, γ) = (X, γ(0)) and ω(X, γ) = (X, γ(1)). The argument of Exercise 3G now shows that the functors π_{1} ∘ α : {pathed topological spaces} → {groups} and π_{1} ∘ ω : {pathed topological spaces} → {groups} are equivalent functors. From this we conclude:

Corollary. Let X be a path-connected topological space. Let x, x′ ∈ X. Then the groups π_{1}(X, x) and π_{1}(X, x′) are isomorphic.

The argument of Exercise 3H shows:

Lemma. Let X and Y be topological spaces, let x ∈ X and let f, g : X → Y be continuous maps. Assume that f and g are homotopic. Let f̃_{∗} : π_{1}(X, x) → π_{1}(Y, f(x)) be the group homomorphism induced by f and let g̃_{∗} : π_{1}(X, x) → π_{1}(Y, g(x)) be the group homomorphism induced by g. Then there is a group isomorphism σ : π_{1}(Y, f(x)) → π_{1}(Y, g(x)) such that σ ∘ f̃_{∗} = g̃_{∗}.

Corollary. Let X and Y be topological spaces. Let f : X → Y and g : Y → X be continuous maps. Assume that g ∘ f is homomotopic to the identity id_{X} : X → X. Let x ∈ X. Let f̃_{∗} : π_{1}(X, x) → π_{1}(Y, f(x)) be the group homomorphism induced by f. Let g̃_{∗} : π_{1}(X, f(x)) → π_{1}(Y, g(f(x))) be the group homomorphism induced by g. Then there is a group isomorphism σ : π_{1}(X, x) → π_{1}(X, g(f(x))) such that f̃_{∗} ∘ g̃_{∗} = σ.

Proof: This follows from the preceding lemma, with Y replaced by X, f replaced by g ∘ f and g replaced by by id_{X}. QED

Corollary. Let X and Y be topological spaces. Assume that X and Y are homotopy equivalent. Then there exist x_{0} ∈ X and y_{0} ∈ Y such that π_{1}(X, x_{0}) is isomorphic to π_{1}(Y, y_{0}).

Proof: Let x ∈ X. Define y := f(x), x′ := g(y) and y′ := f(x′). Let f̃_{∗} : π_{1}(X, x) → π_{1}(Y, y) be induced by f. Let g̃_{∗} : π_{1}(Y, y) → π_{1}(X, x′) be induced by g. Let f′_{∗} : π_{1}(X, x′) → π_{1}(Y, y′) be induced by f. We will show that g̃_{∗} is an isomorphism.

By the preceding corollary, choose group isomorphisms σ : π_{1}(X, x) → π_{1}(X, x′) and τ : π_{1}(Y, y) → π_{1}(Y, y′) such that g̃_{∗} ∘ f̃_{∗} = σ and such that f′_{∗} ∘ g̃_{∗} = τ.

Then g̃_{∗} ∘(f̃_{∗} ∘σ^{−1}) is the identity on π_{1}(X, x′). Moreover, (τ^{−1} ∘f′_{∗})∘g̃_{∗} is the identity on π_{1}(Y, y). We are now done, by applying the next remark, with C replaced by {groups}, with g replaced by g̃_{∗}, with a replaced by τ^{−1} ∘ f′_{∗} and with b replaced by f̃_{∗} ∘ σ^{−1}. QED

Remark. Let C be a category and let g : C → C′ be an arrow in C. Assume that there exist arrows a, b : C′ → C in C such that ag = id_{C} and gb = id_{C′}. Then g is an isomorphism.

We are saying that, if an arrow in a category has a left inverse and a right inverse, then it has a two-sided inverse. In fact, we’ll show that the two one-sided inverses are equal:

Proof: We have a = a(id_{C′}) = a(gb) = (ag)b = (id_{C})b = b. Then ag = id_{C} and ga = id_{C′}, so a is an inverse for g, so g is an isomorphism. QED

We will soon be spending quite a bit of time developing further our ability to compute π_{1} on (pointed) topological spaces, but we break off from that momentarily to make some general comments.

We have now shown that ℝ^{1} and ℝ^{2} are not homeomorphic and that ℝ^{2} and ℝ^{3} are not homeomorphic. Eventually we’ll develop enough algebraic topological tools to show that, if m, n ≥ 0 are integers then: ℝ^{m} is homeomorphic to ℝ^{n} iff m = n. There are, in fact, a number of possible approaches to this, but they mostly rely on a good understanding of homology theory, which is a major topic in the last half of this course.

Eventually, we’ll develop enough tools to show both of the following:

(•) For any C ⊆ ℝ^{2} if C is homeomorphic to S^{1}, then ℝ^{2}\C is not connected. (This

is sometimes called the “Jordan Curve Theorem”.) (•) For any C ⊆ ℝ^{3}, if C is homeomorphic to S^{1}, then ℝ^{3}\C is connected.

This gives an alternative way to show that ℝ^{2} is not homeomorphic to ℝ^{3}, and is somehow in the same spirit as our proof that ℝ is not homeomorphic to ℝ^{2}.

We’ll also be able to show:

(•) For any L ⊆ ℝ^{2} if L is closed in ℝ^{2} and if L is homeomorphic to ℝ^{1}, then ℝ^{2}\L

is not connected. (•) For any L ⊆ ℝ^{3}, if L is closed in ℝ^{3} and if L is homeomorphic to ℝ^{1}, then ℝ^{3}\L

is connected. This also shows that ℝ^{2} is not homeomorphic to ℝ^{3}.

We’ll also be able to show:

(•) For any S ⊆ ℝ^{3} if S is homeomorphic to S^{2}, then ℝ^{2}\S is not connected. (•) For any S ⊆ ℝ^{4}, if S is homeomorphic to S^{2}, then ℝ^{4}\S is connected.

This gives an alternative way to show that ℝ^{3} is not homeomorphic to ℝ^{4}.

Definition. A topological space X is said to be simply connected if it is path-connected, and if, for all x ∈ X, we have that |π_{1}(X, x)| = 1, i.e., if we have that π_{1}(X, x) is the trivial group.

We’ll also be able to show:

(•) For any C ⊆ ℝ^{3} if C is homeomorphic to S^{1}, then ℝ^{3}\C is not simply connected. Here’s a challenge: Show

(•) For any C ⊆ ℝ^{4} if C is homeomorphic to S^{1}, then ℝ^{4}\C is simply connected. Given the last two results, we get yet another proof that ℝ^{3} is not homeomorphic to ℝ^{4}.

To be efficient about it, much of this work needs to wait until the development of homology theory, which comes later. For now, we return our focus to π_{1}.

Our next goal is to show that π_{1}(S^{1}) is isomorphic to the additive group ℤ.

Sketch of Proof: Let I := [0, 1] and let x := (1, 0) ∈ S^{1}. Define π_{0} : ℝ → S^{1} by π_{0}(t) = (cos(2πt), sin(2πt)). Using that π_{0} has unique ({0}, I) lifting, for all γ ∈ P_{x}^{x}(S^{1}), let γ̂ : I → ℝ be the unique π_{0}-lift of γ such that γ̂(0) = 0. We now define a map Φ : π_{1}(S^{1}, x) → ℤ by Φ([γ]) = γ̂(1). We leave it as an unassigned exercise to show that Φ is well-defined. (Hint: Use the theorem following Exercise 4F.) We also leave it as an unassigned exercise to show that Φ is a group isomorphism. QED

After filling in the details, one has the interesting situation where covering space theory has been used to compute π_{1} of a pointed topological space.

It is our intention to generalize the above proof, in the hopes that we may be able to use covering space theory as a general tool for computing π_{1}.

Definition. A topological group is (1) a group G; and

(2) a Hausdorff topology on G such that (x, y) ↦ xy : G × G → G and x ↦ x^{−1} : G → G are both continuous.

The arrows in the category of topological groups are continuous homomorphisms. First example of a topological group is the additive group ℝ. Another example is the

multiplicative group of matrices SL(2, ℝ) := {g ∈ ℝ^{2×2} | det(g) ≠ 0}. (This is given the inherited topology from the vector space topology on the set ℝ^{2×2} of 2 × 2 matrices with real entries.

EXERCISE 5D: Let G be a topological group, let 1 denote the identity in G and let V be an open neighborhood of 1 in G. Show that there exists an open neighborhood U of 1 in G such that UU^{−1} ⊆ V, i.e., such that, for all a, b ∈ U, we have ab^{−1} ∈ V.

Note that we can replace UU^{−1} in the above exercise by various similar expresssions, and the result remains true, with much the same proof. For example, we could use UU, often denoted U^{2}. Or we could use U^{−1}U or U^{2}U^{−1}U^{−2}UU^{−1}. Or any product of postive and negative powers of U.

In this course “action” will always mean “left action” unless otherwise specified.

Definition. Let G be a topological group acting on a topological space X. We say that the action is continuous if the action map (g, x) ↦ gx : G × X → X is continuous.

This is equivalent to saying that the “secondary” action map (g, x) ↦ (g, gx) : G × X → G × X is continuous. (The use of the word “secondary” there is my choice. Either of these two maps is, at times, called the action map, but I want to distinguish between them to avoid confusion.)

A simple example of a continuous action is obtained by taking a topological space G and letting G act on itself by multiplication. That is, the action map is just multiplication (g, x) ↦ gx : G × G → G.

Another equally simple example is as follows: Given a topological group G and a topological space Y, let G act on G × Y via the action g(a, y) = (ga, y).

Definition. Let G be a topological group. A topological G-space is a topological space X, together with a continuous G-action on X.

Recall that if a group G acts on a set X then a subset X_{0} ⊆ X is said to be Ginvariant if GX_{0} ⊆ X_{0}, i.e., if, for all g ∈ G, for all x ∈ X_{0}, we have gx ∈ X_{0}.

In the category of topological G-spaces, the arrows are continuous G-equivariant maps. That is, if X and Y are topological G-spaces, then an arrow from X to Y is a continuous map f : X → Y such that, for all g ∈ G, for all x ∈ X, we have f(gx) = g(f(x)).

Note that, if X is a topological G-space and if X_{0} ⊆ X is nonempty and G-invariant, then X_{0} is a topological G-space (under the inherited topology and restricted G-action) and the inclusion X_{0} → X is an arrow in the category of topological G-spaces.

I remark again that, if G is a topological group, and if Y is a topological space, then G × Y is a topological G-space, where the action is given by g(a, y) = (ga, y).

Definition. Let G be a topological group and X a topological G-space. We say that X is trivially principal if there is a topological space Y such that X is isomorphic, in the category of topological G-spaces, to G × Y. We say that X is principal if, for all x ∈ X, there is a G-invariant open neighborhood X_{0} of x in X such that X_{0} is trivially principal.

So one might use “locally trivially principal” as a synonym for “principal”.

If G is a topological group acting continuously on a topological space X, then we’ll say that the G action is principal if X is a principal G-space. Similarly, we’ll say that the G action is trivially principal if X is a trivially principal G-space. By a principal G-space, we mean a topological G-space that is principal. By a trivially principal G-space, we mean a topological G-space that is trivially principal.

Let the additive group ℤ act on ℝ by: the action of n on t yields n + t. It is not hard to show that, with this action, ℝ becomes a principal ℤ-space which is not trivially principal.

A discrete group is a topological group whose topology is discrete (which means that every subset of the group is open). Note that there is a forgetful functor {discrete groups} → {groups}. Conversely, given any group, we can give it the discrete topology, making it into a discrete group. This gives a functor {groups} → {discrete groups}. These two functors are inverses of one another, and so the category {groups} is isomorphic to the category {discrete groups}. That is, expertise in group theory is exactly the same as expertise in discrete group theory.

Recall some terminology and notation from group theory. Given a group G acting on a set X, we let G\X := {Gx | x ∈ X} be the set of orbits of G on X. The canonical map X → G\X is the map x ↦ Gx.

When G is a topological group and X is a topological G-space, then G\X automatically has a topology, namely, the quotient topology, i.e., the finest topology making the canonical map X → G\X continuous.

EXERCISE 5E: Let Γ be a discrete group, and let X be a principal Γ-space. Show that the canonical map X → Γ\X is a covering map.

Theorem. Let Γ be a discrete group and let X be a principal Γ-space. Assume that X is simply connected. Then, for all y ∈ Γ\X, we have that the groups π_{1}(Γ\X, y) and Γ are isomorphic.

Sketch of Proof: Let Y := Γ\X. We wish to show that π_{1}(Y, y) is isomoprhic to Γ.

Let I := [0, 1]. Let π : X → Γ\X be the canonical map. Then π is surjective. Choose x ∈ X such that π(x) = y. By Exercise 5E, Γ is a covering map, and therefor has unique ({0}, I) lifting. For all α ∈ P_{y}^{y}(Y), let α̂ : I → ℝ be the unique π-lift of α such that α̂(0) = x. Note that π(α̂(1)) = α(1) = y and that π(α̂(0)) = π(x) = y, so there is a unique γ_{α} ∈ Γ such that α̂(1) = γ_{α}(α̂(0)). (Unassigned exercise: Explain existence and uniqueness in the last sentence.) We now define a map Φ : π_{1}(Y, y) → Γ by Φ([α]) = γ_{α}. We leave it as an unassigned exercise to show that Φ is well-defined. (Hint: Use the theorem following Exercise 4F.) We also leave it as an unassigned exercise to show that Φ is a group isomorphism. (Note: Simple connectedness of X is needed in the proof that Φ is injective.) QED

A more colloquial way of saying the above theorem is: If a discrete group Γ acts principally on a simply connected topological space X, then π_{1}(Γ\X) is isomorphic to Γ. Note that, since Γ\X is path-connected, the fundamental group does not depend on the basepoint, up to isomorphism.

Question: Is the preceding theorem true if we drop the assumption that X be simply connected?

Answer: No. Let Γ := ℤ/2ℤ act on X := ℝ/(2ℤ) by: the action of 1 on t + (2ℤ) yields t + 1 + (2/Z). Let y ∈ Y := Γ\X. We have Y = ℝ/Z, so π_{1}(Y, y) = ℤ. So π_{1}(Y, y) is not isomorphic to Γ.

Question: Is the preceding theorem true if we drop the assumption that the Γ-action on X is principal?

We will answer this last question below.

For any group G, we will let 1_{G} denote the identity element of G.

EXERCISE 5F: Let Γ be a discrete group acting continuously on a topological space X. Show that the Γ-action on X is principal iff: for every x ∈ X, there exists an open neighborhood U of x in X such that, for all γ ∈ Γ\{1_{Γ}}, we have (γU) ∩ U = ∅.

EXERCISE 5G: Let G be a topological group and let Γ be a discrete subgroup of G. Let Γ act on G by left multiplication, i.e., the result of γ acting on g is the product γg. Show that this Γ-action on G is principal. (Hint: Use Exercise 5F and Exercise 5D.)

Let the additive group ℤ act on ℝ by the rule: n acting on t yields n + t. By Exercise 5F (adapted for additive groups and with G replaced by ℝ), we see that this ℤ-action on ℝ is principal. The quotient topological space ℤ\ℝ is the same as the topological space ℝ/ℤ of cosets of ℤ in ℝ. By the preceding theorem, we conclude that π_{1}(ℝ/ℤ) = ℤ.

The next three exercises, taken togther, show that ℝ/ℤ is homeomoprhic to S^{1}, so we obtain another proof that π_{1}(S^{1}) ≅ ℤ.

EXERCISE 5H: Show that ℝ/ℤ is compact.

EXERCISE 5I: Show that there is a continous bijection ℝ/ℤ → S^{1}.

EXERCISE 5J: Show that, if X and Y are topological spaces, if X is compact, and if f : X → Y is a continuous bijection, then f : X → Y is a homeomorphism.

Definition. Let G be a group. Then a G-set is a set X, together with a G-action on X.

The arrows in the category of G-sets are G-equivariant functions. Note that, for every group homomorphism ϕ : G → H, we get a functor F_{ϕ}{H−sets} → {G−sets}, given by: For any H-set X, F_{ϕ}(X) is the G-set whose underlying set is X and where the G-action on X is given by: g acting on x yields (f(g))x. (Question: What is the effect of F_{ϕ} on arrows?)

Some notation: When a group G acts on a set X and when x ∈ X, we define Stab_{G}(x) = {g ∈ G | gx = x}. This subgroup of G is called the stabilizer of x in G.

A G-set X is transitive if |G\X| = 1, i.e., if there is exactly one orbit in X. For any subgroup G_{0} of G, we let G act on G/G_{0} by g acting on g′G_{0} yields gg′G_{0}. This action is transitive. It is not hard to show that if X is a transitive G-space, if x ∈ X and if G_{0} := Stab_{G}(x), then X is isomorphic to G/G_{0} in the category {G − sets}. This can be thought of as a “classification of transitive G-sets”.

Note that any G-set breaks up into orbits, and the G-action on each orbit is transitive. Thus any G-set is isomorphic (in the category {G − sets}) to a disjoint union of transitive G-sets. Combined with the earlier classification of transitive G-sets, we have no classified all G-sets, up to isomorphism.

Definition. Let G be a group acting on a set X. The G-action is said to be free if there is a set Y such that, in the category of G-sets, X is isomorphic to G × Y, where the G-action on G × Y is given by g(a, y) = (ga, y).

Fact. Let G be a group acting on a set X. Then the G-action on X is free iff, for all x ∈ X, we have Stab_{G}(x) = {1_{G}}.

More colloquially put, “an action is free iff all stabilizers are trivial”. Or: “an action is free iff nothing stabilizes anything, except for the identity, which stabilizes everything”.

EXERCISE 5K: Show that any principal action is free. That is, if G is a topological group acting continuously and principally on a topological space X, show that the G-action on X is free.

Question: Is the converse true? That is, if G is a topological group acting continuously and freely on a topological space X, does it follow that G-action on X is principal?

Answer: No. Let α ∈ ℝ\ℚ and let ℤ act on S^{1} by: the action of 1 on (cos(2πt), sin(2πt)) yields (cos(2π(t + α)), sin(2π(t + α))). This action is sometimes called the “irrational rotation” action on the circle. We leave it as an unassigned exercise to show that this action is free and that every orbit is dense. Since orbits are dense, it follows from Exercise 5F that the Γ-action on X is not principal.

We are now in a position to answer an earlier question.

Question: Is the preceding theorem true if we drop the assumption that the Γ-action on X is principal?

Answer: Let Γ := Z act on X := S^{1} by irrational rotation, as above. Let Y := Γ\X and let y ∈ Y. Because every orbit is dense, it follows that Y is indiscrete, and therefore that π_{1}(Y, y) is trivial. Then π_{1}(Y, y) is not isomorphic to Γ.

I don’t know a counterexample in which the quotient space is Hausdorff. That is, I don’t know the answer to the following question:

Question: Let X be a simply connected topological space and let Γ be a discrete group. Let Γ act freely on X. Assume that Y := Γ\X is Hausdorff. Let y ∈ Y. Does it follow that π_{1}(Y, y) is isomorphic to Γ?

As we have seen, principal implies free, but free does not always imply principal. However, the next exercise gives one situation in which free does imply principal.

EXERCISE 5L: Let Γ be a finite discrete group acting continuously and freely on a Hausdorff topological space X. Then show that the Γ-action on X is principal. (Hint: You may use, without proof, the fact that, for any integer n ≥ 2, for any n distinct points in a Hausdorff topological space, one may choose open neighborhoods of the n points which are pairwise disjoint. Now use Exercise 5F.)

Let Γ := ℤ/2ℤ. Then Γ is the additive cyclic group with two elements. We label the identity element as 0 and the other as 1. Give Γ the discrete topology, so that Γ is a finite discrete group. Let n ≥ 1 be an integer. Let Γ act on S^{n} by: 1 acting on p yields −p. We define ℝP^{n} := Γ\S^{n}.

By the preceding exercise, the Γ-action on S^{n}, being free, must be principal. We already proved that S^{2} is simply connected, and the same argument works to show that, for all n ≥ 2, S^{n} is simply connected. We can appeal to the preceding theorem to show that, if n ≥ 2, then π_{1}(ℝP^{n}) ≅ Γ = ℤ/2ℤ.

Definition. Let Γ be a group and let X be a Γ-set. Let x ∈ X. The orbit map at x is the map γ ↦ γx : Γ → X.

Let f : Γ → X be the orbit map at x, defined by f(γ) = γx. Note that the image f(Γ) is the orbit Γx of x. Note that f^{−1}(x) = Stab_{Γ}(x).

EXERCISE 6A: Let Γ be a group acting on a set X. Show that the Γ-action on X is free iff every orbit map is injective.

We now return back to the last theorem, and give a more detailed proof.

Theorem. Let Γ be a discrete group and let X be a principal Γ-space. Assume that X is simply connected. Then, for all y ∈ Γ\X, we have that the groups π_{1}(Γ\X, y) and Γ are isomorphic.

Proof: Let Y := Γ\X. Let π : X → Y be the canonical map. Fix x ∈ π^{−1}(y). Then Γx = π^{−1}(y).

Let f : Γ → X be the orbit map at x, defined by f(γ) = γx. By Exercise 6A, we see that f is injective. Moreover, f(Γ) = Γx = π^{−1}(y). Then f : Γ → π^{−1}(y) is a bijection. Let f^{−1} : π^{−1}(y) → Γ be its inverse. (Intuitively, because the Γ-action on X is free, every Γ-orbit “looks like” Γ; moreover the orbits are exactly the fibers of the canonical map. So π^{−1}(y), being a fiber of the canonical map, must “look like” Γ, and f^{−1} is the map that makes this precise.)

For any α ∈ P_{y}^{y}(Y), let [α] denote the endpoint fixed homotopy class of α. Then π_{1}(Y, y) = {[α] | α ∈ P_{y}^{y}(Y)}. Similarly, for any α ∈ P_{x}^{x}(X), let [α] denote the endpoint fixed homotopy class of x, so that π_{1}(X, x) = {[α] | α ∈ P_{x}^{x}(X)}.

By Exercise 5E, we see that π : X → Y is a covering map. Then, by an earlier theorem, π has unique ({0}, I) lifting. For every α ∈ P_{y}^{y}(Y), let α̂ : I → X be the unique π-lift of α such that α̂(0) = x. By an earlier theorem, for all α, β ∈ P_{y}^{y}(Y), if α is endpoint fixed homotopic to β, then α̂(1) = β̂(1). For all [α] ∈ π_{1}(Y, y), let t_{[α]} := α̂(1); then π(t_{[α]}) = π(α̂(1)) = α(1) = y, so t_{[α]} ∈ π^{−1}(y).

Define ψ : π_{1}(Y, y) → Γ by ψ([α]) = f^{−1}(t_{[α]}). We will show that this map is a homomorphism of groups and that it is injective. We will leave the proof of surjectivity as an unassigned exercise.

Proof that ψ is a homomorphism: Fix α, β ∈ P_{y}^{y}(Y). We wish to show that

(ψ([α]))(ψ([β])) = ψ([α][β]).

Let γ := ψ([α]), let δ := ψ([β]). As [α][β] = [α∥β], we must show that γδ = (ψ([α∥β]). We have α̂(1) = t_{[α]} = f(ψ([α])) = f(γ) = γx. Similarly, β̂(1) = δx. Define γβ̂ : I → X by (γβ̂)(t) = γ(β̂(t)). Then (γβ̂)(0) = γ(β̂(0)) = γx = α̂(1), so

α̂∥(γβ̂) is defined. We have π ∘ (γβ̂) = π ∘ β̂ = β, so π ∘ (α̂∥(γβ̂)) = α∥β. Moreover, (α̂∥(γβ̂))(0) = α̂(0) = x. Then α̂∥β = α̂∥(γβ̂). Then t_{[α∥β]} = α̂∥β(1) = (α̂∥(γβ̂))(1) = (γβ̂)(1) = γ(β̂(1)) = γ(δx) = γδx = f(γδ). Then ψ([α∥β]) = f^{−1}(t_{[α∥β]}) = γδ, End of proof that ψ is a homomorphism.

Proof that ψ is injective: Let c_{x} : I → X be defined by c_{x}(t) = x and let c_{y} : I → Y be defined by c_{y}(t) = y. Then [c_{x}] is the identity element of π_{1}(X, x) and [c_{y}] is the identity element of π_{1}(Y, y). Let α ∈ P_{y}^{y}(Y) and assume that ψ([α]) = 1_{Γ}. We wish to show that [α] = [c_{y}].

We have α̂(0) = x, by definition of α̂. We have α̂(1) = t_{[α]} = f(ψ([α])) = f(1_{Γ}) = x. Then α̂ ∈ P_{x}^{x}(X). However, X is simply connected, so π_{1}(X, x) is trivial. Then [α̂] = [c_{x}]. That is, α̂ is endpoint fixed homotopic to c_{x}. Then π ∘ α̂ is endpoint fixed homotopic to π ∘ c_{x}. That is, α is endpoint fixed homotopic to c_{y}. That is, [α] = [c_{y}]. End of proof that ψ is injective. QED

So, given a path-connected topological space Y we have a strategy for computing its fundamental group: Find a principal action of a discrete group Γ on a simply connected topological space whose quotient is (homeomorphic to) Y. If we can do this, then the preceding result immediately gives us that π_{1}(Y) ≅ Γ.

This strategy worked for computing π_{1}(S^{1}) and π_{1}(ℝP^{2}). A reasonable question is: Does it always work? That is:

Question: Given a path-connected topological space Y, can we always find a simply connected topological space X, a discrete group Γ and a principal action of Γ on X such that Γ\X is homeomorphic to Y?

We explore this question next.

Definition. Let X and Y be topological spaces. Let π : X → Y be a covering map. A deck transformation of π is an isomorphism (X, π) → (X, π) in the category

{topological spaces over Y }.

That is, it is a homeomorphism h : X → X such that π ∘ h = π.

EXERCISE 6B: Let X and Y be topological spaces. Let π : X → Y be a covering map. Let Γ := {deck transformations of π}. Show that Γ is a group under composition.

We will always consider {deck transformations of π} to be a topological group: It is a group under composition, and it is given the discrete topology.

EXERCISE 6C: Let X and Y be topological spaces. Let π : X → Y be a covering map. Let Γ := {deck transformations of π}. Show, for all y ∈ Y, that π^{−1}(y) is Γ-invariant.

We give Γ the discrete topology. Note that Γ, by construction, acts continuously on X. Eventually we will show that Γ acts principally on X.

Definition. Let X be a topological space. A topological space under X consists of (1) a topological space Y; and

(2) a continuous map π : X → Y.

What are the arrows in the category {topological spaces under X}?

Note that, if X is a topological space, if Γ is a group and if Γ acts continuously on X, then there is a canonical map c : X → Γ\X, and therefore (Γ\X, c) is a topological space under X. We will often just say that Γ\X is a topological space under X, the canonical map c being understood.

Definition. Let X and Y be topological spaces. Let π : X → Y be a covering map. Let Γ := {deck transformations of π}. We say that π is regular or a regular covering map if Γ\X is isomorphic to (Y, π) in the category {topological spaces under X}.

Are all covering maps regular? More starkly, if π : X → Y is a covering map and π is not a homeomorphism, does it always follow that there is a deck transformation of π that is not equal to the identity map id_{X} : X → X? The answer to both questions is no, and we will get to this, but a full theory of covering spaces is needed to address these questions properly.

For now, (2) in the following theorem asserts that, when the domain is simply connected, the covering map is regular.

Lemma. Let X and Y be path-connected topological spaces. Let π : X → Y be a covering map. Let Γ := {deck transformations of π}. Then

(1) the Γ-action on X is principal; and

(2) if X is simply connected and locally path-connected, then π is regular.

We defer the proof until later.

Definition. Let Y be a path-connected topological space. A universal cover of Y is a simply connected topological space X, together with a covering map X → Y.

By the preceding lemma, if the path-connected, locally path-connected topological space Y has a universal cover (X, π), then there is a discrete group Γ and a principal action of Γ on X such that Γ\X is homeomorphic to Y. One simply lets Γ := {deck transformations of π}, with the discrete topology. In fact, even more is true: Γ\X is isomorphic to (Y, π) in the category of topological spaces under X.

Recall our earlier

Question: Given a path-connected topological space Y, can we always find a simply connected topological space X, a discrete group Γ and a principal action of Γ on X such that Γ\X is homeomorphic to Y?

This is now transformed to:

Question: Does every path-connected topological space have a universal cover?

EXERCISE 6D: Let π_{0} : ℝ → S^{1} be defined by π_{0}(t) = (cos(2πt), sin(2πt)). Find all deck transformations of π_{0}.

EXERCISE 6E: Let X and Y be path-connected topological spaces. Let π : X → Y be a covering map. Let Γ := {deck transformations of π}. Show

(1) the Γ-action on X is free; and

(2) if π is regular, then for all y ∈ Y, the Γ-action on π^{−1}(y) is transitive.

Note: The proof of (1) of Exercise 6E actually appears later in these notes, and you may simply copy what I’ve written, if you wish.

Definition. Let Y be a topological space. We say that Y is locally simply connected if, for any y ∈ Y, for any open neighborhood V of y in Y, there is a simply connected open neighborhood U of y in Y such that U ⊆ V.

More concisely: Y is locally simply connected if every point has arbitrarily small simply connected open neighborhoods.

EXERCISE 6F: Show that any locally simply connected topological space is locally path-connected.

Let A and B be groups and let h : A → B be a homomorphism. Recall that h is said to be trivial if h(A) = {1_{B}}.

EXERCISE 6G: Let Y be a topological space and let U ⊆ Y. Assume that U is path-connected. For all u ∈ U, let i_{u} : (U, u) → (Y, u) be the arrow in the category {pointed topological spaces} defined by the inclusion map U → Y. Assume, for some u_{0} ∈ U, that the map (i_{u_{0}})_{∗} : π_{1}(U, u_{0}) → π_{1}(Y, u_{0}) is trivial. Show, for all u ∈ U, that the map (i_{u})_{∗} : π_{1}(U, u) → π_{1}(Y, u) is trivial.

Definition. Let Y be a topological space and let U ⊆ Y. Assume that U is path-connected. For all u ∈ U, let i_{u} : (U, u) → (Y, u) be the arrow in the category {pointed topological spaces} defined by the inclusion map U → Y. We say that U is relatively simply connected in Y if: for all u ∈ U, the map (i_{u})_{∗} : π_{1}(U, u) → π_{1}(Y, u) is trivial. (Here, “trivial” means that its image consists of the identity alone.)

Example. Let o := (0, 0) ∈ ℝ^{2}. Let D := {p ∈ ℝ^{2} | d(o, p) ≤ 1} be the closed unit disk in the plane. Then S^{1} ⊆ D. Note that S^{1} is not simply connected, but is relatively simply connected in D. On the other hand, S^{1} is not relatively simply connected in D\{o}.

Definition. Let Y be a topological space. We say that Y is semi-locally simply connected if: if, for any y ∈ Y, for any open neighborhood V of y in Y, there is a pathconnected open neighborhood U of y in Y such that U ⊆ V and such that U is relatively simply connected in Y.

More concisely: Y is semi-locally simply connected if every point has arbitrarily small relatively simply connected (path-connected) open neighborhoods.

Theorem. Let Y be a connected, locally path-connected Hausdorff topological space. Then Y has a universal cover iff Y is semi-locally simply connected.

We defer the proof until later. Recall our earlier

Question: Does every path-connected topological space have a universal cover?

Note that a locally path-connected topological space is connected iff it is path-connected. So, for locally path-connected topological spaces, the above question is now transformed to:

Question: Is every connected, locally path-connected topological space semi-locally simply connected?

Exercise 7A below asserts that most of the topological spaces that concern us are, in fact, semi-locally simply connected. On the other hand, part of Exercise 7B below is to show that there do exist topological spaces that are not.

EXERCISE 7A:

(1) Show that if a topological space is locally simply connected, then it is semi-locally simply connected.

(2) Show that any locally path-connected, simply connected topological space is semilocally simply connected.

EXERCISE 7B:

(1) Show that a simply connected topological space is not necessarily locally simplyconnected.

(2) Show that there is a connected, locally path-connected topological space that is not semi-locally simply connected.

(3) Show that a path-connected topological space that is semi-locally simply con-

nected need not be locally simply connected. (Note: I do not require that you prove that your counterexamples work, but they must be stated very carefully. You will probably not find these counterexamples without help. Ask around, consult references. Think about the “Hawaiian earring” and the cone over the Hawaiian earring.)

By (2) of Exercise 7A, your answer to (1) of Exercise 7B, if locally path-connected, will also be an answer to (3) of Exercise 7B.

Note that, any semi-locally simply connected implies locally path-connected, which implies locally connected. Note that, for locally path-connected topological spaces, all the path components are open. Thus, if X is a locally path-connected topological space, then: X is connected iff X is path-connected.

Remark. Let X be a path-connected topological space, let Y be a topological space and let π : X → Y be a covering map. Let Γ := {deck transformations of π}. Then the Γ-action on X is free.

Proof: Fix x_{0} ∈ X and γ ∈ Γ. Assume that γx_{0} = x_{0}. We wish to show that γ = 1_{Γ}. Let x ∈ X. We wish to show that γx = x.

Since γ ∈ Γ, we have π ∘ γ = π. Since X is path-connected, let α̂ ∈ P_{x_{0}}^{x}(X). Let I := [0, 1]. Let β̂:= γ ∘ α̂ : I → X. Let α := π ∘ α̂ : I → Y. Then

π ∘ β̂ = π ∘ (γ ∘ α̂) = (π ∘ γ) ∘ α̂ = π ∘ α̂ = α.

Then both α̂ and β̂ are π-lifts of α. Moreover, β̂(0) = γ(α̂(0)) = γx_{0} = x_{0} = α̂(0).

Since π : X → Y is a covering map, it follows that π has unique ({0}, I) lifting. So, as α̂ and β̂ are both π-lifts of α and as α̂(0) = β̂(0), we conclude that α̂ = β̂. Then x = α̂(1) = β̂(1) = γ(α̂(1)) = γx. QED

EXERCISE 7C: Let D be a discrete topological space. Let V be a topological space. Let d_{0} ∈ D. Let U := V × {d_{0}} ⊆ V × D. Let p : V × D → V be the projection defined by p(v, d) = v. Let γ : V × D → V × D be a deck transformation of p. Show, for all u ∈ (γU) ∩ U, that γu = u.

EXERCISE 7D: Let X and Y be topological spaces. Let π : X → Y be a trivial covering map. Let Γ be a subgroup of {deck transformations of π}. Assume that Γ acts freely on X. Show that Γ acts principally on X. (Hint: Use Exercise 7C and Exercise 5F.)

Corollary. Let X be a path-connected topological space. Let Y be a topological space. Let π : X → Y be a covering map. Let Γ := {deck transformations of π}. Then the Γ-action on X is principal.

Proof: Fix x ∈ X. We will show that there is a Γ-invariant open neighborhood X_{0} of x in X such that the Γ-action on X_{0} is principal.

Let Y_{0} be an open neighborhood of π(x) in Y such that π|[π^{−1}(Y_{0})] : π^{−1}(Y_{0}) → Y_{0} is a trivial covering map.

Let X_{0} := π^{−1}(Y_{0}) and let π_{0} := π|X_{0}. Then π_{0} : X_{0} → Y_{0} is a trivial covering map. Let Γ_{0} := {deck transformations of π_{0}}. Define R : Γ → Γ_{0} by R(γ) = γ|X_{0}. Let Γ_{0} := R(Γ). By the preceding remark, the Γ-action on X is free. Then R : Γ → Γ_{0} is a group isomorphism.

Then the Γ_{0}-action on X_{0} is free. Then, by Exercise 7D, the Γ_{0}-action on X_{0} is principal. Then, as R : Γ → Γ_{0} is a group isomorphism, it follows that the Γ-action on X_{0} is principal. QED EXERCISE 7E: Let X and Y be topological spaces and let π : X → Y be a trivial covering map. Assume that Y is connected. Let Γ be a subgroup of {deck transformations of π}. Assume, for some y_{0} ∈ Y, that the Γ-action on π^{−1}(y_{0}) is transitive. Show, for all y ∈ Y, that the Γ-action on π^{−1}(y) is transitive.

EXERCISE 7F: Let X and Y be topological spaces and let π : X → Y be a covering map. Assume that Y is locally connected. Let Γ := {deck transformations of π}. Let T be the set of all y ∈ Y such that the Γ-action on π^{−1}(y) is transitive. Show that T is a clopen subset of Y. (Hint: Openness follows from Exercise 7E. Openness of the complement follows from a variant of Exercise 7E, in which “transitive” is replaced by “nontransitive”.)

Definition. Let X and Y be topological spaces and let f : X → Y be a function. We say that f is open or that f : X → Y is an open mapping if: for every open subset U of X, we have that f(U) is open in Y. We say that f is a local homeomorphism if, for any point x ∈ X, there is an open neighborhood U of x in X such that f|U : U → f(U) is a homeomorphism.

EXERCISE 7G: Let Γ be a topological group acting continuously on a topological space X. Show that the canonical map X → Γ\X is an open mapping.

EXERCISE 7H: Show that any covering map is an open mapping.

Lemma. Let X be a topological space. Let Y be a connected, locally connected topological space. Let π : X → Y be a covering map and let Γ := {deck transformations of π}. Assume, for some y_{0} ∈ Y, that the Γ-action on π^{−1}(y_{0}) is transitive. Then, for all y ∈ Y, the Γ-action on π^{−1}(y) is transitive.

Proof: Let T denote the collection of all y ∈ Y such that the Γ-action on π^{−1}(Y) is transitive. We wish to show that T = Y.

By Exercise 7F, T is clopen in Y. As Y is connected, we know that ∅ and Y are the only clopen subsets of Y. As y_{0} ∈ T, we see that T ≠ ∅. Thus T = Y. QED

Lemma. Let X and Y be topological spaces. Let π : X → Y be a covering map. Let Γ := {deck transformations of π}. Then π is regular iff, for all y ∈ Y, the Γ-action on π^{−1}(y) is transitive.

Proof: “Only if” is (2) of Exercise 6E. Assume, for all y ∈ Y, that the Γ-action on π^{−1}(y) is transitive. Let W := Γ\X. Let p : X → W be the canonical map. We wish to show that there is a homeomorphism h : W → Y such that h ∘ p = π.

By transitivity on fibers of π, we conclude that {π^{−1}(y) | y ∈ Y } is the set of Γ-orbits in X. Since {p^{−1}(w) | w ∈ W } is also the set of Γ-orbits in X, we see that

{π^{−1}(y) | y ∈ Y } = {p^{−1}(w) | w ∈ W },

which implies that there is a bijective map h : W → Y such that h ∘ p = π. We must show that h is continuous and open.

For any subset Y_{0} ⊆ Y, we have h^{−1}(Y_{0}) = p(π^{−1}(Y_{0})). So, since π : X → Y is continuous and since (by Exercise 7G) p : X → W is open, we conclude that h : W → Y is continuous.

For any subset W_{0} ⊆ W, we have h(W_{0}) = π(p^{−1}(W_{0})). So, since p : X → W is continuous and since (by Exercise 7H) π : X → Y is open, we conclude that h : W → Y is open. QED

Theorem. Let X and Y be topological spaces. Let π : X → Y be a covering map. Let Z be a simply connected, locally path-connected topological space. Let z_{0} ∈ Z. Then π has unique ({z_{0}}, Z) lifting.

Proof: Let f : Z → Y be a continuous map, let y_{0} = f(z_{0}) and let x_{0} ∈ π^{−1}(y_{0}). We wish to show that there is a continuous map f̂ : Z → X such that π ∘ f̂ = f and such that f̂(z_{0}) = x_{0}.

Let I := [0, 1]. For any z ∈ Z, choose α_{z} ∈ P_{z_{0}}^{z}(Z) Then (f ∘ α_{z})(0) = f(z_{0}) = y_{0}, so x_{0} ∈ π^{−1}((f ∘ α_{z})(0)). Since π is a covering map, it follows that π has unique ({0}, I) lifting. Let β_{z} : I → X be the π-lift of f ∘ α_{z} such that β_{z}(0) = x_{0}. Define f̂: Z → X by f̂(z) = β_{z}(1).

Then, for all z ∈ Z, we have π(f̂(z)) = π(β_{z}(1)) = f(α_{z}(1)) = f(z), so π ∘ f̂ = f. Claim: If z ∈ Z, if α ∈ P_{z_{0}}^{z}(Z), if β is a π-lift of f∘α and if β(0) = x_{0}, then f̂(z) = β(1).

Proof of Claim: Since Z is simply connected, for all ρ, σ : I → Z, if ρ(0) = σ(0) and if ρ(1) = σ(1), then ρ and σ are endpoint fixed homotopic. (We leave this as an unassigned exercise.) It follows that α is endpoint fixed homotopic to α_{z}. Then f ∘ α is endpoint fixed homotopic to f ∘α_{z}. Then, by the theorem following Exercise 4F, we see that β(1) = β_{z}(1). As f̂(z) = β_{z}(1), we are done. End of proof of Claim.

Let α_{∗} : I → Z be the constant map at z_{0}, defined by α_{∗}(t) = z_{0}. Let β_{∗} : I → X be the constant map at x_{0}, defined by β_{∗}(t) = x_{0}. Then α_{∗} ∈ P_{z_{0}}^{z_{0}}(Z), β_{∗} is a π-lift of f ∘ α_{∗} and β_{∗}(0) = x_{0}. By the claim, we conclude that f̂(z_{0}) = β_{∗}(1) = x_{0}.

It remains to prove that f̂: Z → X is continuous. Fix z ∈ Z, let x := f̂(z) and let X_{1} be an open neighborhood of x in X. We wish to show that there is an open neighborhood Z_{1} of z in Z such that f̂(Z_{1}) ⊆ X_{1}.

We leave it as an unassigned exercise to show that any covering map is a local homeomorphism. Then, replacing X_{1} by a smaller open neighborhood of x in X, we may assume that π : X_{1} → π(X_{1}) is a homeomorphism. By Exercise 7H, Y_{1} := π(X_{1}) is an open neighborhood of y in Y. Let π_{1} := π|X_{1} : X_{1} → Y_{1}. Then π_{1} : X_{1} → Y_{1} is a homeomorphism.

Since f : Z → Y is continuous, since Z is locally path-connected and since f(z) = π(f̂(z)) = π(x) ∈ π(X_{1}) = Y_{1}, let Z_{1} be a path-connected open neighborhood of z in Z such that f(Z_{1}) ⊆ Y_{1}. We wish to show that f̂(Z_{1}) ⊆ X_{1}. Fix z_{1} ∈ Z_{1}. We wish to show that f̂(z_{1}) ∈ X_{1}.

Choose α ∈ P_{z}^{z_{1}}(Z_{1}). Let β := π_{1}^{−1} ∘ f ∘ α : I → X_{1}. Then β ∈ P_{x}^{x_{1}}(X) is a π-lift of f ∘ α. Recall that β_{z} ∈ P_{x_{0}}^{x}(X) is a π-lift of f ∘ α_{z}. Then β_{z}∥β is a π-lift of f ∘ (α_{z}∥α) and (α_{z}∥α)(0) = α_{z}(0) = z_{0} and (α_{z}∥α)(1) = α(1) = z_{1}. so, by the claim, we have f̂(z_{1}) = (β_{z}∥β)(1) = β(1) ∈ β(I). Since β : I → X_{1}, we conclude that f̂(z_{1}) ∈ X_{1}, as desired. QED

Corollary. Let X be a simply connected, locally path-connected topological space. Let Y be a topological space. Let π : X → Y be a covering map. Then π is regular.

Proof: Let Γ := {deck transformations of π}. Fix y ∈ Y. By the preceding lemma, we wish to show that the Γ-action on π^{−1}(y) is transitive. Fix x, x′ ∈ π^{−1}(y). We wish to show that there exists γ ∈ Γ such that γx = x′.

Let Z := X, let z_{0} := x. By the preceding theorem, π has unique ({z_{0}}, Z) lifting. Let f := π : Z → X. Let x_{0} = x′. Then f(x_{0}) = f(x′) = π(x′) = y = π(x). Therefore there is a unique π-lift f̂: Z → X of f : Z → Y such that f̂(z_{0}) = x_{0}.

We have f̂ : X → X and f̂(x) = f̂(z_{0}) = x_{0} = x′. Similarly, let f̂^{∗} : X → X be the π-lift of f satisfying f̂^{∗}(x′) = x. Let I := f̂^{∗} ∘ f̂. Then I : X → X is the unique π-lift of f satisfying I(x) = x. Since the identity map X → X is another, we conclude from uniqueness that I : X → X is the identity, i.e., that f̂^{∗} ∘ f̂: X → X is the identity. Similarly f̂∘ f̂^{∗} : X → X is the identity. Then f̂: X → X is a homeomorphism.

We have π ∘ f̂ = f = π. Then f̂ is a deck transformation of π, i.e., f̂ ∈ Γ. Let γ := f̂. Then γx = f̂(x) = x′. QED

EXERCISE 7I: Let Y be a connected, locally path-connected topological space. Assume, for some simply connected topological space X, that there exists a covering map X → Y. Show that Y is semi-locally simply connected.

We are now ready to prove:

Theorem. Let Y be a connected, locally path-connected topological space. Then: Y has a universal cover iff Y is semi-locally simply connected.

Proof: “Only if” is given in Exercise 7I. Assume that Y is semi-locally simply connected. We wish to show, for some simply connected topological space X, that there is a covering map π : X → Y.

For any topological space T, for any t ∈ T, let P_{t}(T) := ⋃_{t′∈T} P_{t}^{t′}(T).

Fix y_{0} ∈ Y. Let X̂ := P_{y_{0}}(Y). For any α ∈ X̂, let [α] denote the endpoint fixed homotopy class of α. Let X := {[α] | α ∈ X̂}. Define π : X → Y by π([α]) = α(1). Note that, if α, α′ ∈ X̂ and if α is endpoint fixed homotopic to α′, then α(1) = α′(1), so π is a well-defined function. We now wish to show that there is a simply connected topology on X with respect to which π : X → Y is a covering map.

For any y ∈ Y, let V_{y} denote the collection of all open neighborhoods of y in Y that are relatively simply connected in Y. For any α ∈ X̂, for any V ∈ V_{α(1)}, let

N_{[α]}^{V} := {[α∥β] | β ∈ P_{α(1)}(V)} ⊆ X;

note that [α] ∈ N_{[α]}^{V}.

EXERCISE 7J: For any α ∈ X̂, for any V ∈ V_{α(1)}, show that π|N_{[α]}^{V} : N_{[α]}^{V} → Y is injective.

EXERCISE 7K: Let α ∈ X̂ and let V ∈ V_{α(1)}. Let β ∈ P_{α(1)}(V) and let W ∈ V_{β(1)}. Assume that W ⊆ V. Show that N_{[α∥β]}^{W} ⊆ N_{[α]}^{V}.

EXERCISE 7L: Show that {N_{[α]}^{V} | α ∈ X̂, V ∈ V_{α(1)}} is a basis of a topology on X. (Hint: Use Exercise 7K and Exercise 2H.)

We now give X the topology described in Exercise 7L.

Claim 1: π : X → Y is continuous. Proof of Claim 1: Let α ∈ X̂ and let W be a open neighborhood of π([α]) in Y. We wish to show that there is a open neighborhood N of [α] in X such that π(N) ⊆ W.

Since Y is semi-locally simply connected, choose V ∈ V_{α(1)} such that V ⊆ W. Let N := N_{[α]}^{V}. Then, by definition of N_{[α]}^{V} and of π, we have π(N) ⊆ V. Then π(N) ⊆ W. End of proof of Claim 1.

Claim 2: π : X → Y is open. Proof of Claim 2: Recall that {N_{[α]}^{V} | α ∈ X̂, V ∈ V_{α(1)}}

is a basis for the topology on X. For all α ∈ X̂, for all V ∈ V_{α(1)}, we have π(N_{[α]}^{V}) = V. Thus, the image of every basic open set is open, and it follows that the image of every open set is open. End of proof of Claim 2.

Claim 3: For all α ∈ X̂, for all V ∈ V_{α(1)}, the map π|N_{[α]}^{V} : N_{[α]}^{V} → V is a homeomorphism. Proof of Claim 3: By Exercise 7J, π|N_{[α]}^{V} is injective. By Claim 1, this map is continuous. By Claim 2, π|N_{[α]}^{V} : N_{[α]}^{V} → V is open. So, as π|N_{[α]}^{V} : N_{[α]}^{V} → V is surjective, we are done. End of proof of Claim 3.

EXERCISE 7M: Let y ∈ Y and let α, α′ ∈ P_{y_{0}}^{y}(Y) ⊆ X̂. Let V ∈ V_{y}. Assume that N_{[α]}^{V} ∩ N_{[α′]}^{V} ≠ ∅. Show that [α] = [α′].

For any topological space A, a pairwise disjoint open covering of A, is a subset U ⊆ {open subsets of A} such that ∪U = A and such that, for any U, U′ ∈ U, we have: if U ∩ U′ ≠ ∅, then U = U′. For any topological spaces A and B, for any continuous p : A → B, we have: p : A → B is a trivial covering map iff there is an pairwise disjoint open covering U of A such that, for all U ∈ U, p|U : U → B is a homemorphism. We leave the proof of this last statement as an unassigned exercise.

Then, by Claim 3 and Exercise 7M, we see, for any y ∈ Y, for any V ∈ V_{y}, that π|[π^{−1}(V)] : π^{−1}(V) → V is a trivial covering map; we use the pairwise disjoint open cover {N_{[α]}^{V} | α ∈ P_{y_{0}}^{y}} of π^{−1}(V). We conclude that π : X → Y is a covering map.

It remains to show that X is simply connected. Let c : I → Y be the constant map defined by c(t) = y_{0} Let x_{0} := [c]. We wish to show that π_{1}(X, x_{0}) is trivial. Let ĉ: I → X be the constant map defined by ĉ(t) = x_{0}. Note that π ∘ ĉ = c. We wish to show that any element of P_{x_{0}}^{x_{0}}(X) is endpoint fixed homotopic to ĉ.

Claim 4: For all x ∈ X, for all σ ∈ P_{x_{0}}^{x}(X), we have [π∘σ] = x. Proof of Claim 4: Let α := π ∘ σ. Let I := [0, 1]. For all s ∈ I, let α_{s} : I → Y be defined by α_{s}(t) = α(st). Then α_{0} = c and α_{1} = α. Define α̂ : I → X by α̂(s) = [α_{s}]. We leave it as an unassigned exercise to show that α̂ is continuous. For all s ∈ I, we have π(α̂(s)) = π([α_{s}]) = α_{s}(1) = α(s), so π ∘ α̂ = α. So, since α = π ∘ σ, we conclude that both α̂ and σ are π-lifts of α. Moreover, we have α̂(0) = [α_{0}] = [c] = x_{0} = σ(0). By the proposition preceding Exercise 4F, we see that π has unique ({0}, I)-lifting. It follows that α̂ = σ. Then [π ∘ σ] = [α] = [α_{1}] = α̂(1) = σ(1) = x. End of proof of Claim 4.

Let σ ∈ P_{x_{0}}^{x_{0}}(X). We wish to show that σ is endpoint fixed homotopic to ĉ.

By Claim 4, we see that [π ∘ σ] = x_{0}. Then [π ∘ σ] = [c] = [π ∘ ĉ]. That is, π ∘ σ is endpoint fixed homotopic to π ∘ ĉ. We also have σ(0) = x_{0} = ĉ(0). It then follows, from the theorem following Exercise 4F, that σ is endpoint fixed homotopic to ĉ. QED

We now return to the question we asked before: Are all covering maps are regular? Our approach will be to develop, for any connected, semi-locally simply connected Y, a “classification” of covering maps to Y, and to calculate for each one, its group of deck transformations, and to analyze, for each one, whether it is regular. In the process, we will discover many non-regular covering maps.

First, though, we need some basic results. The following is our main lifting theorem:

Theorem. Let X, Y and Z be topological spaces. Assume that Z is connected and locally path-connected. Let π : X → Y be a covering map. Let f : Z → Y be continuous. Let x_{0} ∈ X and z_{0} ∈ Z. Assume that π(x_{0}) = f(z_{0}). Then the following are equivalent:

(1) there is a unique π-lift f̂ of f such that f̂(z_{0}) = x_{0}; and

(2) f_{∗}(π_{1}(Z, z_{0})) ⊆ π_{∗}(π_{1}(X, X_{0})).

Proof: By Exercise 8A below, (1) implies (2).

EXERCISE 8A: Show: If there is a π-lift f̂ of f such that f̂(z_{0}) = x_{0}, then (2) is true.

Now assume that (2) is true. We wish to prove (1).

EXERCISE 8B: Let γ ∈ P_{z_{0}}^{z_{0}}(Z). Let γ̂ be the π-lift of f ∘ γ such that γ̂(0) = x_{0}. Show that γ̂(1) = x_{0}.

EXERCISE 8C: Let z ∈ Z. Let α, β ∈ P_{z_{0}}^{z}(Z). Let α̂ and β̂ be the π-lifts of f ∘ α, f ∘ β, respectively, such that α̂(0) = x_{0} = β̂(0). Show that α̂(1) = β̂(1).

By the axiom of choice, for all z ∈ Z, let α_{z} ∈ P_{z_{0}}^{z}(Z); then α_{z}(0) = z_{0} and α_{z}(1) = z. For all z ∈ Z, let α̂_{z} be the unique π-lift of f ∘α_{z} such that α̂_{z}(0) = x_{0}. Then π∘α̂_{z} = f ∘α_{z}. Define f̂: Z → X by f̂(z) = α̂_{z}(1).

Let I := [0, 1]. Let c : I → Z be the constant loop at z_{0}, defined by c(t) = z_{0}. Let ĉ: I → X be the constant loop at x_{0}, defined by ĉ(t) = x_{0}. Then ĉ is the unique π-lift of f ∘ c such that ĉ(0) = x_{0}. By Exercise 8C, we have α̂_{z_{0}}(1) = ĉ(1). Then f̂(z_{0}) = α̂_{z_{0}}(1) = ĉ(1) = x_{0}. For all z ∈ Z, we have (π ∘ f̂)(z) = π(f̂(z)) = π(α̂_{z}(1)) = (π ∘ α̂_{z})(1) = (f ∘ α_{z})(1) = f(α_{z}(1)) = f(z).

EXERCISE 8D: Show that f̂ is continuous. That is show the following: Let z ∈ Z. Let V be an open neighborhood of f̂(z) in X. Show that there is an open neighborhood U of z in Z such that f̂(U) ⊆ V.

It remains to prove uniqueness. Let ĝ be a π-lift of f such that ĝ(z_{0}) = x_{0}. We wish to show that ĝ = f̂. Fix z ∈ Z. We wish to show that ĝ(z) = f̂(z).

Because π ∘ ĝ = f, we see that π ∘ (ĝ ∘ α_{z}) = f ∘ α_{z}. Moreover, (ĝ ∘ α_{z})(0) = ĝ(α_{z}(0)) = ĝ(z_{0}) = x_{0}. Then, by definition of α̂_{z}, we see that ĝ ∘ α_{z} = α̂_{z}. Then ĝ(z) = ĝ(α_{z}(1)) = (ĝ ∘ α_{z})(1) = α̂_{z}(1) = f̂(z), as desired. QED

EXERCISE 8E: Let Y be a path-connected, locally path-connected topological space and let (W, p) be a universal cover of Y.

(1) Show that W is locally path-connected.

(2) Let (W′, p′) be another universal cover of Y. Show that (W, p) is isomorphic to (W′, p′) in the category of topological spaces over Y. That is, show that there is a homeomorphism h : W → W′ such that p′ ∘ h = p.

Remark. Let Γ be a group and let F be a free, transitive Γ-set. Let f ∈ F. Then the orbit map γ ↦ γf : Γ → F based at f is a bijection. Consequently, there is a uniqe group structure on F such that γ ↦ γf : Γ → F is a group isomorphism. With this group structure on F, note that f becomes the identity element.

Corollary. Let X and Y be topological spaces. Let π : X → Y be a regular covering map. Let Γ := {deck transformations of π}. Let y_{0} ∈ Y and let F := π^{−1}(y_{0}). Then, for any x_{0} ∈ F, there is a unique group structure on F such that γ ↦ γx_{0} : Γ → F is an isomorphism.

Proof: Since Γ acts freely on X, it follows that Γ acts freely on F. Since π is regular, it follows that Γ acts transitively on F. The result then follows from the preceding remark. QED

The preceding corollary is summarized by the statement: “In a regular covering, any fiber with a chosen point has a unique group structure making the orbit map at that point is an isomorphism with the group of deck transformations.” Note that, under that group structure, the chosen point becomes the identity of the group. Thus we may think of the fibers of a regular covering map as groups isomorphic to the group of deck transformations. However, once again, obtaining the group structure requires first picking a particular point in the fiber that you’ll make the identity.

Recall that we have proved

Fact. Let Y be a path-connected topological space and let (X, π) be a universal cover of Y. Let x_{0} ∈ X and let y_{0} := π(x_{0}). For all α ∈ P_{y_{0}}^{y_{0}}(Y), let α̂ be the unique π-lift of α satisfying α̂(0) = x_{0}. Then the map [α] ↦ α̂(1) : π_{1}(Y, y_{0}) → π^{−1}(y_{0}) is a well-defined bijection.

Let Y be a connected, locally path-connected topological space. Let (X, π) be a universal cover of Y. Let x_{0} ∈ X and let y_{0} := π(x_{0}). Let Γ := {deck transformations of π}. Define ϕ : π_{1}(Y, y_{0}) → π^{−1}(y_{0}) as in the preceding fact. Let ψ : Γ → π^{−1}(y_{0}) be the orbit map defined by ψ(γ) = γx_{0}. Then, by the preceding corollary and fact, we obtain a bijection ψ^{−1} ∘ ϕ : π_{1}(Y, y_{0}) → Γ, and it is an exercise to see that this map is a group isomorphism. Thus we see that π_{1}(Y, y_{0}) and Γ are isomorphic (as we have noted before), but the point we’re making now is that this isomorphism is not “natural” until we have picked a point in the fiber over y_{0}.

To make this more precise, let C be the category whose objects are (1) a connected, locally path connected topological space Y; together with

(2) a universal cover (X, π) of Y; together with

(3) a point x_{0} ∈ X. (What are the arrows in C?) Let F, G : C → {groups} be functors defined by

F(Y, X, π, x_{0}) = π_{1}(Y, π(x_{0}))

and G(Y, X, π, x_{0}) = {deck transformations of π}. Then the paragraph following the preceding fact asserts that F and G are equivalent functors.

We now return to classifying covering map and determining which covering maps are regular.

Setup for covering maps: Let Y be a connected, locally path-connected topological space. Let (X, p) be a universal cover of Y. Let Γ := {deck transformations of p}. For all Δ ≤ Γ, let X_{Δ} := Δ\X and let c_{Δ} : X → X_{Δ} be the canonical map and let p_{Δ} : X_{Δ} → Y be the unique function such that p = p_{Δ} ∘ c_{Δ}.

It is an unassigned exercise to show that such a function p_{Δ} exists and is unqiue. It is a further unassigned exercise to show that it is a covering map.

Theorem. Assume the “Setup for covering maps”. Let Z be a connected topological space. Let q : Z → Y be a covering map. Then there exists Δ ≤ Γ such that (Z, q) is isomorphic to (X_{Δ}, p_{Δ}) in the category of topological spaces over Y.

We defer the proof momentarily.

The preceding theorem raises the question of whether (X_{Δ}, p_{Δ}) and (X_{Λ}, p_{Λ}) might be isomorpic in the category of topological spacdes over Y for different Δ, Λ ≤ Γ.

Remark. Let Δ, Λ ≤ Γ. Then: (X_{Δ}, p_{Δ}) is isomorphic to (X_{Λ}, p_{Λ}) in the category of topological spaces over Y iff there exists γ ∈ Γ such that γΔγ^{−1} = Λ.

We defer the proof momentarily.

Proof of the theorem:

EXERCISE 9A: Show that there is a covering map r : X → Z such that q ∘ r = p. (Hint: First argue that Z is locally path-connected and semi-locally simply connected. Let (Z̃, r_{0}) be a universal cover of Z. Fix y_{0} ∈ Y. Choose x_{0} ∈ p^{−1}(y_{0}), z_{0} ∈ q^{−1}(y_{0}) and z̃_{0} ∈ r_{0}^{−1}(z_{0}). Using the lifthing theorem which precedes Exercise 8A to construct a continuous map a : Z̃ → X such that p ∘ a = q ∘ r_{0} and such that a(z̃_{0}) = x_{0}. Using the lifting theorem again, construct a continuous map r : X → Z such that q ∘ r = p and such that r(x_{0}) = z_{0}. Using the lifting theorem again, construct a continuous map b : X → Z̃ such that r_{0} ∘ b = r and such that r(x_{0}) = z̃_{0}. We have p ∘ a ∘ b = q ∘ r_{0} ∘ b = q ∘ r = p. Then, by uniqueness in the lifting theorem, argue that a ∘ b : X → X is the identity. We have q ∘ r ∘ a = p ∘ a = q ∘ r_{0}. Then, by uniqueness in the lifting theorem, argue that r ∘ a = r_{0}. We have r_{0} ∘ b ∘ a = r ∘ a = r_{0}. Then, by uniqueness in the lifting theorem, argue that b ∘ a : Z̃ → Z̃ is the identity. Since a ∘ b : X → X and b ∘ a : Z̃ → Z̃ are both identity maps, we conclude that a : Z̃ → X and b : X → Z̃ are both homeomorphisms. Then, as r_{0} is a covering map, so is r_{0} ∘ b. However r_{0} ∘ b = r.)

Fix such a map r. Let Δ := {deck transformations of r}.

EXERCISE 9B: Show that Δ ≤ Γ.

As X is simply connected and Z is locally path-connected, it follows that r is regular. Then (Z, r) is isomorphic to (X_{Δ}, c_{Δ}) in the category of topological spaces under X. That is, there is a homeomorphism h : Z → X_{Δ} such that h ∘ r = c_{Δ}. We wish to show that (Z, q) is isomorphic to (X_{Δ}, p_{Δ}) in the category of topological spacaces over Y, so it suffices to show that p_{Δ} ∘ h = q. Fix z ∈ Z. We wish to show that p_{Δ}(h(z)) = q(z).

Choose x ∈ X such that r(x) = z. Then h(z) = h(r(x)) = c_{Δ}(x). So, since p_{Δ} ∘ c_{Δ} = p = q ∘ r, we have p_{Δ}(h(z)) = p_{Δ}(c_{Δ}(x)) = q(r(x)) = q(z). QED

Proof of the remark:

EXERCISE 9C: Show the easy “if” direction of the remark. That is, show that if there is some γ ∈ Γ, such that γΔγ^{−1} = Λ, then there is a homeomorphism h : X_{Δ} → X_{Λ} such that p_{Λ} ∘ h = p_{Δ}.

We now prove “only if”. Let h : X_{Δ} → X_{Λ} be a homomorphism such that p_{Λ}∘h = p_{Δ}. We wish to show, for some γ ∈ Γ, that γΔγ^{−1} = Λ.

By the theorem preceding Exercise 8A, let ĥ : X → X be a c_{Λ}-lift of h∘c_{Δ} : X → X_{Λ}. Then c_{Λ} ∘ ĥ = h ∘ c_{Δ}. Then p_{Λ} ∘ (c_{Λ} ∘ ĥ) = p_{Λ} ∘ (h ∘ c_{Δ}). Then (p_{Λ} ∘ c_{Λ}) ∘ ĥ = (p_{Λ} ∘ h) ∘ c_{Δ}. Then p ∘ ĥ = p_{Δ} ∘ c_{Δ} = p. Then ĥ ∈ Γ. Let γ := ĥ. It remains to show that γ^{−1}Δγ = Λ.

EXERCISE 9D: Show that Δ is the group of deck transformations of c_{Δ}.

EXERCISE 9E: Show, for all δ ∈ Δ, that γδγ^{−1} ∈ Λ.

According to Exercise 9C, γΔγ^{−1} ⊆ Λ, so it remains only to show that Λ ⊆ γΔγ^{−1}. A similar argument to that of Exercise 9C shows that γ^{−1}Λγ ⊆ Δ. Then Λ =

γγ^{−1}Λγγ^{−1} ⊆ γΔγ^{−1}. QED

Recall that, for any group Γ, for any subgroup Δ of Γ, the normalizer in Γ of Δ is N_{Γ}(Δ) := {γ ∈ Γ | γΔγ^{−1} = Δ}; it is the largest subgroup of Γ in which Δ is contained as a normal subgroup. In particular, N_{Γ}(Δ) = Γ iff Δ is a normal subgroup of Γ.

Recall that the kernel of an action of a group G on a set X is {g ∈ G | ∀x ∈ X, gx = x}; it is the intersection of all the stabilizers of elements of X. Denoting this kernel by K, we have that K is normal in G and that, if c : G → G/K denotes the canonical homomorphism, then there is a unique action of G/K on X such that, for all g ∈ G, for all x ∈ X, we have gx = (c(g))x. That is, any action induces an action of the group modulo this kernel. Moreover, if X is a topological space and if the G-action on X is continuous, then the resulting action of G/K on X is continuous as well.

We say that an action is faithful or effective if its kernel is trivial. Note that the induced action of G/K on X is faithful. That is, “once you mod out by the kernel of the action, the resulting action has no kernel”.

Fact. Assume the “Setup for covering maps”. Then, for any subgroup Δ of Γ, the action of N_{Γ}(Δ) on X factors to an action of N_{Γ}(Δ) on X_{Δ}. (That is, there is a unique action of N_{Γ}(Δ) on X_{Δ} such that the map p_{Δ} becomes (N_{Γ}(Δ))-equivariant.) The kernel of this action of N_{Γ}(Δ) on X_{Δ} is exactly Δ. Thus (N_{Γ}(Δ))/Δ acts continuously and faithfully on X_{Δ}. The resulting collection of maps

{x ↦ τx : X_{Δ} → X_{Δ} | τ ∈ (N_{Γ}(Δ))/Δ}

is the group of deck transformations of p_{Δ}. In particular, the group of deck transformations of p_{Δ} is isomorphic to N_{Γ}(Δ))/Δ.

We leave the proof of this fact as an exercise for the interested reader. Here’s another unassigned exercise:

Unassigned exercise. Assume the “Setup for covering maps”. Let Δ be a subgroup of Γ. Then, for all y ∈ Y, we have |(p_{Δ})^{−1}(y)| = |Γ/Δ|.

Now define S_{0} := {1, 2, 3} and let Γ_{0} be the set of all bijections S_{0} → S_{0}. That is, Γ_{0} is the set of all permutations of {1, 2, 3}, i.e., it is the symmetric group on three letters. Let Δ_{0} := {γ ∈ Γ_{0} | γ(1) = 1}. We leave it as an unassigned exercise to show that N_{Γ_{0}}(Δ_{0}) = Δ_{0}. Note that |Γ_{0}| = 3! and |Δ_{0}| = 2!, so |Γ_{0}/Δ_{0}| = (3!)/(2!) = 3.

In the theorem below, combined with Exercise 9F below, we will show that there is a locally path-connected, contractible, principal topological Γ_{0}-space X := E_{Γ_{0}}. Let Y := Γ_{0}\X. Then the group Γ of deck transformations of the canonical map p : X → Y is, by covering space theory, isomorphic to Γ_{0}, and therefore contains a subgroup Δ such that N_{Δ}(Γ) = Δ and such that |Γ/Δ| = 3. Note that (N_{Δ}(Γ))/Δ is the trivial group.

Define X_{Δ} := Δ\X and define p_{Δ} : X_{Δ} → Y as in the “Setup for covering maps”. Recall that the fibers of p_{Δ} all have cardinalty |Γ/Δ|. So every fiber of p_{Δ} has three elements. By the fact stated above, the group of deck transformations of p_{Δ} is isomorphic to (N_{Δ}(Γ))/Δ, which is the trivial group. Thus the group of deck transformations of p_{Δ}, being trivial, cannot act transitively on the fibers of p_{Δ}, as they all have three elements. Thus, p_{Δ} is not a regular covering, which answers a question posed long ago.

This entire discussion was based on:

Theorem. Let I := [0, 1]. Let Γ be a discrete group. Then there is a locally path-connected, principal topological Γ-space E_{Γ} and a continuous map h_{Γ} : E_{Γ} × E_{Γ} × I → E_{Γ} such that:

(1) for all γ ∈ Γ, for all x, y ∈ E_{Γ}, we have h_{Γ}(γx, γy, t) = γ(h_{Γ}(x, y, t)); and

(2) for all x, y ∈ E_{Γ}, we have h_{Γ}(x, y, 0) = x and h_{Γ}(x, y, 1) = y.

EXERCISE 9F: Show that E_{Γ} is contractible.

We define B_{Γ} := Γ\E_{Γ} and note that, because of the covering space theory that we have developed, π_{1}(B_{Γ}) is isomorphic to Γ. Thus, a consequence of the preceding theorem and exercise is the fact that every discrete group is the fundamental group of some locally path-connected, semi-locally simply connected topological space.

Proof of the preceding theorem: Let ℕ := {0, 1, 2, …}. For any c ∈ I^{ℕ}, let

supp(c) := {i ∈ ℕ | c_{i} ≠ 0}.

Let Ê_{Γ} := {(c, γ) ∈ I^{ℕ} × Γ^{ℕ} | |supp(c)| < ∞, ∑ c_{i} = 1}. We define an equivalence relation ∼ on Ê_{Γ} by: (c, γ) ∼ (c′, γ′) iff, for all i ∈ ℕ, we have

(1) c_{i} = c′_{i}; and

(2) if c_{i} ≠ 0 ≠ c′_{i}, then γ_{i} = γ′_{i}.

Let E_{Γ} := Ê_{Γ}/ ∼. Let q : Ê_{γ} → E_{Γ} be the canonical map. For each (c, γ) ∈ Ê_{Γ}, we denote q(c, γ) by c_{0}γ_{0} + c_{1}γ_{1} + · · ·. (Warning: This + is not commutative. Moreover, for any i ∈ ℕ, if c_{i} = 0, then we may and will replace the term c_{i}γ_{i} by 0, since γ_{i} is irrelevant; however, we cannot omit this term 0.)

For all j ∈ ℕ, define t_{j} : E_{j} → I by t_{j}(c_{0}γ_{0} + c_{1}γ_{1} + ⋯) = c_{j}, let E_{Γ}^{j} := t_{j}^{−1}((0, 1]) and define x_{j} : E_{Γ}^{j} → Γ by x_{j}(c_{0}γ_{0} + c_{1}γ_{1} + ⋯) = γ_{j}. Give E_{Γ} the coarsest topology such that all these maps t_{0}, t_{1}, …, x_{0}, x_{1}, … are continuous. That is, give E_{Γ} the topology generated by

{t_{j}^{−1}(U) | j ∈ ℕ, U open in I}  ∪  {x_{j}^{−1}(V) | j ∈ ℕ, V ⊆ Γ}.

We leave it as an unassigned exercise to show that E_{Γ} is locally path connected.

For any x = ∑ c_{i}γ_{i} ∈ E_{Γ} and any x′ = ∑ c′_{i}γ′_{i} ∈ E_{Γ}, if,

for all i ∈ (supp(c)) ∩ (supp(c′)), we have γ_{i} = γ′_{i},

then we define the straight path from x to x′ to be the path τ_{x}^{x′} ∈ P_{x}^{x′}(E_{Γ}) defined as follows: Define γ′′ ∈ Γ^{ℕ} by

(1) for all i ∈ supp(c), γ′′_{i} := γ_{i};

(2) for all i ∈ supp(c′), γ′′_{i} := γ′_{i}; and
(3) for all i ∈ ℕ\[(supp(c)) ∪ (supp(c′))], γ′′_{i} := 1_{Γ}.

Finally, define τ_{x}^{x′} : I → E_{Γ} by τ_{x}^{x′}(t) = ∑[(1 − t)c_{i} + tc′_{i}]γ′′_{i}.

Let

E′_{Γ} := {c_{0}γ_{0} + 0 + c_{1}γ_{1} + 0 + c_{2}γ_{2} + 0 + ⋯ | c_{0}, c_{1}, c_{2}, … ∈ I, γ_{0}, γ_{1}, γ_{2}, … ∈ Γ}.

For all x = c_{0}γ_{0} + 0 + c_{1}γ_{1} + 0 + c_{2}γ_{2} + 0 + ⋯ ∈ E′_{Γ}, let x_{0} := x; let

x_{1} := c_{0}γ_{0} + c_{1}γ_{1} + 0 + c_{2}γ_{2} + 0 + c_{3}γ_{3} + ⋯

be obtained from x by dropping the second term (and no others); let

x_{2} := c_{0}γ_{0} + c_{1}γ_{1} + c_{2}γ_{2} + 0 + c_{3}γ_{3} + 0 + ⋯

be obtained from x by dropping the second and fourth terms (and no others); let

x_{3} := c_{0}γ_{0} + c_{1}γ_{1} + c_{2}γ_{2} + c_{3}γ_{3} + 0 + c_{4}γ_{4} + ⋯

be obtained from x by dropping the second and fourth and sixth terms (and no others); etc. For all x = c_{0}γ_{0} + 0 + c_{1}γ_{1} + 0 + c_{2}γ_{2} + 0 + ⋯ ∈ E′_{Γ}, let

x_{∞} := c_{0}γ_{0} + c_{1}γ_{1} + c_{2}γ_{2} + c_{3}γ_{3} + ⋯

be obtained from x by dropping all even terms (and no odd terms). For all x ∈ E′_{Γ}, for all n ∈ ℕ, let α_{x}^{n} := τ_{x_{n}}^{x_{n+1}} be the straight path from x_{n} to x_{n+1}.

For all n ∈ ℕ, let s_{n} := 1 − 2^{−n}. For all n ∈ ℕ, let I_{n} := [s_{n}, s_{n+1}]. For all n ∈ ℕ, let l_{n} : I_{n} → I be the linear map such that l_{n}(s_{n}) = 0 and l_{n}(s_{n+1}) = 1; it is defined by l_{n}(t) = (1 − t)s_{n} + ts_{n+1}.

For all x ∈ E′_{Γ}, define α_{x} : I → E_{Γ} by the rules:

(1) for all n ∈ ℕ, for all t ∈ I_{n}, α_{x}(t) = α_{x}^{n}(l_{n}(t)); and
(2) α_{x}(1) = x_{∞};

we leave it as an unassigned exercise to verify that α_{x} is continuous. Then, for all x ∈ E′_{Γ}, we have α_{x} ∈ P_{x}^{x_{∞}}(E_{Γ}).

For all x = c_{0}γ_{0} + c_{1}γ_{1} + ⋯ ∈ E_{Γ}, define

x^{∗} := c_{0}γ_{0} + 0 + c_{1}γ_{1} + 0 + c_{2}γ_{2} + 0 + ⋯ ∈ E′_{Γ};

note that (x^{∗})_{∞} = x. For all x ∈ E_{Γ}, let β_{x} ∈ P_{x}^{x^{∗}}(E_{Γ}) be defined by β_{x}(t) = α_{x^{∗}}(1 − t). For all x = c_{0}γ_{0} + c_{1}γ_{1} + ⋯ ∈ E_{Γ}, define

x^{#} := 0 + c_{0}γ_{0} + 0 + c_{1}γ_{1} + 0 + c_{2}γ_{2} + 0 + ⋯ ∈ E_{Γ}.

Note, for all x, y ∈ E_{Γ}, that (supp(x^{∗})) ∩ (supp(y^{#})) = ∅.

For all integers k ∈ [1, 4], let J_{k} := [(k − 1)/4, k/4] and let r_{k} : J_{k} → I be the linear map such that r_{k}((k − 1)/4) = 0 and r_{k}(k/4) = 1; then r_{k} : J_{k} → I is defined by r_{k}(t) = (1 − t)((k − 1)/4) + t(k/4). For all x, y ∈ E_{Γ} we define a path λ_{x}^{y} ∈ P_{x}^{y}(E_{Γ}) by

(1) for t ∈ J_{1}, λ_{x}^{y}(t) = β_{x}(r_{1}(t)); then λ_{x}^{y}(1/4) = x^{∗};

(2) for t ∈ J_{2}, λ_{x}^{y}(t) = τ_{x^{∗}}^{y^{#}}(r_{2}(t)); then λ_{x}^{y}(1/2) = y^{#};
(3) for t ∈ J_{3}, λ_{x}^{y}(t) = τ_{y^{#}}^{y^{∗}}(r_{3}(t)); then λ_{x}^{y}(3/4) = y^{∗}; and
(4) for t ∈ J_{4}, λ_{x}^{y}(t) = α_{y^{∗}}(r_{4}(t)); then λ_{x}^{y}(1) = (y^{∗})_{∞} = y.

We now define h_{Γ} : E_{Γ} × E_{Γ} × I → E_{Γ} by h(x, y, t) = λ_{x}^{y}(t). Note, for all x, y ∈ E_{Γ}, that h(x, y, 0) = λ_{x}^{y}(0) = x and h(x, y, 1) = λ_{x}^{y}(1) = y. We leave it as an unassigned exercise to show that it has the required properties of continuity and equivariance, as stated in the theorem. QED

For any arrow f : C → C′ in a category C, we will define dom(f) = C and tar(f) = C′. A diagram in C is a set D consisting of objects and arrows in C such that: for any arrow f ∈ D, we have dom(f) ∈ D and tar(f) ∈ D. We denote, by Obj(D), the set of all objects in D. We denote, by Arr(D), the set of all arrows in D

Given a diagram D in a category C, given D, D′ ∈ Obj(D), a path in D from D to D′ in D is a finite sequence p = (f_{1}, …, f_{m}) of arrows in D such that

(1) dom(f_{1}) = D, tar(f_{m}) = D′; and

(2) for any integer i ∈ [2, m], we have: dom(f_{i}) = tar(f_{i−1}).

In this case, we define ∘p := f_{m} ∘ ⋯ ∘ f_{1} : D → D′.

Let D be a diagram in a category C. Then we say D is commutative if, for all D, D′ ∈ Obj(D), for any two paths p, q from D to D′ in D, we have: ∘p = ∘q.

Definition. Let D be a commutative diagram in a category C. A “receiver” of D in C consists of (1) an object R in C; and

(2) a set of arrows A in C such that

(A) for all f ∈ A, we have dom(f) ∈ Obj(D) and tar(f) = R; (B) for all D ∈ Obj(D), there is a unique f ∈ A such that dom(f) = D; and (C) D ∪ {R} ∪ A is a commutative diagram in C.

Definition. Let D be a commutative diagram in a category C and let (R, A) be a receiver of D in C. Then (R, A) is a direct limit of D in C means that: for any receiver (S, B) of D in C, there is a unique arrow h : R → S such that B = {h ∘ f | f ∈ A}.

EXERCISE 10A: For all integers n ≥ 1, let ι_{n} : S^{n} → S^{n+1} be the function defined by ι_{n}(x_{0}, …, x_{n}) = (x_{0}, …, x_{n}, 0). Let D := {S^{1}, S^{2}, …} ∪ {ι_{1}, ι_{2}, …}. Construct a direct limit of D in the category {sets}. (Show that it is a direct limit.)

Definition. We say that a category C has direct limits if, for any commutative diagram D in C, there exists a direct limit in C of D.

Fact. The following categories have direct limits: {sets}, {groups}, T S = {topological spaces}, {pointed topological spaces}. Somehow, “most” basic categories do seem to have direct limits.

To give some idea about this, let D be a diagram in the category {sets}, and we will describe how to construct a direct limit of D. To simplify matters assume, for all A, B ∈ Obj(D), that either A ∩ B = ∅ or A = B; otherwise the construction is made slightly more difficult. Let R_{0} := ∪[Obj(D)]. For all A, B ∈ Obj(D), let S_{AB} be the set of all (a, b) ∈ A × B such that, for some path p in D from A to B, we have (∘p)(a) = b. Let ∼ be the smallest equivalence relation on R_{0} containing ∪{S_{AB} | A, B ∈ Obj(D)}. Then R_{0}/ ∼ is the direct limit:

EXERCISE 10B: Show that R := R_{0}/ ∼ is a direct limit of D in {sets}.

In the other categories mentioned in the preceding fact, the construction of direct limits is at a similar level of difficulty.

Be aware that there is some ambiguity in the meaning of the term “direct limit”:

Definition. Let D be a commutative diagram in a category C. An object R in C is said to be a direct limit of D in C if: there exists a set of arrows A in C such that (R, A) is a direct limit of D in C.

EXERCISE 10C: Let D be a commutative diagram in a category C. Let R and R′ be direct limits of D in C. Show that R and R′ are isomorphic in C.

As a result of Exercise 10C, one sometimes talks about the direct limit of a diagram, instead of a direct limit.

Given any set S, let ⟨S⟩ denote the free group on S. Let S_{0} be a two element set and let a and b be the distinct elements of S_{0}. Define A := ⟨{a}⟩, B := ⟨{b}⟩ and G := ⟨{a, b}⟩. Let f : {1} → A and g : {1} → B be the trivial maps. Let p : A → G and q : B → G be the inclusion maps. We leave it as an unassigned exercise to show that (G, {p, q}) is a direct limit of the diagram {{1}, A, B, f, g} in {groups}.

Incidentally, to save some writing, one often simply writes ⟨a, b⟩ instead of the more technically correct ⟨{a, b}⟩.

The type of direct limit described in the preceding paragraph is sometimes called a “pushout”:

Definition. Let D be a diagram in a category C. Suppose that (1) there are two arrows and three objects in D;

(2) for all f, g ∈ Arr(D), we have dom(f) = dom(g); and

(3) for all f, g ∈ Arr(D), if f ≠ g, then tar(f) ≠ tar(g). In this case, a direct limit of D in C is sometimes called a pushout of D in C.

Let I be a set. For all i ∈ I, let S_{i} be a set, and let R_{i} ⊆ ⟨S_{i}⟩ be a subset of the free group generated by S_{i}. For all i ∈ I, let G_{i} := ⟨S_{i} | R_{i}⟩ be the group with generators S_{i} and relations R_{i}. (That is, G_{i} is the quotient of ⟨S_{i}⟩ by the smallest normal subgroup of ⟨S_{i}⟩ which contains R_{i}. This “smallest normal subgroup” is exactly the subgroup generated by

⋃_{g∈⟨S_{i}⟩} gR_{i}g^{−1}.)

For a given i ∈ I, the pair (S_{i}, R_{i}) is sometimes called a “presentation” of G_{i}; it completely determines G_{i}, up to isomorphism.

Let K be a set. For all k ∈ K, let i_{k}, j_{k} ∈ I and let f_{k} : G_{i_{k}} → G_{j_{k}} be a group homomorphism. Let D := {S_{i} | i ∈ I} ∪ {f_{k} | k ∈ K}. Then D is a diagram in the category {groups} and we set ourselves to the task of computing the direct limit of D.

To simplify matters, let’s assume, for all i, j ∈ I, that: if i ≠ j, then S_{i} ∩ S_{j} = ∅. For all i ∈ I, let c_{i} : ⟨S_{i}⟩ → G_{i} be the canonical map. For all k ∈ K, for all s ∈ S_{i_{k}}, choose w_{s,k} ∈ S_{j_{k}} such that f_{k}(c_{i_{k}}(s)) = c_{j_{k}}(w_{s,k}).

For a given k ∈ K, the mapping s ↦ w_{s,k} : S_{i_{k}} → S_{j_{k}} is sometimes called a “presentation” of f_{k}; it completely determines f_{k}.

Let S := ⋃_{i∈I} S_{i} and let R := ⋃_{i∈I} R_{i}. For all k ∈ K, let

R′_{k} := {s^{−1}w_{s,k} |s ∈ S_{i_{k}}} ⊆ ⟨S_{i_{k}} ∪ S_{j_{k}}⟩ ⊆ ⟨S⟩.

Let R′ := ⋃_{k∈K} R′_{k}. It is an unassigned exercise to demonstrate that ⟨S | R ∪ R′⟩ is a direct

limit of D.

The point of the preceding is simply that, if a diagram in the category of groups is presented completely in terms of generators and relations, then it is straightforward to write down a presentation of its direct limit. Let’s refer to this as the “group direct limit algorithm”.

If D is a set of groups (i.e., a diagram in {groups} with no arrows), then the direct limit of D is often denoted ∗D and is called the free product of D. When D is finite, one often writes down the elements of D with “∗”s separating them. Thus, for example, the free product of {A, B} is denoted A ∗ B, and is often referred to as “the free product of A and B”.

A notational comment: Consider the group

⟨{a, b, c, p, q, r, s, y, z} | {a^{−1}bc, pqrs, yzy^{−1}z^{−1}}⟩.

It would be typical to write this in a different way: ⟨a, b, c, x, y | a = bc, pqrs = 1, yz = zy}. On the left and right of the vertical bar (|), one removes the braces. On the right, for each relator, one sets the relator to 1 and then tries to manipulate the resulting equation to be more readable. For example, yzy^{−1}z^{−1} = 1 can be manipulated to yz = zy.

Let A, B and C be groups. Let f : C → A and g : C → B be group homomorphisms. Then the pushout of {A, B, C, f, g} in {groups} is often denoted A ∗_{C} B. Note that f and g are suppressed from the notation, although they really should not be. One might write A ∗_{C}^{f,g} B if there were any confusion. This group is called the almagamated product of A and B, with amalgamation over C. It can be described exactly in an algebraic way: It is the quotient of the free product A ∗ B modulo the smallest normal subroup of A ∗ B containing {(f(c))(g(c))^{−1} | c ∈ C}.

Consider the special case where C = ⟨c | c^{2} = 1⟩, where A = ⟨a | a^{4} = 1⟩ and where B = ⟨b | b^{6} = 1⟩. Let f : C → A be the homomorphism which sends (the image of) c to (the image of) a^{2}. Let g : C → B be the homomorphism which sends (the image of) c to (the image of) b^{3}. Then, according to group direct limit algorithm, one can immediately write down a presentation of A ∗_{C} B, namely:

⟨a, b, c | c^{2} = 1, a^{4} = 1, b^{6} = 1, c = a^{2}, c = b^{3}⟩.

The generator c is not really needed; we leave it to the interested reader to verify that A ∗_{C} B is isomorphic to

⟨a, b | a^{4} = 1, b^{6} = 1, a^{2} = b^{3}⟩.

Definition. Let P be the category of pointed topological spaces. Let (X, x), (Y, y) ∈ P. Let f, g : (X, x) → (Y, y) be arrows in P. Let f_{0}, g_{0} : X → Y be the continuous maps underlying f, g, respectively. (Then f_{0}(x) = g_{0}(x) = y.) Let I := [0, 1]. A pointed homotopy from f to g is a continuous map h : I × X → Y such that

(1) h(0, ·) = f_{0}, h(1, ·) = g_{0}; and

(2) for all s ∈ I, h(s, x) = y.

Let X be a topological space. An open cover U of X is said to be locally finite if, for any x ∈ X, there is an open neighborhood V of x in X such that |{U ∈ U | U ∩V ≠ ∅}| < ∞. A refinement of an open cover V of X is an open cover U of X with the property that: for all U ∈ U, there exists V ∈ V such that U ⊆ V. We say that X is paracompact if every open cover of X has a locally finite refinement.

Let F : C → D be a functor. Let S be a set of objects and arrows in C. Then, by F(S), we mean {F(s) | s ∈ S}.

Theorem (van Kampen’s Theorem). Let T S^{∗} be the category of pointed, path-connected, locally path-connected, semi-locally simply connected, Hausdorff, regular, second countable topological spaces whose arrows are pointed homotopy classes of pointed continuous maps. Then π_{1} : T S^{∗} → {countable groups} preserves direct limits. That is, if D is a commutative diagram in T S^{∗} and if (R, A) is a direct limit of D in T S^{∗}, then (π_{1}(R), π_{1}(A)) is a direct limit of π_{1}(D) in {countable groups}.

We defer the proof, and proceed to an application of van Kampen’s Theorem:

Let S be the genus two orientable surface obtained from a closed solid octagon \overline{K} in ℝ^{2} after the (a, b, \overleftarrow{a}, \overleftarrow{b}, c, d, \overleftarrow{c}, \overleftarrow{d})-identification of boundary points. We set ourselves to the task of computing π_{1}(S).

Let K ⊆ S denote the interior in ℝ^{2} of \overline{K} Let U_{0} be an open disk in ℝ^{2} such whose closure is contained in K, i.e., Cl_{ℝ^{2}}(U_{0}) ⊆ K. Let D_{0} be a closed disk in ℝ^{2} such that D_{0} ⊆ U_{0}. Let U and D be the images in S of U_{0} and D_{0}. Let V := S\D. Let s ∈ U ∩ V. For any two subsets P, Q ⊆ S, if s ∈ P ⊆ Q, then we let ι_{P}^{Q} : (P, s) → (Q, s) denote pointed homotopy class of the pointed inclusion map.

Then (S, s) is a pushout in T S^{∗} of {(U ∩ V, s), (U, s), (V, s), ι_{U∩V}^{U} , ι_{U∩V}^{V} }. Let

A := π_{1}(U ∩ V, s),  B := π_{1}(U, s),  C := π_{1}(V, s).

Then, by van Kampen’s Theorem, we see that π_{1}(S, s) = A ∗_{C} B, where the maps A → B and A → C are π_{1}(ι_{U∩V}^{U}) and π_{1}(ι_{U∩V}^{V}). Consequently, if we can calculate A, B, C, f and g, then the problem of calculating π_{1}(S, s) is reduced to the algebraic problem of understanding pushouts in the category {countable groups}.

We come back to this in a moment but pause now to note that, quite generally, if X is a paracompact topological space, if U is an open cover of X with nonempty intersection, if x ∈ ∩U and if, for all U, V ∈ U, we have that U ∩ V is path-connected,

then one can form a diagram D in T S^{∗} whose set of objects is {(U ∩ V, x) | U, V ∈ U}, whose arrows come from inclusions and whose direct limit is (X, x).

Van Kampen’s Theorem then reduces the calculation of π_{1}(X, x) to the calculation of π_{1}(D) combined with the algebraic problem of computing the direct limit of π_{1}(D) in the category {countable groups}.

Returning to the specific case of the genus two orientable surface S, we have a presentation of the fundamental grou π_{1}(U ∩ V, s) = ⟨ϵ⟩, where ϵ is the endpoint fixed homotopy class of a parameterized circle going once around the annulus U ∩ V in the clockwise direction. Let s′ be the image in S of a vertex on the boundary of the octagon. (Note that the vertices all have the same image.) Let ∂K be the boundary of the octagon \overline{K}. Let α_{∗} be a parametrization running along one of the sides of ∂K labeled a. Let α′ ∈ P_{s′}^{s′}(S) be the image of α_{∗} in S. Using sides b, c and d, we similarly obtain loops β′, γ′, δ′ ∈ P_{s′}^{s′}(S). Then a simple application of van Kampen’s theorem shows that π_{1}(V, s′) = ⟨α′, β′, γ′, δ′⟩. That is, π_{1}(V, s′) is a free group on four letters. Let ρ be a path in S from s to s′. Let α := (ρ∥α′)∥\overleftarrow{ρ} ∈ P_{s}^{s}(S). Similarly, define β, γ, δ ∈ P_{s}^{s}(S). Then π_{1}(V, s) = ⟨α, β, γ, δ⟩. The inclusion (U ∩ V, s) → (V, s) induces the map g : π_{1}(U ∩ V, s) → π_{1}(V, s) defined by ϵ ↦ αβα^{−1}β^{−1}γδγ^{−1}δ^{−1}. According to standard notation, [α, β] := αβα^{−1}β^{−1} and [γ, δ] := γδγ^{−1}δ^{−1}, so ϵ ↦ [α, β][γ, δ].

As U is contractible, π_{1}(U, s) ≅ ⟨η | η = 1⟩. The map f : π_{1}(U ∩ V, s) → π_{1}(U, s) induced by inclusion is then the trivial map, defined by ϵ ↦ 1.

Then π_{1}(S, s) is the pushout in the category {countable groups} of the diagram

{π_{1}(U ∩ V, s), π_{1}(U, s), π_{1}(V, s), f, g}.

We can therefore use the group direct limit algorithm to see that π_{1}(S, s) is isomorphic to

⟨α, β, γ, δ, ϵ, η | η = 1, ϵ = [α, β][γ, δ], ϵ = 1⟩.

It is then an unassigned group theory exercise to see that ϵ and η are not needed. In fact, we have that π_{1}(S, s) is isomorphic to

⟨α, β, γ, δ | [α, β][γ, δ] = 1⟩.

We now turn to preparations for the proof of van Kampen’s theorem. The fundamental category theoretic notion we will need is that of an “adjoint pair”.

Given two categories C and E, define C × E to be the category such that an object in C × E is

(α) a pair (C, E), where C is an object in C and E is an object in E; and such that an arrow (C, E) → (C′, E′) in C × E is

(β) a pair of arrows (f, g), where f : C′ → C is an arrow in C and g : E → E′ is an arrow in E.

Note, above, that the arrow f goes from C′ to C, and not the other way around.

Definition. Let C and E be categories and let F : C → E and G : E → C be functors. We say that (F, G) is an adjoint pair the following two functors C × E → {sets} are equivalent:

(∗) (C, E) ↦ Hom_{C}(C, GE); and (∗∗) (C, E) ↦ Hom_{E}(FC, E).

Let η be an equivalence between the functors (∗) and (∗∗). Then, for any (C, E), η_{(C,E)} is a bijection between Hom_{C}(C, GE) and Hom_{E}(FC, E). This gives us four ways of moving arrows between C and E:

(1) for any arrow a : C → C′ in C, Fa : FC → FC′ is an arrow in E;

(2) for any arrow a : E → E′ in E, Ga : GE → GE′ is an arrow in C;

(3) for any arrow C → GE in C; there is an η-corresponding arrow FC → E in E;

(4) for any arrow FC → E in E; there is an η-corresponding arrow C → GE in C. Since η is an equivalence between (∗) and (∗∗), we have:

(A) for any objects C, C′ in C, for any object E in E, for any arrows f : C′ → C and g : C → GE in C, if g′ : FC → E corresponds to g under η, then g′ ∘ (Ff) corresponds to g ∘ f under η.

Since η is an equivalence between (∗) and (∗∗), we also have:

(B) for any objects E, E′ in E, for any object C in C, for any arrows f : FC → E and g : E → E′ in E, if f′ : C → GE corresponds to f under η, then (Gg) ∘ f′ corresponds to g ∘ f under η.

We now come to the main result we’ll be using on adjoint pairs:

Theorem. Let C and E be categories. Let F : C → E and G : E → C be functors. Assume that (F, G) is an adjoint pair. Then F preserves direct limits. That is, if D is a diagram in C and if (L, f) is a direct limit for D in C, then (FL, Ff) is a direct limit for FD in E.

Proof: Let (R, g) be a receiver of FD in E. We wish to show that there is a unique h : FL → R such that {h ∘ (Ff_{0}) | f_{0} ∈ f} = g.

Let η be an equivalence from

(C, E) ↦ Hom_{C}(C, GE) : C × E → {sets}

to

(C, E) ↦ Hom_{E}(FC, E) : C × E → {sets}.

For each D ∈ Obj(D), let g_{D} be the unique arrow in g ∩ (Hom(FD, R)). Then set g′ := {η_{(D,R)}^{−1}(g_{D}) | D ∈ Obj(D)}. Then (GR, g′) is a receiver for D.

Since (L, f) a direct limit of D in C, choose h′ : L → GR such that g′ = {h′∘f_{0} | f_{0} ∈ f}. Now set h := η_{(L,R)}(h′) ∈ Hom_{E}(FL, R). Then, by the naturality properties of η, we see that g = {h ∘ (Ff_{0}) | f_{0} ∈ f}, as desired.

We leave the uniqueness of h as an exercise. QED

EXERCISE 11A: Let C and E be categories. Let F : C → E and G : E → C be functors. Assume that (F, G) is an adjoint pair. Let D be a diagram in C. Let (L, f) be a direct limit of D in C. Let (R, g) be a receiver of FD in G. Let h, h̃ ∈ Hom_{E}(FL, R). Assume that {h ∘ (Ff_{0}) | f_{0} ∈ f} = g = {h̃ ∘ (Ff_{0}) | f_{0} ∈ f}. Show that h = h̃.

Recall, for any group Γ, that E_{Γ} denotes the set of all

c_{0}γ_{0} + c_{1}γ_{1} + ⋯

such that c_{0}, c_{1}, … ∈ [0, 1], such that γ_{0}, γ_{1}, … ∈ Γ, such that |{i ∈ ℕ | c_{i} ≠ 0}| < ∞ and such that c_{0} + c_{1} + ⋯ = 1, appropriately topologized. For any group Γ, let 1 · 1_{Γ} + 0 + 0 + 0 + ⋯ be the basepoint of E_{Γ}. For any group Γ, we have that Γ acts on E_{Γ} via γ(c_{0}γ_{0} + c_{1}γ_{1} + ⋯) = c_{0}γγ_{0} + c_{1}γγ_{1} + · · ·. Recall, for any group Γ, that B_{Γ} := Γ\E_{Γ}; let the basepoint of B_{Γ} be the Γ-orbit of the basepoint 1 · 1_{Γ} + 0 + 0 + 0 + ⋯ of E_{Γ}.

Then Γ ↦ B_{Γ} : {countable groups} → T S^{∗} is a functor, which, in the following proposition, we will denote by B. (What does B do to arrows in {countable groups}?)

The functor π_{1} : T S → {groups} also induces a functor T S^{∗} → {countable groups}, which we will also denote by π_{1} in the following proposition.

In view of the preceding theorem, van Kampen’s Theorem follows from:

Proposition. The pair (π_{1}, B) is an adjoint pair.

Sketch of proof: Fix X ∈ T S^{∗} and fix a group Γ. We wish to construct functions

η : Hom(X, B_{Γ}) → Hom(π_{1}X, Γ)

and

ζ : Hom(π_{1}X, Γ) → Hom(X, B_{Γ}).

After we complete this construction below, we then leave it as an unassigned exercise to the reader to show that η and ζ are inverses. We also leave it as an unassigned exercise to prove the naturality requirements of η and ζ, as (X, Γ) varies.

Since we have a chosen basepoint in E_{Γ}, this determines an isomorphism between π_{1}B_{Γ} and the group of deck transformations of the canonical map E_{Γ} → B_{Γ}. However, this group is exactly Γ, so we have an isomorphism τ : π_{1}B_{Γ} → Γ. Similarly, we define an isomorphism μ : π_{1}B_{π_{1}X} → π_{1}X.

We define

η : Hom(X, B_{Γ}) → Hom(π_{1}X, Γ)

by η(f) = τ ∘ (π_{1}f). It remains to construct ζ.

Claim: There is a unique α ∈ Hom(X, B_{π_{1}X}) such that μ ∘ (π_{1}α) ∈ Hom(π_{1}X, π_{1}X) is the identity. Proof of Claim: Let Ξ := π_{1}(X) and let I := [0, 1]. Let (X̃, q) be a pointed universal cover of X. Let r : E_{Ξ} → B_{Ξ} be the canonical map.

For any pointed space S, let p_{S} denote the basepoint of S. The underlying topological space of S will also be denoted by S. Let e := p_{E_{Ξ}} be the basepoint of E_{Ξ}.

We first prove uniqueness. Let α, β ∈ Hom(X, B_{Ξ}) and assume that both μ ∘ (π_{1}α) and μ ∘ (π_{1}β) are equal to the identity arrow π_{1}X → π_{1}X. We wish to show that α = β. Fix α_{0} ∈ α and β_{0} ∈ β. We wish to show that α_{0} is pointed homotopic to β_{0}.

Let X̃ be a pointed universal cover of X. By lifting theory, there exist Ξ-equivariant pointed maps α̂_{0}, β̂_{0} : X̃ → E_{Ξ} such that r ∘ α̂_{0} = α_{0} ∘ q and r ∘ β̂_{0} = β_{0} ∘ q

We wish to show that there exists a continuous map H : X̃ × I → E_{Ξ} such that

(1) for all x ∈ X̃, we have H(x, 0) = α̂_{0}(x); and H(x, 1) = β̂_{0}(x);

(2) for all ξ ∈ Ξ, for all x ∈ X̃ for all t ∈ I, we have H(ξx, t) = ξ(H(x, t)); and

(3) for all t ∈ I, we have H(p_{X}, t) = e.

EXERCISE 11B: Show that there exists a continuous map k : E_{Ξ} × E_{Ξ} × I → E_{Ξ} such that

(A) for all x, y ∈ E_{Ξ}, we have k(x, y, 0) = x and k(x, y, 1) = y;

(B) for all ξ ∈ Ξ, for all x, y ∈ E_{Ξ}, for all t ∈ I, we have k(ξx, ξy, t) = ξ(k(x, y, t));

and (C) for all t ∈ I, we have k(e, e, t) = e.

Fix k : E_{Ξ} × E_{Ξ} × I → E_{Ξ} as in Exercise 11B. Now define H : X̃ × I → E_{Ξ} by H(x, t) = k(α̂_{0}(x), β̂_{0}(x), t). This ends the proof of uniqueness.

We now prove existence. It suffices to show that there exists a continuous Ξ-invariant map α̂_{0} : X̃ → E_{Ξ} such that α_{0}(p_{X}) = e.

Let \overline{V}_{0}, \overline{V}_{1}, \overline{V}_{2}, … be a locally finite open cover of X such that, for all integers i ≥ 0, we have that V_{i} := q^{−1}(\overline{V}_{i}) is a trivial principal topological Ξ-space. Reordering, if ncessary, we may assume that p_{X} ∈ \overline{V}_{0}. For each integer i ≥ 1, by, if necessary, replacing \overline{V}_{i} by \overline{V}_{i}\\{p_{X}\}, we may assume that p_{X} ∉ \overline{V}_{i}.

For all integers i ≥ 0, choose a topological space U_{i} and an isomorphism of topological Ξ-spaces ϕ_{i} : V_{i} → Ξ×U_{i}. We may assume that ϕ_{0}(p_{X}) ∈ {1_{Ξ}}×U_{i}. For all integers i ≥ 0, let ψ_{i} : Ξ × U_{i} → Ξ be projection onto the first coordinate, and let λ_{i} := ψ_{i} ∘ ϕ_{i} : V_{i} → Ξ; then λ_{i} is Ξ-equivariant, i.e., for all v ∈ V_{i}, for all ξ ∈ Ξ, we have λ_{i}(ξv) = ξ(λ_{i}(v)). For all integers i ≥ 0, let κ_{i} : X̃ → Ξ be any function such that κ_{i}|V_{i} = λ_{i}. Note that

κ_{0}(p_{X}) = 1_{Ξ}.

By paracompactness, let \overline{V} ′_{0}, \overline{V} ′_{1}, \overline{V} ′_{2}, … be a shrinking of \overline{V}_{0}, \overline{V}_{1}, \overline{V}_{2}, …; this means that \overline{V} ′_{0}, \overline{V} ′_{1}, \overline{V} ′_{2}, … is an open cover of X such that: for all integers i ≥ 0, the closure in X of \overline{V} ′_{i} is contained in \overline{V}_{i}.

For all integers i ≥ 0, by Urysohn’s Lemma, let \overline{f}′_{i} : X → [0, 1] be a continuous map

such that \overline{f}′_{i}(\overline{V} ′_{i}) = \{1\} and such that \overline{f}′_{i}(X\\overline{V}_{i}) = \{0\}. Let \overline{f}′ := ∑_{i=0}^{∞} \overline{f}′_{i}. For all integers

i ≥ 0, define \overline{f}_{i} := \overline{f}′_{i}/\overline{f}.

For all integers i ≥ 0, let f_{i} := \overline{f}_{i} ∘ q : X̃ → [0, 1]. Now define α̂_{0} : X̃ → E_{Ξ} by:

α̂_{0}(x) = [f_{0}(x)][κ_{0}(x)] + [f_{1}(x)][κ_{1}(x)] + [f_{2}(x)][κ_{2}(x)] + ⋯

This ends the proof of existence. End of proof of claim.

Fix α ∈ Hom(X, B_{π_{1}X}) as in the claim. We now define

ζ : Hom(π_{1}X, Γ) → Hom(X, B_{Γ})

by ζ(f) = B_{f} ∘ α. QED

For all integers n ≥ 0, let e_{0}^{n}, …, e_{n}^{n} be the standard basis of ℝ^{n+1}, let ν_{n} := {e_{0}^{n}, …, e_{n}^{n}} and let

σ_{n} := {c_{0}e_{0}^{n} + ⋯ + c_{n}e_{n}^{n} | c_{0}, …, c_{n} ∈ [0, 1], c_{0} + ⋯ + c_{n} = 1}

be the closed conved hull of ν_{n}.

Let S be a finite set and let n := |S| − 1. A labeling of S is a bijective map S → ν_{n}. Let S be a set and let R ⊆ S. Let m := |R| − 1 and n := |S| − 1. Let ϕ : R → ν_{m} and

ψ : S → ν_{n} be labelings. Let χ := ψ ∘ (ϕ^{−1}). Then we define I_{ϕ}^{ψ} : σ_{m} → σ_{n} by

I_{ϕ}^{ψ}(c_{0}e_{0}^{m} + ⋯ + c_{m}e_{m}^{m}) = c_{0}[χ(e_{0}^{m})] + ⋯ + c_{m}[χ(e_{m}^{m})].

Definition. An (abstract) simplicial complex consists of (1) a nonempty set S; and

(2) F ⊆ {finite subsets of S} such that

(A) for all s ∈ S, we have {s} ∈ F; and (B) for all f ∈ F, for all f′ ⊆ f, we have f′ ∈ F.

For any integer n ≥ 0, an n-face of a simplicial complex (S, F) is an element f ∈ F such that |f| = n + 1. For any integer n ≥ 0, a labeled n-face of a simplicial complex (S, F) consists of

(α) an n-face f of (S, F); and (β) a labeling of f.

We define the dimension, dim(S, F), of a simplicial complex (S, F) to be the supremum of the set of integers n ≥ 0 satisfying: there exists an n-face of (S, F). This supremum may equal ∞.

Definition. Let (S, F) be a simplicial complex. For all integers n ≥ 0, let F_{n} be the

set of all labeled n-faces of (S, F). Let X := ⋃_{n=0}^{∞} F_{n} × σ_{n}. For each integer n ≥ 0, give

F_{n} the discrete topology, give each σ_{n} the inherited topology from the standard topology on ℝ^{n+1} and let τ_{n} denote the product topology on F_{n} × σ_{n}. Let X have the topology

generated by ⋃_{n=0}^{∞} τ_{n}. Define a relation ↦ on X by: ((f, ϕ), x) ↦ ((g, ψ), y) iff both f ⊆ g

and I_{ϕ}^{ψ}(x) = y. Let ∼ be the equivalence relationg generated by ↦. Then the realization of (S, F) is X/ ∼.

Example. Let S := {1, 2, 3} and let F := {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}}. Then (S, F) is a simplical complex. Its realization is a “triangle”, so that it is homeomorphic to S^{1}. Consequently, the fundamental group of its realization in isomorphic to ℤ. Let F′ := F ∪ {{1, 2, 3}}. Then the realization of (S, F′) is a solid triangle, and is therefore homeomorphic to a disk. It is therefore contractible and its fundamental group is trivial.

Definition. A tree is a simply connected topological space which is homeomorphic to the realization of a simplicial complex of dimension ≤ 1. A graph is a locally path-connected topological space with a universal cover that is a tree.

Let x := (1, 0) ∈ S^{1} be a point on the circle. Let K be a set. Give K the discrete topology. Then a bouquet of K-many circles is the topological space

B := (S × K)/({x} × K)

obtained by identifying {x} × K to a point in S × K. Then B is a graph. Van Kampen’s theorem shows that the fundamental group of this space is homeomorphic to the free group ⟨K⟩.

Proposition. Let G be a graph. Then there exists T ⊆ G such that T is a tree, such that G/T is a bouquet of circles and such that π_{1}(G) is isomorphic to π_{1}(G/T).

We omit proof except to comment that one takes, in G, a so-called “maximal tree” T, and the proposition follows.

Corollary. For any graph G, we have: π_{1}(G) is a free group.

Our next goal is to use topology to prove the algebraic fact that a subgroup of a free group is free.

However, we need first to address the question of whether a composite of covering maps is again a covering map. In fact, I’m unsure if this is true in complete generality, but I can prove it when the spaces in question are connected and semi-locally simply connected. For the moment, however, let’s make the following ad-hoc definition:

Definition. Let P and Q be topological spaces and let α : P → Q be a function. Then we say that α is a composite covering map if there is an integer k ≥ 1 and there are topological spaces R_{0}, …, R_{k} and there are covering maps

β_{1} : R_{0} → R_{1},  …,  β_{k} : R_{k−1} → R_{k}

such that P = R_{0}, Q = R_{k} and α = β_{k} ∘ ⋯ ∘ β_{1}.

That is, a map is a compostite covering map if it is a composition of covering maps.

EXERCISE 12A: Let P, Q and X be connected, locally path-connected topological spaces. Let α : P → Q be a composite covering map. Let f : X → Q be continuous. Let q ∈ Q, let p ∈ α^{−1}(q) and let x ∈ f^{−1}(q). Assume that f_{∗}(π_{1}(X, x)) ⊆ α_{∗}(π_{1}(P, p)). Show that there exists a unique continuous map f̂: X → P such that α ∘ f̂ = f and such that f̂(x) = p.

Definition. Let Y be a path-connected topological space. A universal composite cover of Y is a simply connected topological space X, together with a composite covering map X → Y.

EXERCISE 12B: Let Y be a connected, locally path-connected topological space. Let (X, ϕ) and (X′, ϕ′) be universal composite covers of Y. Show that (X, ϕ) and (X′, ϕ′) are isomorphic in the category {topological spaces over Y }.

Fact. Let X and Y be connected, locally path-connected topological spaces. Let ϕ : X → Y be a composite covering map. Let (X̃, ψ) be a universal cover of X. Then (X̃, ϕ ∘ ψ) is a universal cover of Y.

Proof: Since X has a universal cover, X is semi-locally simply connected. Then Y is semilocally simply connected. Let (Ỹ, χ) be a universal cover of Y. Then, by Exercise 12B, (X̃, ϕ ∘ ψ) is isomorphic to (Ỹ, χ) in the category {topological spaces over Y }. So, since (Ỹ, χ) is a universal cover of Y, (X̃, ϕ ∘ ψ) must be as well. QED

If X and Y are topological spaces, then we say that X is a covering of Y if there exists a covering map X → Y. With this definition, the preceding fact asserts that a universal cover of a cover of Y is a universal cover of Y.

Proposition. Let W, X and Y be connected, semi-locally simply connected topological spacs. Let ϕ : W → X and ψ : X → Y be covering maps. Then ψ ∘ ϕ : W → Y is a covering map.

Proof: Let (W̃, χ) be a universal cover of W. Then (W̃, ϕ ∘ χ) is a universal cover of X and (W̃, ψ ∘ ϕ ∘ χ) is a universal cover of Y. Then χ, ϕ ∘ χ and ψ ∘ ϕ ∘ χ are all regular covering maps.

Let Γ be the group of deck transformations of χ. Let Λ be the group of deck transformations of ϕ ∘ χ. Let Δ be the group of deck transformations of ψ ∘ ϕ ∘ χ. Note that Γ ⊆ Λ ⊆ Δ. Then W, X and Y are isomorphic to Γ\W̃, Λ\W̃ and Δ\W̃, respectively, in the category {topological spaces under W̃}.

Let π : Γ\W̃ → Δ\W̃ be the canonical map. A short diagram chase then shows that ψ ∘ ϕ is isomorphic to π in the arrow category of {topological spaces}. Consequently, it suffices to show that π : Γ\W̃ → Δ\W̃ is a covering map.

We leave this as an unassigned exercise. QED

Lemma. Any covering of a graph is a graph.

Proof: Let Y be a graph and let X be a covering of Y. Then, by the preceding fact, any universal covering of X is a universal covering of Y, and is therefore a tree. We conclude that X is a graph. QED

Theorem. Any subgroup of a free group is free.

Proof: Let S be a set. We wish to show that any subgroup of ⟨S⟩ is free.

Let Y be a bouquet of S-many circles. Then, by van Kampen’s Theorem, π_{1}(Y) is isomporphic to ⟨S⟩. Let (X, ϕ) be a universal cover of Y. Let Γ be the group of deck transformations of ϕ. Then Γ is isomorphic to π_{1}(Y). Let Γ_{0} be a subgroup of Γ. We wish to show that Γ_{0} is free.

Since Γ\X is homeomorphic to Y, it follows that Γ\X is a graph. Since Γ_{0}\X is a covering of Γ\X, we see, from the preceding lemma, that Γ_{0}\X is a graph. Then π_{1}(Γ_{0}\X) is free. However, π_{1}(Γ_{0}\X) is isomorphic to Γ_{0}, so we are done. QED

Last updated: Wednesday 2 February at 5:00 pm