Mathematics 4530-Fall 2013

Mathematics 4530

Tychonoff’s Theorem

Ken Brown, Cornell University, October 2013

1. Introduction

Tychonoff’s theorem asserts that the product of an arbitrary family of compactspaces is compact. This is proved in Chapter 5 of Munkres, but his proof is notvery straightforward. The proof I’ll give below follows a paper by David Wright(Proceedings of the American Mathematical Society 120 (1994), 985–987). As awarmup, let’s start with two factors.

2. The Baby Tychonoff Theorem

Theorem. If X and Y are compact, then so is X × Y .

Note that this immediately extends to arbitrary finite products by induction onthe number of factors. This yields:

Corollary. A subset of Rn is compact if and only if it closed and bounded.

Proof of the corollary. If X ⊆ Rn is compact, then it is closed and bounded by thesame proof we used in class when n = 1. Conversely, if X is closed and bounded,then X is a closed subset of a rectangle R = I1× · · · × In, where each Ii is a closedinterval in R. Since each Ii is compact, so is the product R; the closed subset Xof R is therefore also compact. �

To prove a space X is compact, one usually proves that if U is a family of opensets that covers X, then a finite subcollection covers X. Sometimes, however, it ismore convenient to prove the contrapositive: If U is a family of open sets such thatno finite subcollection covers X, then U does not cover X. This is what we will dobelow.

Proof of the theorem. Let W be a collection of open subsets of X ×Y such that nofinite subcollection covers X × Y ; we will show that W does not cover X × Y .

Claim 1. There exists x0 ∈ X such that no open tube U ×Y with x0 ∈ U is finitelycovered by W.

Proof. If this is false, then every x ∈ X has a neighborhood Ux such that Ux×Y isfinitely covered. By compactness of X, finitely many of these sets Ux cover X, sofinitely many of the tubes Ux × Y cover X × Y . This contradicts the assumptionthat X × Y is not finitely covered.

Claim 2. There exists y0 ∈ Y such that no open rectangle U×V containing (x0, y0)is finitely covered by W.

Proof. If this is false, then for every y ∈ Y there is a finitely covered open rectangleUy × Vy containing (x0, y). By compactness of Y , there is a finite subset F ⊆ Ysuch that Y =

⋃y∈F Vy. Set U :=

⋂y∈F Uy. Then U is a neighborhood of x0, and

the tubeU × Y =

⋃y∈F

U × Vy ⊆⋃y∈F

Uy × Vy

is finitely covered, contradicting Claim 1.1

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Claim 2 immediately implies the theorem. Indeed, we have a point z := (x0, y0)such that no basic open set containing z is finitely covered by W. In particular, nobasic open set containing z can be contained in a set W ∈ W, so z /∈

⋃W∈WW .

This shows that W does not cover X × Y . �

Remark. We did not have to use the contrapositive of the condition for compactness.In fact, it is more straightforward to just use the original definition, and to base theproof on the tube lemma; see Munkres, pp. 167–168. (The tube lemma asserts thatin a product X × Y with Y compact, any neighborhood of a slice x × Y containsa tube U × Y , where U is a neighborhood of x; the proof of this is very similar tothe proof of Claim 2.) The advantage of our convoluted proof, however, is that itextends easily to infinitely many factors.

3. Countably Many Factors

In preparation for treating infinitely many factors, we record a slight generaliza-tion of the argument used in the proof of Claim 2 above.

Lemma. Let W be a family of open sets in a product X×Y ×Z. Assume there is apoint x0 ∈ X such that no open set U ×Y ×Z with x0 ∈ U is finitely covered by W.If Y is compact, then there is a point y0 ∈ Y such that no open set U ×V ×Z with(x0, y0) ∈ U × V is finitely covered by W.

Proof. Suppose no such y0 exists. Then for every y ∈ Y there is a finitely coveredopen set Uy × Vy × Z with x0 ∈ Uy and y ∈ Vy. By compactness of Y , there isa finite subset F ⊆ Y such that Y =

⋃y∈F Vy. Set U :=

⋂y∈F Uy. Then U is a

neighborhood of x0, and the set

U × Y × Z =⋃y∈F

U × Vy × Z ⊆⋃y∈F

Uy × Vy × Z

is finitely covered, contradicting the hypothesis. �

We can now easily generalize the “baby” argument to treat countably manyfactors:

Theorem. Let X =∏∞i=1Xi, where each Xi is compact. Then X is compact.

Proof. Let W be a family of open sets that does not finitely cover X; we willconstruct a point x = (x1, x2, . . . ) such that no neighborhood of x is finitely covered.Note first that there is a point x1 ∈ X1 such that no open tube U ×X2 ×X3 × · · ·with x1 ∈ U is finitely covered. The proof of this assertion is the same as the proofof Claim 1 above, with X2 × X3 × · · · playing the role of Y . Next, we can findx2 ∈ X2 such that no open rectangle U×V ×X3×X4×· · · with (x1, x2) ∈ U×V isfinitely covered. This follows from the lemma if we view X as X1×X2×(X3×· · · ).Continuing in this way, we inductively define x1, x2, x3, . . . such that for each n, nobasic open set of the form U1×· · ·×Un×Xn+1×· · · , with xi ∈ Ui for i ≤ n, is finitelycovered. For the inductive step, view X as (X1× · · · ×Xn−1)×Xn× (Xn+1× · · · )and apply the lemma.

We now have a point x = (x1, x2, . . . ) ∈ X such that no basic neighborhood of xis finitely covered by W. Thus x /∈

⋃W∈WW . �

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4. Grown-up Tychonoff

Finally, here is the full-fledged Tychonoff theorem:

Theorem. Given an arbitrary family (Xα)α∈J of compact spaces, their productX :=

∏α∈J Xα is compact.

Proof. Once again, we letW be a family of open sets that does not finitely cover X,and we construct x = (xα)α∈J such that no neighborhood of x is finitely covered.We may assume by the well-ordering theorem that the index set J is well-ordered,and we construct xα by transfinite induction so that no basic open set of the form∏

β≤α

Uβ ×∏β>α

Xβ ,

with xβ ∈ Uβ for all β ≤ α, is finitely covered. For the inductive step, assume thatxβ has been defined with the desired property for all β < α, and apply the lemma,viewing X as

(∏β<αXβ

)×Xα×

(∏β>αXβ

). The details are essentially the same

as in the countable case. �

Remark. The proof used the axiom of choice, in the form of the well-ordering the-orem. This is unavoidable. Indeed, logicians have shown that Tychonoff’s theoremcannot be proved without the axiom of choice.

5. Application: Invariant Means

Having gone to all the trouble of proving the well-ordering theorem and thegeneral form of Tychonoff’s theorem, we now give a (somewhat weird) applicationof it. As is typical of results that require the axiom of choice, it proves the existenceof something that one could never hope to actually construct concretely. We beginwith a naıve question:

Is there a sensible way of associating an average value to every bounded, doublyinfinite sequence of real numbers, such as

(1) . . . , 0, 1, 0, 1, 0, 1, . . .?

To phrase the question precisely, consider sequences a = (an)n∈Z, i.e., elements ofRZ. We denote by B the set of all such sequences that are bounded. It is a realvector space. We will not make use of any topology on B, but we recall in passingthat there is a standard one, different from the product topology, that is useful inanalysis: One puts a norm on B by setting ‖a‖ = supn∈Z|an|, and this yields ametric d(a, b) := ‖a− b‖ and hence a topology.

A mean on B is a linear map µ : B → R such that

infnan ≤ µ(a) ≤ sup

nan

for all a ∈ B. Means exist in great abundance. For example, we can take a finitesubset F ⊆ Z and define a mean by averaging over F :

µ(a) :=1

|F |∑n∈F

an.

More generally, we can assign a weight wn ≥ 0 to each n ∈ Z, with∑n wn = 1,

and use the weighted average

µ(a) :=∑n

anwn.

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The shift operator a 7→ as on B is defined by (as)n = an+1. A mean is said to beinvariant, or shift-invariant, if µ(as) = µ(a) for all a ∈ B. It is reasonable to expecta sensible averaging procedure to be shift-invariant; this just says that we can talkabout the average of a sequence like the one in (1) above, without having to choosearbitrarily which element is considered to be in postion 0. But no one knows howto construct an invariant mean. (In particular, the weighted averages above arenot invariant). The best we can do is to construct approximately invariant meansand then deduce, in a nonconstructive way, that invariant means exist. Here is theconstructive part:

Lemma. There is a sequence (µn)n≥1 of means such that for all a ∈ B,

limn→∞

|µn(as)− µn(a)| = 0.

Proof. Let µn(a) := (1/n)∑ni=1 ai. Then

|µn(as)− µn(a)| = |an+1 − a1|/n ≤ 2‖a‖/n,

which tends to 0 as n→∞. �

There are several known methods for doing the nonconstructive part of the proofthat invariant means exist. The one we will use is to topologize the set M of allmeans in such a way that it becomes a compact space (via Tychonoff). It will thenfollow that the approximately invariant sequence µn has a cluster point µ, which isinvariant. Here are the details.

View M as a subset of RB by identifying a mean µ with the indexed family(µ(a))a∈B. We then give RB the product topology and M the subspace topology.Thus our topology is cooked up to make the evaluation map µ 7→ µ(a) a continuousmap M → R for each a ∈ B. Note that the definition of “mean” actually makesM a subset of a product of closed intervals:

M⊆∏a∈B

[m(a),M(a)],

where m(a) = infn an and M(a) = supn an.

Proposition. M is compact.

Proof. We know that X :=∏a∈B[m(a),M(a)] is compact, so it suffices to show

that M is a closed subset of X. Now an element µ ∈ X is a mean if and only if itsatisfies

µ(a+ b)− µ(a)− µ(b) = 0

for all a, b ∈ B and

µ(λa)− λµ(a) = 0

for all a ∈ B and λ ∈ R. This exhibits M as an intersection of sets of the formF−1(0) for various continuous maps F : X → R, so M is closed. �

We now appeal to the “Bolzano–Weierstrass” property of compact spaces:

Proposition. If X is a compact space, then every sequence (xn)n≥1 in X has acluster point, i.e., there is a point x ∈ X such that every neighborhood of x containsxn for infinitely many n.

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Proof. For each x that is not a cluster point, there is a neighborhood Ux of x thatcontains xn for only finitely many n. Clearly no finite subcollection of the Ux cancover X, so the family of all Ux does not cover X. Since

⋃x Ux contains all points

that are not cluster points, it follows that there must exist at least one clusterpoint. �

We can now prove our promised existence theorem:

Theorem. There exists an invariant mean µ : B → R.

Proof. Take a sequence (µn) as in the lemma, and let µ be a cluster point. To showthat µ is invariant, let a ∈ B be arbitrary. We will argue that for a suitable large n,

(2) µ(as) ≈ µn(as) ≈ µn(a) ≈ µ(a).

More precisely, fix ε > 0 and let

U :={ν ∈M

∣∣ |ν(a)− µ(a)| < ε and |ν(as)− µ(as)| < ε}.

Then U is a neighborhood of µ in M. [In fact, it is a basic neighborhood, definedby imposing conditions on two “coordinates” of a general ν.] We therefore haveµn ∈ U for infinitely many n. Take such an n large enough that |µn(as)−µn(a)| < ε.Then the three approximations in (2) are all valid, with error less than ε, whence|µ(as)− µ(a)| < 3ε. Since ε is arbitrary, it follows that µ(as) = µ(a). �

Mathematics 4530 Prelim 1 October 22, 2013

Do any 4 of the following 6 problems, and show clearly which 4 you have chosen. Youmay do additional problems for fun and a tiny amount of extra credit. Your proofsmay use any result proved in class, in the assigned reading, or in the homework,unless you are explicitly asked to prove such a result.

1. Recall that a metric space X is said to be bounded if there is a number M suchthat d(x, y) ≤ M for all x, y ∈ X. If X is a finite union of balls, show that X isbounded.

2. Define each of the following terms; verbosity is not necessary.

(a) topology on a set X(b) topological space(c) subspace topology(d) product topology (on an arbitrary product, possibly infinite)(e) connected topological space(f) compact topological space

3. Let X be a topological space. Given B ⊆ A ⊆ X, topologize A as a subspaceof X, and then topologize B as a subspace of A. Show that the resulting topologyon B is the same as the topology that B inherits as a subspace of X.

4. Let {Aα}α∈J be a family of subsets of a metric space X. Define fα : X → R by

fα(x) := d(x,Aα),

and let f : X → RJ be the map with components fα, where RJ :=∏α∈J R is given

the product topology. Is f continuous? Justify your answer.

5. Give a proof or counterexample for each of the following statements.

(a) If A and B are connected subspaces of a topological space and A∩B 6= ∅, thenA ∪B is connected.

(b) If A and B are connected subspaces of a topological space, then A ∩ B isconnected.

6. Give a proof or counterexample for each of the following statements.

(a) If A and B are compact subspaces of a topological space, then A∪B is compact.(b) If A and B are compact subspaces of a Hausdorff space, then A∩B is compact.

Brief solutions to Prelim 1

1. A ball B of radius r is bounded because d(x, y) ≤ 2r for all x, y ∈ B by thetriangle inequality. So it suffices to show that a finite union of bounded sets isbounded. A straightforward induction reduces this to the case of two bounded setsA1, A2. Let M1 and M2 be bounds for A1 and A2, respectively. We may assumethat A1 and A2 are both nonempty, in which case we can choose points ai ∈ Ai. Iclaim that M := M1 +d(a1, a2) +M2 is a bound for A1∪A2: Given x, y ∈ A1∪A2,if they are both in A1, then d(x, y) ≤ M1 ≤ M , so the claim holds in this case.Similarly, the claim holds if they are both in A2. Finally, if they are in differentsets, say x ∈ A1 and y ∈ A2, then two applications of the triangle inequality give

d(x, y) ≤ d(x, a1) + d(a1, y)

≤ d(x, a1) + d(a1, a2) + d(a2, y)

≤M1 + d(a1, a2) + M2

= M.

2. See Munkres.

3. This was a homework problem, but here’s the solution: The given topology onB has {(U ∩A) ∩B} as its open sets, where U ranges over the open sets of X.Since (U ∩A)∩B = U ∩ (A∩B) = U ∩B, this is the same as the subspace topologythat B inherits from X.

4. Yes, f is continuous because each component function is continuous.

5. (a) This is true; see Munkres, Theorem 23.3.(b) This is false. A counterexample is provided by the circle, which is the union of

two closed arcs intersecting in two points.

6. Recall that compactness for a subspace A of a space X can be formulated asfollows (see Munkres, Lemma 26.1): If U is a family of open subsets of X thatcovers A, then a finite subset of U covers A.

(a) This is true. If U is a family of open subsets of X that covers A ∪ B, thenfinitely many cover A and finitely many cover B; hence finitely many coverA ∪B.

(b) This is also true. A and B are closed in X because X is Hausdorff, so A ∩ Bis closed. Therefore A ∩ B is compact, being a closed subset of the compactspace A.

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Mathematics 4530

Prelim 2, due December 5, 2013

What follows is a generalization of the calculation of π1(S1). It is essentially ahandout, but with all the proofs removed. Your task is to fill in the proofs inSections 1–6. [You may read Section 7 for fun if you want, but there is nothingin it for you to write up.] I have given a clear indication of the places where youare expected to fill in a proof. I have also labeled some proofs as “extra credit” tokeep the exam from being too long; they are not necessarily more difficult than theother proofs.

You may use any result proved in class, in the assigned reading, or in yourhomework. You may not use any sources other than your textbook and class notes.You may not use results from unassigned homework problems or from sections ofthe book that you have not been told to read. Andrew and I will be glad to clearup any ambiguities; you may not discuss the exam with anyone else.

1. Introduction

A crucial feature of the exponential covering p : R→ S1 is that two points in thesame fiber differ by an integral multiple of 2π. This was used in the definition of thedegree homomorphism π1(S1)→ Z, which was then proved to be an isomorphism.We wish to generalize this to other covering maps, the so-called regular ones. As aconsequence, one can calculate many fundamental groups with no more effort thanwas required for S1.

2. Group actions

We have defined a group to be a set G together with a binary operation ∗satisfying certain axioms. This may give you the wrong impression of group theory,in that it gives you no clue as to how groups arise in nature. The way groups arisein nature is that they act on things, thereby exhibiting the symmetry that thosethings have.

Definition 1. Let G be a group and X a topological space. By an action of Gon X we mean a function G×X → X, denoted (g, x) 7→ g ∗ x, satisfying:

(1) g ∗ (h ∗ x) = (g ∗ h) ∗ x for all g, h ∈ G and x ∈ X.(2) 1 ∗ x = x for all x ∈ X, where 1 is the identity element of G.(3) For each g ∈ G, the map x 7→ g ∗ x is continuous.

It follows that the map x 7→ g ∗ x, which we will call the action of g, is actually ahomeomorphism. Its inverse is given by the action of g−1. Group actions are studiedextensively in algebra courses, such as Math 4340. For our purposes, however, itsuffices to have some examples.

Example 1. The additive group Z acts on R by (n, x) 7→ 2πn + x for n ∈ Z andx ∈ R. (We say “Z acts on R by translation.”) This is the action that’s implicit inour work on the exponential covering map, which led to the calculation π1(S1) ∼= Z.

Example 2. The multiplicative group {±1} acts on S2 by scalar multiplication.This action is related to the double covering S2 → P 2 mentioned in class, whereP 2 is the projective plane.

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Example 3. More generally, whenever we have a 2-fold covering map p : X → Y ,there is an associated action of a group of order 2 on X. This is essentially whatyou proved in additional problem 6 on Assignment 10.

Example 4. Let Zn be the additive group of integers mod n. There is an action ofZn on S1, in which the integer k mod n acts as rotation through 2πk/n radians.Alternatively, if we replace Zn by the isomorphic group consisting of the nth rootsof unity in C, then we can describe the action using multiplication of complexnumbers.

Example 5. The concept of “Cayley graph,” which I will explain in Section 7, leadsto many examples of group actions. For example, the snowflake that covers thefigure 8 is the Cayley graph of the free group F2; as a result, there is an action ofF2 on the snowflake.

To simplify the notation in what follows, we will always use multiplicative no-tation for our groups (i.e., we will write gh instead of g ∗ h), and similarly we willwrite (g, x) 7→ gx for group actions.

Every one of the examples above has the property that it is fixed-point free, i.e.,the action of any g 6= 1 has no points x such that gx = x. The significance of thisfor us is that if two points x, x′ are in the same G-orbit, i.e., if there is a g ∈ Gsuch that gx = x′, then the element g is unique. [If gx = x′ and hx = x′, thengx = hx =⇒ h−1gx = x =⇒ h−1g = 1 =⇒ g = h.] We can think of g assomething like the “difference” between x and x′. For the case of Z acting on R asin Example 1, it really is the difference (divided by 2π).

3. Regular G-covers

Definition 2. Let G be a group. A regular G-cover is a covering map p : X → Ytogether with a fixed-point-free action of G on X whose orbits are the fibers of p.

Let’s spell this out explicitly:

Lemma 1. Let p : X → Y be a covering map with an action of a group G on X.Then p is a regular G-cover if and only if it satisfies the following two conditions:

(a) p(gx) = p(x) for all g ∈ G, x ∈ X.(b) If p(x) = p(x′) then there is a (unique) g ∈ G such that x′ = gx.

Proof. *** Your proof goes here. *** �

Remark 1. The group action is useful in connection with path liftings. Namely, ifα is a path in Y and α is the lift starting at some point in the fiber over α(0), thenwe can construct all the other liftings of α by applying the group action.

*** Your explanation goes here. ***

We have already seen many examples of regular G-covers. Indeed, every examplein the previous section has an associated regular G-cover, which was mentionedexplicitly except in Example 4. [You can ask me if you can’t figure out the coverthat goes with that example.] On the other hand, there are many covering mapsthat are not regular because they do not have enough symmetry. Indeed, conditions(a) and (b) imply, intuitively, that a regular G-cover looks the same no matter wherewe stand in a given fiber. For example:

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Lemma 2. Let p : X → Y be a regular G-cover for some group G, and let α be aclosed curve in Y that admits a closed lift α. Then every lift of α is closed.


One can deduce from this that not all covering maps are regular.

Example 6. *** Your example goes here. ***

[To make it interesting, make sure that X and Y are both connected.]

4. The generalized degree homomorphism

Recall that we defined the degree of a closed curve α in S1 by lifting it to R andmeasuring the difference between the starting and ending points of the lift. We cando something similar for any regular cover.

Let p : X → Y be a regular G-cover. Pick a basepoint y ∈ Y and a basepointx ∈ p−1(y). Given a path homotopy class [α] ∈ π1(Y, y), let α be the lift of αstarting at x. Then α ends in the fiber p−1(y), so there is a unique g ∈ G such thatα(1) = gx. We set

δ([α]) := g.

In case p is the exponential covering of S1, this is precisely the degree defined inclass. Imitating the four steps that occurred in our calculation of π1(S1), we willprove:

Theorem 1. Suppose that X is simply connected. Then

δ : π1(Y, y)→ G

is an isomorphism.

The four steps are given in the following four lemmas, valid for any regularG-covering.

Lemma 3. δ is well defined.


Lemma 4. δ is a group homomorphism.


Lemma 5. If X is path connected, then δ is surjective.


Lemma 6. If X is simply connected, then δ is bijective.


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5. Examples

Example 7. Combining Theorem 1 and Example 2, we obtain:

*** Your conclusion goes here, with full justification. ***

Example 8. Applying Theorem 1 to a suitable covering of the torus, we obtain:

*** Your conclusion goes here, with full justification. ***

Example 9. I promised that I would prove that the fundamental group of the figure 8is the free group F2 on two generators. One way to do this is to apply Theorem 1 tothe snowflake cover of the figure 8. I will sketch how this can be done in Section 7.A different proof will be given in class.

6. Existence of regular G-covers

It is natural to ask how widely applicable Theorem 1 is. In this section we willsee that, in principle, Theorem 1 can be used to calculate the fundamental groupof every reasonable space. Here “reasonable” has a somewhat technical definition:

Definition 3. A space Y is said to be semilocally simply connected (SLSC) ifevery point y ∈ Y has a neighborhood U such that the inclusion U ↪→ Y inducesthe trivial homomorphism π1(U, y)→ π1(Y, y).

This condition may seem strange; but in fact, most spaces that one meets “innature” satisfy the even stronger condition of being locally contractible. [Feel freeto ask for examples if you have trouble seeing this.] Here’s more evidence that theSLSC condition is reasonable:.

Proposition 1. If Y admits a simply connected covering space, then Y is SLSC.


Our goal in this section is to prove the following theorem:

Theorem 2. Suppose Y is path connected, locally path connected, and SLSC. Thenthere exists a group G and a regular G-cover p : X → Y such that X is simplyconnected.

[Necessarily, G will be isomorphic to π1(Y ).]

The first step is to figure out what X has to look like.

Lemma 7. (a) If U satisfies the condition in Definition 3, then every smallerneighborhood of y also satisfies the condition.

(b) Suppose Y is SLSC and locally path connected. Let B be the collection of path-connected open sets U such that π1(U) → π1(Y ) is the trivial homomorphismfor some (or every) basepoint in U . Then B is a basis for Y .

(c) For any U ∈ B let β1, β2 be paths in U with β1(0) = β2(0) and β1(1) = β2(1).Then [β1] = [β2] in the fundamental groupoid of Y .

Proof. *** Extra credit. *** �

Let Y be a path-connected space with basepoint y0, let G be its fundamentalgroupoid, and let G0 ⊆ G be the subset consisting of path classes starting at y0.Thus G0 is a set whose elements are equivalence classes a = [α], where α is a pathin Y with α(0) = y0. Let p : X → Y be a covering map, and let x0 ∈ p−1(y0) be a

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basepoint. Then one can use path lifting to define a function h : G0 → X. Namely,given [α] ∈ G0, we lift α to a path α in X starting at x0 and set

h([α]) = α(1).

Lemma 8. (a) h is well defined.(b) h is surjective if X is path connected and bijective if X is simply connected.


From now on we assume that Y is path connected, locally path connected, andSLSC. We have just seen that if Y admits a simply connected covering space X,then the points of X must be in 1–1 correspondence with the elements of the set G0.With this as motivation, we will prove Theorem 2 by starting with G0 and puttinga suitable topology on it. Let B be the basis for Y given in Lemma 7(b).

Let a = [α] be an element of G0. For any U ∈ B with α(1) ∈ U , let Ua ⊆ G0 bethe set of path classes of the form a ∗ b, where b = [β] for some path β in U with

β(0) = α(1). We will call Ua a basic neighborhood of a, and we will call a subsetof G0 open if it contains a basic neighborhood of each of its points.

Lemma 9. The open sets just defined form a topology on G0, with the sets Ua asa basis.


[Hint: It might help to show that if a1 ∈ Ua, then Ua1= Ua.]

As an aid to the intuition and a reminder of what we’re trying to do, we willdenote by X the set G0 with the topology that we have just defined. Let p : X → Ybe defined by p([α]) = α(1).

Lemma 10. p is continuous and open and maps Ua homeomorphically onto U .


Lemma 11. Suppose a1 and a2 are classes of paths starting at y0 and having acommon endpoint y. Let U be a member of B such that y ∈ U . If a1 6= a2, then

Ua1and Ua2

are disjoint.


Lemma 12. p is a covering map.


Given a path α starting at y0 and given t ∈ [0, 1], let αt be the path given byαt(s) = α(ts) for 0 ≤ s ≤ 1. You may find it helpful to describe by means of wordsand/or pictures the path t 7→ [αt] in X. [You don’t need to write this down.]

Lemma 13. The path t 7→ [αt] in X is indeed a path, i.e. it is continuous.


Note that this lemma shows you how to lift any path α in Y starting at y0 to apath α in X starting at x0 := [ey0

].

Lemma 14. X is path connected.

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Lemma 15. X is simply connected.


[Hint: If [α] ∈ π1(Y, y0) is nontrivial, show that its lift α is not a closed curve;deduce the result from this by covering space theory.]

Lemma 16. There is an action of π1(Y, y0) on X that makes p a regular π1(Y, y0)-cover.


This completes the proof of Theorem 2.

Remark 2. It is possible to show that X covers every other path-connected coveringspace of Y . [One can use Lemma 8.] For this reason X is often called the universalcover of Y .

This is the end of the exam unless you want to read the next sectionfor fun.

7. Cayley graphs and the figure 8

This (optional) section assumes a little more knowledge of group theory than Ihave assumed previously.

Let G be a group with a generating set S. The Cayley graph of (G,S) is thegraph Γ = Γ(G,S) with G as its vertex set and with an edge joining g and gs foreach g ∈ G and s ∈ S. For example, the Cayley graph of the free group F2 withits standard generating set is the snowflake. A second example is shown in thefollowing picture; see if you can guess what G and S are. [Hint 1: The number ofvertices is the order of the group. Hint 2: Each generator s ∈ S is assigned a color;so the number of colors of edges is the number of generators.]

The left-translation action of G on itself induces an action of G on the graph Γand hence an action of G on the geometric realization X := |Γ|. This action isfixed-point free and makes X a regular G-cover of Y := X/G. If G = F2 with itsstandard generating set, X is a tree and hence is contractible. In particular, it issimply connected. And Y is the figure 8 in this case. Theorem 1 therefore impliesthat the fundamental group of the figure 8 is free of rank 2, as claimed.

Mathematics 4530

Prelim 2

Brief solutions

1. Introduction

A crucial feature of the exponential covering p : R→ S1 is that two points in thesame fiber differ by an integral multiple of 2π. This was used in the definition of thedegree homomorphism π1(S1)→ Z, which was then proved to be an isomorphism.We wish to generalize this to other covering maps, the so-called regular ones. As aconsequence, one can calculate many fundamental groups with no more effort thanwas required for S1.

2. Group actions

We have defined a group to be a set G together with a binary operation ∗satisfying certain axioms. This may give you the wrong impression of group theory,in that it gives you no clue as to how groups arise in nature. The way groups arisein nature is that they act on things, thereby exhibiting the symmetry that thosethings have.

Definition 1. Let G be a group and X a topological space. By an action of Gon X we mean a function G×X → X, denoted (g, x) 7→ g ∗ x, satisfying:

(1) g ∗ (h ∗ x) = (g ∗ h) ∗ x for all g, h ∈ G and x ∈ X.(2) 1 ∗ x = x for all x ∈ X, where 1 is the identity element of G.(3) For each g ∈ G, the map x 7→ g ∗ x is continuous.

It follows that the map x 7→ g ∗ x, which we will call the action of g, is actually ahomeomorphism. Its inverse is given by the action of g−1. Group actions are studiedextensively in algebra courses, such as Math 4340. For our purposes, however, itsuffices to have some examples.

Example 1. The additive group Z acts on R by (n, x) 7→ 2πn + x for n ∈ Z andx ∈ R. (We say “Z acts on R by translation.”) This is the action that’s implicit inour work on the exponential covering map, which led to the calculation π1(S1) ∼= Z.

Example 2. The multiplicative group {±1} acts on S2 by scalar multiplication.This action is related to the double covering S2 → P 2 mentioned in class, whereP 2 is the projective plane.

Example 3. More generally, whenever we have a 2-fold covering map p : X → Y ,there is an associated action of a group of order 2 on X. This is essentially whatyou proved in additional problem 6 on Assignment 10.

Example 4. Let Zn be the additive group of integers mod n. There is an action ofZn on S1, in which the integer k mod n acts as rotation through 2πk/n radians.Alternatively, if we replace Zn by the isomorphic group consisting of the nth rootsof unity in C, then we can describe the action using multiplication of complexnumbers.

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Example 5. The concept of “Cayley graph,” which I will explain in Section 7, leadsto many examples of group actions. For example, the snowflake that covers thefigure 8 is the Cayley graph of the free group F2; as a result, there is an action ofF2 on the snowflake.

To simplify the notation in what follows, we will always use multiplicative no-tation for our groups (i.e., we will write gh instead of g ∗ h), and similarly we willwrite (g, x) 7→ gx for group actions.

Every one of the examples above has the property that it is fixed-point free, i.e.,the action of any g 6= 1 has no points x such that gx = x. The significance of thisfor us is that if two points x, x′ are in the same G-orbit, i.e., if there is a g ∈ Gsuch that gx = x′, then the element g is unique. [If gx = x′ and hx = x′, thengx = hx =⇒ h−1gx = x =⇒ h−1g = 1 =⇒ g = h.] We can think of g assomething like the “difference” between x and x′. For the case of Z acting on R asin Example 1, it really is the difference (divided by 2π).

3. Regular G-covers

Definition 2. Let G be a group. A regular G-cover is a covering map p : X → Ytogether with a fixed-point-free action of G on X whose orbits are the fibers of p.

Let’s spell this out explicitly:

Lemma 1. Let p : X → Y be a covering map with an action of a group G on X.Then p is a regular G-cover if and only if it satisfies the following two conditions:

(a) p(gx) = p(x) for all g ∈ G, x ∈ X.(b) If p(x) = p(x′) then there is a (unique) g ∈ G such that x′ = gx.

Sketch of proof. (a) says that every orbit is contained in a fiber. (b) [except for theuniqueness part] says that, conversely, every fiber is contained in an orbit. So theorbits are the fibers. It remains to sort out the uniqueness part of (b). We showedabove that if the action is fixed-point free, then uniqueness holds in (b). Conversely,suppose the uniqueness assertion in (b) holds. If gx = x, then gx = 1x =⇒ g = 1,so the action is fixed-point free. �

Remark 1. The group action is useful in connection with path liftings. Namely, ifα is a path in Y and α is the lift starting at some point in the fiber over α(0), thenwe can construct all the other liftings of α by applying the group action. Indeed,if α starts at x, let x′ be any other point in the same fiber. Then x′ = gx for someg ∈ G, and gα is the lift of α starting at x′. Here (gα)(t) := gα(t) for t ∈ I; it is alift of α by Lemma 1(a).

We have already seen many examples of regular G-covers. Indeed, every examplein the previous section has an associated regular G-cover, which was mentionedexplicitly except in Example 4. On the other hand, there are many covering mapsthat are not regular because they do not have enough symmetry. Indeed, conditions(a) and (b) imply, intuitively, that a regular G-cover looks the same no matter wherewe stand in a given fiber. For example:

Lemma 2. Let p : X → Y be a regular G-cover for some group G, and let α be aclosed curve in Y that admits a closed lift α. Then every lift of α is closed.

Sketch of proof. This is immediate from Remark 1, since gα is closed if α is closed.�

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One can deduce from this that not all covering maps are regular.

Example 6. In class we saw a 3-fold cover of the figure 8 by a union of 4 circles. Itwas explicitly pointed out that there are closed curves in the figure 8 with a closedlift at one point but a non-closed lift at another. So that cover is not regular.

4. The generalized degree homomorphism

Recall that we defined the degree of a closed curve α in S1 by lifting it to R andmeasuring the difference between the starting and ending points of the lift. We cando something similar for any regular cover.

Let p : X → Y be a regular G-cover. Pick a basepoint y ∈ Y and a basepointx ∈ p−1(y). Given a path homotopy class [α] ∈ π1(Y, y), let α be the lift of αstarting at x. Then α ends in the fiber p−1(y), so there is a unique g ∈ G such thatα(1) = gx. We set

δ([α]) := g.

In case p is the exponential covering of S1, this is precisely the degree defined inclass. Imitating the four steps that occurred in our calculation of π1(S1), we willprove:

Theorem 1. Suppose that X is simply connected. Then

δ : π1(Y, y)→ G

is an isomorphism.

The four steps are given in the following four lemmas, valid for any regularG-covering.

Lemma 3. δ is well defined.

Sketch of proof. If two closed curves α, β at y are path-homotopic, then their liftsstarting at x end at the same point by the homotopy lifting theorem. It followsthat δ(α) = δ(β), so δ is well-defined on path-homotopy classes. �

Lemma 4. δ is a group homomorphism.

Sketch of proof. Given [α], [β] ∈ π1(Y, y), let γ := α ∗ β. To compute the lift ofγ at x, we need to lift α starting at x and then lift β starting where the first liftfinished. Let α, β be the lifts starting at x. Then α(1) = gx and β(1) = hx,

where g := δ(α) and h := δ(β). The lift of β starting at the end of α is gβ (see

Remark 1), so the lift of γ starting at x is α ∗ gβ. This ends at gβ(1) = ghx. Thusδ(γ) = gh = δ(α)δ(β). �

Lemma 5. If X is path connected, then δ is surjective.

Sketch of proof. Given g ∈ G, choose a path in X from x to gx. Composing with p,we obtain a closed curve α in Y whose lift α at x is the original path from x to gx.Then δ([α]) = g. �

Lemma 6. If X is simply connected, then δ is bijective.

Sketch of proof. Suppose δ([α]) = δ([β]). Then the lifts α, β starting at x end at

the same point. Since X is simply connected, it follows that α 'p β. Composingwith p, we obtain α 'p β, i.e., [α] = [β]. �

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5. Examples

Example 7. Combining Theorem 1 and Example 2, we deduce that the fundamentalgroup of P 2 is the group of order 2, since we know that S2 is simply connected.

Example 8. Applying Theorem 1 to the regular Z × Z covering of the torus byR× R, we deduce that the fundamental group of the torus is Z× Z.

Example 9. We proved in class, using the Seifert–van Kampen theorem, that thefundamental group of the figure 8 is the free group F2 on two generators. A differentproof can be obtained by applying Theorem 1 to the snowflake cover of the figure 8.I will sketch how this can be done in Section 7.

6. Existence of regular G-covers

It is natural to ask how widely applicable Theorem 1 is. In this section we willsee that, in principle, Theorem 1 can be used to calculate the fundamental groupof every reasonable space. Here “reasonable” has a somewhat technical definition:

Definition 3. A space Y is said to be semilocally simply connected (SLSC) ifevery point y ∈ Y has a neighborhood U such that the inclusion U ↪→ Y inducesthe trivial homomorphism π1(U, y)→ π1(Y, y).

This condition may seem strange; but in fact, most spaces that one meets “innature” satisfy the even stronger condition of being locally contractible. Here’smore evidence that the SLSC condition is reasonable:

Proposition 1. If Y admits a simply connected covering space, then Y is SLSC.

Sketch of proof. Let p : X → Y be a covering map with X simply connected. Giveny ∈ Y , let U be an evenly-covered neighborhood of y and let V be one of the slicesof p−1(U). Then the inclusion i : U ↪→ Y factors as a composite

U → V ↪→ X → Y,

so i∗ factors through the trivial group and is therefore trivial. �

Our goal in this section is to prove the following theorem:

Theorem 2. Suppose Y is path connected, locally path connected, and SLSC. Thenthere exists a group G and a regular G-cover p : X → Y such that X is simplyconnected.

[Necessarily, G will be isomorphic to π1(Y ).]

The first step is to figure out what X has to look like.

Lemma 7. (a) If U satisfies the condition in Definition 3, then every smallerneighborhood of y also satisfies the condition.

(b) Suppose Y is SLSC and locally path connected. Let B be the collection of path-connected open sets U such that π1(U) → π1(Y ) is the trivial homomorphismfor some (or every) basepoint in U . Then B is a basis for Y .

(c) For any U ∈ B let β1, β2 be paths in U with β1(0) = β2(0) and β1(1) = β2(1).Then [β1] = [β2] in the fundamental groupoid of Y .

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[Extra credit.]

Let Y be a path-connected space with basepoint y0, let G be its fundamentalgroupoid, and let G0 ⊆ G be the subset consisting of path classes starting at y0.Thus G0 is a set whose elements are equivalence classes a = [α], where α is a pathin Y with α(0) = y0. Let p : X → Y be a covering map, and let x0 ∈ p−1(y0) be abasepoint. Then one can use path lifting to define a function h : G0 → X. Namely,given [α] ∈ G0, we lift α to a path α in X starting at x0 and set

h([α]) = α(1).

Lemma 8. (a) h is well defined.(b) h is surjective if X is path connected and bijective if X is simply connected.

Sketch of proof. (a) Use the homotopy-lifting theorem. (b) Suppose X is path-connected. Given x ∈ X, choose a path from x0 to x, and let α be its image in Y .Then α is the original path from x0 to x, hence h([α]) = x and h is surjective.

Suppose X is simply connected. If h([α]) = h([β]), then α(1) = β(1), hence α 'p β.This implies that α 'p β, i.e., [α] = [β]. �

From now on we assume that Y is path connected, locally path connected, andSLSC. We have just seen that if Y admits a simply connected covering space X,then the points of X must be in 1–1 correspondence with the elements of the set G0.With this as motivation, we will prove Theorem 2 by starting with G0 and puttinga suitable topology on it. Let B be the basis for Y given in Lemma 7(b).

Let a = [α] be an element of G0. For any U ∈ B with α(1) ∈ U , let Ua ⊆ G0 bethe set of path classes of the form a ∗ b, where b = [β] for some path β in U with

β(0) = α(1). We will call Ua a basic neighborhood of a, and we will call a subsetof G0 open if it contains a basic neighborhood of each of its points.

Lemma 9. The open sets just defined form a topology on G0, with the sets Ua asa basis.

[Extra credit.]

As an aid to the intuition and a reminder of what we’re trying to do, we willdenote by X the set G0 with the topology that we have just defined. Let p : X → Ybe defined by p([α]) = α(1).

Lemma 10. p is continuous and open and maps Ua homeomorphically onto U .

[Extra credit.]

Lemma 11. Suppose a1 and a2 are classes of paths starting at y0 and having acommon endpoint y. Let U be a member of B such that y ∈ U . If a1 6= a2, then

Ua1 and Ua2 are disjoint.

[Extra credit.]

Lemma 12. p is a covering map.

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[Extra credit.]

Given a path α starting at y0 and given t ∈ [0, 1], let αt be the path given byαt(s) = α(ts) for 0 ≤ s ≤ 1. You may find it helpful to describe by means of wordsand/or pictures the path t 7→ [αt] in X. [You don’t need to write this down.]

Lemma 13. The path t 7→ [αt] in X is indeed a path, i.e. it is continuous.

[Extra credit.]

Note that this lemma shows you how to lift any path α in Y starting at y0 to apath α in X starting at x0 := [ey0

].

Lemma 14. X is path connected.

[Extra credit.]

Lemma 15. X is simply connected.

Sketch of proof. If [α] ∈ π1(Y, y0) is nontrivial, then the lift α of α starting at x0is given by t 7→ [αt]. This ends at [α], which is not x0 since [α] is nontrivial. Thusα is not closed. This proves that the homomorphism p∗ : π1(X,x0) → π1(Y, y0) isthe trivial homomorphism. But we know that this homomorphism is injective, sothe domain π1(X,x0) must be the trivial group. �

Lemma 16. There is an action of π1(Y, y0) on X that makes p a regular π1(Y, y0)-cover.

[Extra credit.]

This completes the proof of Theorem 2.

Remark 2. It is possible to show that X covers every other path-connected coveringspace of Y . [One can use Lemma 8.] For this reason X is often called the universalcover of Y .

7. Cayley graphs and the figure 8

This (optional) section assumes a little more knowledge of group theory than Ihave assumed previously.

Let G be a group with a generating set S. The Cayley graph of (G,S) is thegraph Γ = Γ(G,S) with G as its vertex set and with an edge joining g and gs foreach g ∈ G and s ∈ S. For example, the Cayley graph of the free group F2 withits standard generating set is the snowflake. A second example is shown in thefollowing picture; see if you can guess what G and S are. [Hint 1: The number ofvertices is the order of the group. Hint 2: Each generator s ∈ S is assigned a color;so the number of colors of edges is the number of generators.]

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The left-translation action of G on itself induces an action of G on the graph Γand hence an action of G on the geometric realization X := |Γ|. This action isfixed-point free and makes X a regular G-cover of Y := X/G. If G = F2 with itsstandard generating set, X is a tree and hence is contractible. In particular, it issimply connected. And Y is the figure 8 in this case. Theorem 1 therefore impliesthat the fundamental group of the figure 8 is free of rank 2, as claimed.

Mathematics 4530 Final Exam December 18, 2013

Do any 5 of the following 8 problems, and indicate clearly which 5 you have chosen.You may use any result proved in class, in the assigned reading, or in the homework,unless the problem asks you to prove such a result.

1. True or false. Justify each answer with a proof or counterexample.

(a) If X is a simply connected space and A is a path connected subspace, then Ais simply connected.

(b) If α : I → X is a path in a Hausdorff space X, then the image of α is a closedsubset of X.

(c) Let X be a topological space and A a subspace, and let i : A ↪→ X be theinclusion map. Then the induced homomorphism i∗ : π1(A, a0) → π1(X, a0) isinjective for any choice of basepoint a0 ∈ A.

(d) If U and V are path connected open subsets of a space X, then their intersectionis path connected.

2. Let p : X → X be a covering map. Let x0 be a basepoint in X, and let x0 :=p(x0). Suppose the following condition holds: For any closed curve α in X at x0such that [α] is nontrivial in π1(X,x0), its lift α starting at x0 is not closed. Show

that the group π1(X, x0) is trivial.

3. Given a topological space X, we define the connected components of X to bethe equivalence classes determined by the following relation: x ∼ y if and only ifthere is a connected subset of X containing both x and y.

(a) Prove that this relation is in fact an equivalence relation.(b) Prove that every connected component is connected.

4. Let q : X → Y be a quotient map. State and prove a universal mapping propertythat characterizes continuous maps from Y to an arbitrary space Z.

5. Let X be a locally compact Hausdorff space whose one-point compactification ismetrizable. Show that X is the union of an increasing sequence of compact subsetsK1 ⊆ K2 ⊆ K3 ⊆ · · · .

6. (a) What does it mean to say that a topological space is regular?(b) What does it mean to say that a topological space is normal?(c) State Urysohn’s lemma and Tietze’s extension theorem.(d) Let X be a metric space and let (xn)n≥1 be an infinite sequence in X such that

d(xi, xj) ≥ 1 for i 6= j. Prove that there is a continuous map f : X → R suchthat f(xn) = n for all n.

7. (a) Define covering map.

(b) Suppose p : X → X is a covering map with X and X path connected. If X issimply connected, show that p is a homeomorphism.

8. (a) State the Seifert–van Kampen theorem, with hypotheses and conclusionspelled out precisely.

(b) Give an example of a space whose fundamental group is a free group on 3generators. Justify your answer as completely as you can; in particular, spellout explicitly three closed curves that represent the 3 generators.

Very Brief Solutions to Final Exam

1. (a) False.(b) True.(c) False.(d) False.

2. See the take-home prelim.

3. (a) Reflexivity and symmetry are obvious. Transitivity follows from the factthat the union of two connected sets with nonempty intersection is connected.

(b) The connected component containing x is the union of all the connected setscontaining x.

4. See Munkres.

5. The point at infinity has a countable neighborhood base.

6. (a) See Munkres.(b) See Munkres.(c) See Munkres.(d) The set A := {x1, x2, . . . } is closed and has the discrete topology. Apply the

Tietze extension theorem to the continuous map f : A→ R given by f(xn) := n.

7. (a) See Munkres.(b) p is an open surjection, so it suffices to prove that it’s injective. If p(x1) = p(x2),

choose a path α from x1 to x2. Its image α := p(α) is path homotopic to aconstant path, whose lifts are closed, so the lift α is also closed. Thus x1 = x2.

8. (a) See Munkres.(b) See additional problem 1 on assignment 13. Or consider a rose with three

petals.

Mathematics 4530-Fall 2013

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