Homotopy theory lecture notes

Dr. Thomas Baird

Fall 2011

Contents

1 Review of Point-Set Topology 21.1 New spaces from old. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Covers and Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Connectedness and Path-Connectedness . . . . . . . . . . . . . . . . . . . 51.4 Hausdorff spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.5 Metric spaces and the Lebesgue number lemma . . . . . . . . . . . . . . . 6

2 The Fundamental Group 72.1 The Fundamental Group of a Circle . . . . . . . . . . . . . . . . . . . . . . 9

2.1.1 Fundamental Theorem of Algebra . . . . . . . . . . . . . . . . . . . 112.2 Induced Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.2.1 The Brouwer Fixed Point Theorem . . . . . . . . . . . . . . . . . . 132.3 The Fundamental Group of Sn . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.3.1 The Borsuk-Ulam Theorem . . . . . . . . . . . . . . . . . . . . . . 14

3 General Concepts in Homotopy Theory 153.1 Homotopy and Homotopy Equivalence . . . . . . . . . . . . . . . . . . . . 15

3.1.1 Homotopy invariance of π1 . . . . . . . . . . . . . . . . . . . . . . . 163.1.2 Deformation Retractions . . . . . . . . . . . . . . . . . . . . . . . . 17

4 Van Kampen’s Theorem 184.1 Free Products of Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

4.1.1 Free Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.1.2 Presentations of Groups . . . . . . . . . . . . . . . . . . . . . . . . 19

4.2 Van Kampen’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.3 The effect of attaching cells . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5 Cell complexes 255.1 Homotopy extension property and subcomplexes . . . . . . . . . . . . . . . 275.2 Cell complex propaganda . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305.3 Higher homotopy groups and Whitehead’s Theorem . . . . . . . . . . . . . 31

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6 Covering Spaces 336.1 The Classification of Covering Spaces . . . . . . . . . . . . . . . . . . . . . 37

1 Review of Point-Set Topology

In this section we collect some basic facts from general topology that will be required inthis course. Proofs of these results can be found in any introductory textbook on generaltopology (e.g. Munkres’ Topology) or in the point-set topology notes I have posted onD2L.

Definition 1. A topological space (or simply space) (X, τ) is a set X and a collectionτ of subsets of X, called the open sets, satisfying the following conditions:

i) ∅ and X are open,ii) Any union of open sets is open,iii) Any finite intersection of open sets is open.

A set is called closed if its complement is open. Usually, we will denote the topologicalspace (X, τ) simply by X.

Example 1 (Euclidean Topology). An open ball in Rn is a set of the form

B = Bε(p) := x ∈ Rn| ||x− p|| < ε

for some p ∈ Rn and ε > 0. A subset U ⊆ Rn is called open if it is a union of openballs. Equivalently, U is open if for every p ∈ U , there exists an open ball B such thatp ∈ B ⊂ U .

In the example above, we say that open balls form a basis for the Euclidean topology.More generally, a collection of open sets B in a topological space X is called a basis ifevery other open set in X is a union of sets in B.

Definition 2. A continuous map f : X → Y between topological spaces is a map ofsets for which pre-images of open sets are open. I.e.

U ⊆ Y is open ⇒ f−1(U) := x ∈ X|f(x) ∈ U ⊆ X is open

Definition 3. A homeomorphism is a continuous bijection f : X → Y such that theinverse f−1 is also continuous. This is the notion of isomorphism for topological spaces.

We will only rarely need use the abstract Definition 2 explicitly. More often, we willmake use of certain properties of continuous functions, including the following.

Proposition 1.1. Let X, Y and Z be topological spaces.

• The identity map IdX : X → X is continuous.

• If f : X → Y and g : Y → Z are continuous, then the composition g f : X → Zis continuous.

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• Any constant map f : X → Y is continuous.

The first two conditions above make topological spaces + continuous maps into acategory. We will speak more about categories later.

1.1 New spaces from old.

Almost all of the topological spaces we encounter in this course are constructed from Rn

using the operations below.

Definition 4. Let X be a topological space and A ⊂ X a subset. The subspace topol-ogy on A is the topology for which V ⊂ A is open if and only if V = A ∩ U for someopen set U in X.

Example 2. Any subset of Rn acquires a subspace Euclidean topology. Unless otherwisestated, we will always assume subsets of Rn to have this topology.

The inclusion map i : A → X is continuous (with respect to the subspace topology).In fact, we have the following special property: A map f : Y → A from a topologicalspace Y is continuous if and only if the composition i f : Y → X is continuous.

Definition 5. The product space X×Y of two spaces X and Y is the Cartesian productof sets X × Y , with a basis of open sets of the form U × V where U ⊂ X and V ⊂ Y areboth open.

The above definition iterates to define products of any finite number of spaces (infiniteproducts require a different definition).

Example 3. The n-fold product R×R×...×R is homeomorphic to Rn with the Euclideantopology.

The key property of product spaces is that a map

F : Z → X × Y

is continuous if and only if the coordinate functions F = (F1, F2) are continuous as mapsfrom Z to X and to Y respectively.

Definition 6. Let Xα be a (possibly infinite) collection of spaces indexed by α. Thecoproduct space or disconnected union

∐αXα is the disjoint union of the sets Xα

with U ⊆∐

αXα is open if and only if U ∩Xα is open for all α.

The inclusions iα0 : Xα0 →∐Xα are all continuous. A map F :

∐Xα → Y is

continuous if and only if the composition F iα : Xα → Y are continuous for all α.

Definition 7. An equivalence relation on a set X is a relation ∼ satisfying, for allx, y ∈ X

(i) x ∼ x

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(ii) x ∼ y implies y ∼ x

(iii) x ∼ y and y ∼ z implies x ∼ z.

Given any relation R on X, we can generate the “smallest” equivalence relation ∼Rsuch that xRy implies x ∼R y. Explicitly, we define x ∼R y if and only if there exists afinite sequence xi ∈ Xni=0 for n ≥ 0 satisfying

x0 = x,

xn = y and,

xiRxi−1 or xi−1Rxi for all i = 1, ..., n.

Given x ∈ X, the equivalence class of x is

[x] := y ∈ X|x ∼ y

Notice that [x] = [y] if and only if x ∼ y. The equivalence classes determine a partitionof X into disjoint sets. Let E := [x]|x ∈ X be the set of equivalence classes (we willsometimes denote E = X/ ∼ ) . There is a canonical map

Q : X → E, x 7→ [x]

called the quotient map.

Definition 8. Let X be a topological space and let ∼ be an equivalence relation on theset underlying X. The quotient topology on E is the topology for which U ⊂ E isopen if and only if Q−1(U) is open in X.

Observe that Q : X → E is continuous and that a map f : E → Y is continuous ifand only if f Q : X → Y is continuous.

Example 4. Suppose X and Y are topological spaces, A ⊆ X is a subspace, and f : A→Y is a continuous map. Define an equivalence relation on the coproduct X

∐Y generated

by f(a) ∼ a for all a ∈ A. We say that the quotient space (X∐Y )/ ∼ is obtained by

attaching X to Y along A using f .

Example 5. Suppose X is a topological space, and G is a group that acts on X viahomeomorphisms. Define two points in X equivalent if they lie in the same orbit of G.The quotient space in this case is called the orbit space and is denoted X/G.

1.2 Covers and Compactness

Definition 9. An open (closed) cover of a topological space X is a collection of open(closed) sets Uα such that the union ∪αUα = X.

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Proposition 1.2. Let Uα be either an open cover or a finite closed cover of X. A mapof sets

f : X → Y

between topological spaces is continuous if and only if the restrictions f |Uα : Uα → Y arecontinuous for all α (where Uα has the subspace topology).

The preceding proposition will be used in two ways: to test if a map f is continuous byconsidering the restrictions, and also to construct a map f by gluing together continuousmaps defined on the Uα that agree on overlaps.

Definition 10. A space X is called compact if every open cover Uα of X containsa finite subcover. I.e., there exists a finite collection U1, ..., Un ⊆ Uα such that∪ni=1Ui = X.

Proposition 1.3. A subspace of Rn is compact if and only if it is closed and bounded.

Compactness is preserved under the following:

• A closed subspace of a compact space is compact.

• A finite union of compact spaces is compact.

• A product of compact spaces is compact.

• If f : X → Y is continuous and X is compact, then the image f(X) ⊆ Y is compact.

1.3 Connectedness and Path-Connectedness

Let I denote the unit interval [0, 1] ⊂ R with the Euclidean topology.

Definition 11. A space X is called path-connected if for any two points p, q ∈ X thereexists a continuous map γ : I → X such that γ(0) = p and γ(1) = q.

Definition 12. A space X is called connected if there is no proper subset A ⊂ Xwhich is both open and closed. (proper means other than X or ∅, which are always bothopen and closed).

Observe that if A ⊂ X is both open and closed, then the complement Ac is also bothopen and closed, and there is a natural isomorphism A

∐Ac ∼= X. Thus spaces that are

not connected can be decomposed into a disconnected union of nonempty spaces.

Proposition 1.4. Path-connected spaces are connected.

The converse of Proposition 1.4 is not true in general. However all the connectedspaces we encounter in this course will also be path-connected.

Connectedness and path-connectedness are preserved under the following operations

• A product of (path-)connected spaces is (path-)connected.

• The continuous image of a (path-)connected space is (path-)connected.

• Let Uα be a covering of X such that each Uα is (path-)connected and the inter-section ∩αUα is non-empty. Then X is (path-)connected.

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1.4 Hausdorff spaces

Definition 13. A space X is called Hausdorff if for any pair of distinct points p, q ∈ X,there exist open sets U, V such that p ∈ U , q ∈ V and U ∩ V = ∅.

Proposition 1.5. Any metric space is Hausdorff. In particular, any subset of Rn isHausdorff.

The Hausdorff property is preserved under the following:

• Products of Hausdorff spaces are Hausdorff.

• Subspaces of Hausdorff spaces are Hausdorff.

• Coproducts of Hausdorff spaces are Hausdorff.

Definition 14. A space X is locally compact, if every point p ∈ X is contained in anopen neighbourhood p ∈ U such that the closure U is compact.

For example, compact spaces are locally compact. Also, Rn is locally compact.

Proposition 1.6. Let X be a locally compact space and let Y be a Hausdorff space. Ifφ : X → Y is a continuous map that is also a bijection of sets, then φ is a homeomorphism.

1.5 Metric spaces and the Lebesgue number lemma

Definition 15. Let X be a set. A metric on X is a function

d : X ×X → R≥0

called the distance or metric function, satisfyingi) d(x, x′) = 0 ⇔ x = x′ (d separates points)ii) d(x, x′) = d(x′, x) (d is symmetric)iii) d(x, x′′) ≤ d(x, x′) + d(x′, x′′) (the triangle inequality)

A metric space (X, d) determines a metric topology on X, which is generated bythe basis of open balls Bε(p) = x ∈ X|d(x, p) < ε. If A is a subset of a metric space Xthen A becomes a metric space by restriction. The metric topology on A is the same asthe subspace topology on A.

The following result will come up repeatedly.

Lemma 1.7 (Lebesgue number Lemma). Let A be an open covering of a compact metricspace X. There exists δ > 0, called the Lebesgue number, such that for all p ∈ X, theopen ball Bδ(p) is contained in some U ∈ A.

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2 The Fundamental Group

Definition 16. LetX be a space and I the unit interval [0, 1] ⊂ R. A path is a continuousmap f : I → X. A homotopy of paths is a family of paths ft : I → X, with parametert ∈ I = [0, 1], such that

1. The endpoints ft(0) = x0 and ft(1) = x1 are independent of t.

2. The associated map F : I × I → X, defined by F (s, t) := ft(s) is continuous.

When two paths f0 and f1 are related by a homotopy ft, we say that f0 and f1 arehomotopic.

Example 6. Let X ⊂ Rn be a convex set (i.e. X contains the line segment joiningany two points in X). Then any two paths f0 and f1 with common end-points in X arehomotopic, via the homotopy ft = (1− t)f0 + tf1.

Proposition 2.1. The relation of homotopy on the set of paths with fixed end-pointsx0, x1 ∈ X is an equivalence relation.

Proof. Clearly any path f is homotopic to itself, via the constant homotopy ft = f . Also,if f0 is homotopic to f1 via ft, then f1 is homotopic to f0 via f1−t.

More interesting is transitivity. Let ft be a homotopy between f0 and f1, let gt be ahomotopy between g0 and g1 and suppose that f1 = g0. Then we must prove that f0 ishomotopic to g1. Define a homotopy

ht :=

f2t for t ∈ [0, 1/2]

g2t−1 for t ∈ [1/2, 1]

Then ht is a well-defined family of paths joining f0 to g1 and determines a continuousmap H : I × I → X because it restricts to continuous functions on the closed subsetsI × [0, 1/2] and I × [1/2, 1] (see Proposition 1.2).

Definition 17. Let f and g be two paths in X such that f(1) = g(0). The compositionor product of paths f · g is the path

(f · g)(s) :=

f(2t) for s ∈ [0, 1/2]

g(2s− 1) for s ∈ [1/2, 1]

Observe that composition respects homotopies, in the sense that if ft and gt arehomotopies, with ft(1) = gt(0), then ft · gt determines a homotopy between f0 · g0 andf1 · g1 (prove this).

Composition is particularly interesting for paths f with a single end-point:

f(0) = f(1) = x0.

Such a path is called a loop based at x0. Let π1(X, x0) denote the set of equivalenceclasses of loops based at x0, modulo homotopy.

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Proposition 2.2. The set π1(X, x0) forms a group under composition of loops [f ][g] =[f · g]. This is called the fundamental group of X at x0.

Proof. Composition of loops based at x0 is well defined, because end-points always agree.That this is well-defined on homotopy equivalence classes has already been explained.Thus composition defines a well-defined operation

π1(X, x0)× π1(X, x0)→ π1(X, x0)

and it remains to check the group axioms.Given a path f , define a reparametrization of f to be a composition f φ, where φ

is any continuous map φ : I → I such that φ(0) = 0 and φ(1) = φ(1). Then a path f ishomotopic to f φ by the homotopy

ft(s) = f(tφ(s) + (1− t)s).

First we show that the product is associative, i.e that ([f ][g])[h] = [f ]([g][h]). Thisfollows from the fact that (f · g) · h is sent to f · (g · h) under the reparametrization

φ(s) :=

s/2 for s ≤ 1/2

3s/2− 1/2 for s ≥ 1/2

The identity element is represented by the constant map e(s) = x0, because for any basedloop f ,

(f · e) = f φ

under the reparametrization

φ(s) :=

2s for s ≤ 1/2

1 for s ≥ 1/2.

A similar argument works to show [e][f ] = [f ] as well.Finally, given a path f define f−1 by f−1(s) = f(1 − s). For a loop f consider the

homotopy ft · f−1t where ft(s) = f(s(1 − t)). Then ft · f−1

t determines a continuoushomotopy between f0 · f−1

0 = f · f−1 and f1 · f−11 = e. Thus

[f ][f−1] = [f · f−1] = [e],

so π1(X, x0) contains inverses.

Example 7. If X ⊂ Rn is convex, then π1(X, x0) = 0 is trivial, because every loop ishomotopic to the constant loop (see Example 6).

The next result says that for path-connected spaces, the fundamental group is inde-pendent of the base point.

Proposition 2.3. Let h be a path in X with h(0) = x0 and h(1) = x1. Then the mapβh : π1(X, x0)→ π1(X, x1) defined by βh([f ]) = [h · f · h−1] is an isomorphism of groups.

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Proof. If ft is a homotopy of loops based at x0, then h · ft · h−1 is a homotopy of loopsbased at x1, so βh is well defined. Furthermore, βh is a homomorphism, because

βh([f ])βh([g]) = [h · f ·h−1][h · g ·h−1] = [h · f ·h−1 ·h · g ·h−1] = [h · f · g ·h−1] = βh([f ][g]).

Finally, βh is invertible with inverse βh−1 because

βh−1(βh([f ])) = [h−1 · h · f · h−1 · h] = [f ]

A consequence of Proposition 2.3 is that for a path connected space X, π1(X, x0) inindependent of the base point x0, and it is common to denote π1(X) = π1(X, x0) in thissituation. A space X is called simply connected if it is path-connected and π1(X) = 0is trivial.

Example 8. Convex sets in Rn are simply connected.

The terminology “simply connected” is motivated by the following fact.

Proposition 2.4. A space X is simply connected if and only if there is a unique homotopyclass of paths connecting any pair of points in X.

Proof. Path-connectedness is necessary for either side. Now suppose that X is simplyconnected and let γ and η be two paths joining x0 to x1. Then [γ] = [γ · η−1 · η] = [η],the later equality follows from the fact that γ · η−1 is a loop and so must be homotopic tothe constant map. The converse is immediate.

2.1 The Fundamental Group of a Circle

We denote by Sn the unit sphere in Rn+1. In particular the unit circle in R2 is denotedS1. The aim of this section is to prove the following result.

Theorem 2.5. The map ψ : Z → π1(S1) sending n to (the homotopy class of) the loopωn(s) := (cos(2πns), sin(2πns)) is an isomorphism. We call n the winding number of[ωn].

The proof relies on the “covering map” p : R→ S1 defined by p(s) = (cos(2πs), sin(2πs)).We begin with a lemma.

Lemma 2.6. Given a space Y , a map F : Y × I → S1 and a map F : Y × 0 → Rsuch that p F = F |Y×0, then there is a unique map F : Y × I → R extending F and

satisfying p F = F .

Observe that we are abusing notation by denoting any partial lift of F by F . We hopethis doesn’t cause confusion.

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Remark 1. Lemma 2.6 is an example of a homotopy lifting property. We will call Fa lift of F , because of the commuting diagram

Rp

Y × I F //

F

;;wwwwwwwwwS1

.

We may rephrase Lemma 2.6 as: A lift F of F defined on Y ×0 can be extended uniquelyto a lift F defined on Y × I.

Proof of Lemma 2.6. Choose an open covering Uα of S1 such that for each α, p−1(Uα)decomposes into a disjoint union of sets each of which maps homeomorphically onto Uαby p (for example take two arcs covering S1). We begin by showing that for any pointy0 ∈ Y , there is an open neighbourhood y0 ∈ N ⊆ Y for which we can construct a liftF : N × I → R .

By the continuity of F and the definition of the product topology, for every t ∈ I,there exists an open set Nt × (at, bt) containing (y0, t) such that F (Nt × (at, bt)) ⊂ Uαfor some α. These form an open cover of the compact set y0 × I, so there existsa finite subcover Nj × (aj, bj)nj=1. Setting N := ∩ni=1Ni, we may choose a partition0 < t1 < t2 < ... < tm = 1 such that for each i = 1, ...,m,

F (N × [ti−1, ti]) ⊂ Uα

for some α.Now assume inductively that a lift F has been constructed on N × [0, ti] and sup-

pose F (N × [ti, ti+1]) ⊂ Uα. Then by assumption, there exists a subset Uα of R con-taining F (y0, ti) which maps homeomorphically onto Uα by p. After replacing N withN ∩ F−1

N×ti(Uα), we have also that F (N ×ti) ⊂ Uα. We can now extend F by defining

F |N×[ti,ti+1] := p−1 F,

where p−1 is the inverse homeomorphism from Uα to Uα. Clearly this is the uniquelift of F |N×[ti,ti+1] taking values in Uα so it agrees with F |N×[0,ti] on the overlap and itis continuous since it is continuous on the closed cover [0, ti], [ti, ti+1]. Iterating thisprocess completes the construction of F |N×I .

To prove uniqueness, it suffices to consider the case that Y is a point (so Y can beomitted from the notation). Thus, suppose F : I → S1 and F and F ′ are two lifts ofF with F (0) = F ′(0). Then there exists α such that F ([0, t1]) ⊂ Uα. Because [0, t1] isconnected, the images F ([0, t1]) and F ′([0, t1]) are connected sets. Since they intersect,they must lie in must lie in the same open set Uα ⊂ p−1(Uα) that maps homeomorphicallyto Uα. Since p|Uα is injective and pF = pF ′, we deduce that F |[0,t1] = F ′|[0,t1]. Iterating

this process, we get F = F ′.Finally we prove existence for general Y × I. We have shown that it is possible to find

an cover of Y by open sets N such there exists a lift F |N×I , and by uniqueness we knowthat these lifts agree on overlaps (N1 × I) ∩ (N2 × I). Thus these lifts paste together todefine a unique lift F : Y × I → R (by Proposition 1.2).

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Proof of Theorem 2.5. We reinterpret the map ωn : I → S1 as a composition ωn = p ωnwhere

ωn : I → Ris the map ωn(s) = ns, the straight line path from 0 to n. This ωn is the unique lift ofωn sending 0 ∈ I, to 0 ∈ R. All other lifts of ωn have the form ωn + m, for some integerm. Since R is simply connected we have ωn · (ωm + n) is homotopic to ωm+n (since theyhave the same end-points). It follows that

ωn · ωm = p(ωn) · p(ωm + n)) = p(ωn · (ωm + n)) ∼ p(ωm+n) = ωm+n

where ∼ means homotopic, and we deduce that ψ : Z→ π1(S1) is a homomorphism.To see that ψ is surjective, observe that any loop f : I → S1 that is based at (0, 1)

lifts uniquely to a path f : I → R with f(0) = 0 and f(1) = n for some integer n. Thusf is homotopic to ωn so f is homotopic to ωn.

To see that ψ is injective, suppose that ωm is homotopic to ωn for some m 6= n. Thenthere exists a homotopy F : I × I → S1 with F (s, 0) = ωm(s) and F (s, 1) = ωn(s).By Lemma 2.6, this homotopy lifts uniquely to F : I × I → R, with F (s, 0) = ω(s).But because F (1, t) and F (0, t) are constants, this implies that F (0, t) and F (1, t) areconstants, which contradicts the necessity that F (1, 0) − F (0, 0) = m and F (1, 1) −F (0, 1) = n.

2.1.1 Fundamental Theorem of Algebra

We now apply Theorem 2.5 to prove a famous result.

Theorem 2.7 (Fundamental Theorem of Algebra). Every nonconstant polynomialwith coefficients in C has a root in C.

Proof. It suffices to consider monic polynomials of the form p(z) = zn+an−1zn−1 + ...+a0.

Suppose that p(z) has no roots. Then for any real number r ≥ 0, the formula

fr(s) =p(re2πis)/p(r)

|p(re2πis)/p(r)|determines a homotopic family of loops fr : I → S1 based at (1, 0). Observe that f0 isthe constant loop, so fr has winding number 0 for all r ≥ 0.

More generally, consider the family of polynomials the polynomial

pt := zn + t(an−1zn−1 + ...+ a0)

for 0 ≤ t ≤ 1. Choose a real number r0 ≥ 1 + |an−1|+ ...+ |a0|. Then for any z with norm|z| = r0 and t ∈ [0, 1], we have

|zn| = r0|z|n−1 > (t|an−1 + ...+ a0|)|z|n−1 ≥ |tan−1zn−1 + ...+ ta0| = |pt(z)− zn|

so pt(z) 6= 0 on the circle |z| = r0. Thus the formula

fr,t(s) :=pt(re

2πis)/pt(r)

|pt(re2πis)/pt(r)|determines a homotopy of loops between fr,1 = fr and fr,0 = zn/|zn| = ωn. So fr haswinding number n and we conclude that n = 0 and p(z) is constant.

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The fundamental group is well behaved with respect to products spaces.

Proposition 2.8. Let X and Y be topological spaces with base points x0 and y0 respec-tively. Then there is a natural isomorphism of groups

π1(X × Y, (x0, y0)) ∼= π1(X, x0)× π1(Y, y0).

Proof. Recall that a map f : Z → X ×Y is continuous iff the coordinate maps f = (g, h)are continuous as maps g : Z → X and h : Z → Y . Thus a loop in X × Y is equivalentto a pair of loops in X and Y respectively. Similarly a homotopy of loops is the same asa pair of homotopies. Thus we have a natural bijection between π1(X × Y, (x0, y0)) andπ1(X, x0)× π1(Y, y0) which is easily seen to be a group isomorphism.

Example 9. The fundamental group of a n-torus (S1)n = S1 × ...× S1 is Zn.

Example 10. The fundamental group of a cylinder S1 × I is Z.

Example 11. The punctured plane R2 \ (0, 0) is homeomorphic to S1 × R (use polarcoordinates), so π1(R2 \ (0, 0)) ∼= Z.

2.2 Induced Homomorphisms

Let φ : X → Y be a map sending x0 ∈ X to y0 ∈ Y . We use a notational short handφ : (X, x0)→ (Y, y0) in this situation. Then φ induces a homomorphism of groups

φ∗ : π1(X, x0)→ π1(Y, y0)

defined by φ∗([f ]) = [φ∗f ]. This map is well defined, because if ft is a homotopy ofloops based at x0, then φ ft is a homotopy of loops based at y0. We see that φ is ahomomorphism because φ(f · g) = φ(f) · φ(g).

The two basic properties of φ∗ are:

• (φ ψ)∗ = φ∗ ψ∗ for a composition (X, x0)φ→ (Y, y0)

ψ→ (Z, z0).

• The identity map Id : X → X induces the identity map on π1(X).

The proof of these claims is immediate from the definition (i.e. they are true at the levelof loops, even without passing to homotopy classes). These properties can be restated as:π1 defines a functor from the category of “pointed” topological spaces and continuousmaps, to the category of groups and homomorphisms.

Definition 18. Let X be a topological space, and let i : A → X be a subspace. Aretraction of X onto A is a continuous map r : X → A such that r i is the identity onA.

Proposition 2.9. Let i : A → X be a subspace and let r : X → A be a retraction. Ifx0 ∈ A ⊂ X, then i∗ : π1(A, x0)→ π1(X, x0) must be injective.

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Proof. By functoriality, the composition

π1(A, x0)i∗→ π1(X, x0)

r∗→ π1(A, x0) (1)

is the identity map on π1(A, x0) so necessarily i∗ is injective.

Remark 2. We may much stronger consequences from A ⊂ X being a retract. If we let

Γ = ker(r∗) C π1(X),

then (1) implies that π1(X) is a semi-direct product of Γ and π1(A) for which i∗ is factorinclusion.

2.2.1 The Brouwer Fixed Point Theorem

Let Dn := x ∈ Rn|||x|| ≤ 1 denote (closed) unit disk. One of the earliest results inalgebraic topology is the following.

Theorem 2.10 (Brouwer Fixed Point Theorem). Every continuous map h : D2 →D2 has a fixed point. I.e., there exists p ∈ D2 such that h(p) = p.

Proof. We argue by contradiction. Suppose the exists a map h : D2 → D2 without fixedpoints. Then we can define a continuous map r(x) : D2 → S1 defined by setting r(x)equal to the intersection with S1 of the ray based at h(x) and passing through x.

We can see (informally) that r is continuous because small perturbations of x mustproduce small perturbation of r(x). It is clear from the definition that r(x) = x for pointsx ∈ S1 ⊂ D2, so r : D2 → S1 is an example of a retraction. Thus by Proposition 2.9,i∗ : π1(S1)→ π1(D2) is injective. However π1(S1) = Z and π1(D2) = 0 is trivial so this isimpossible.

2.3 The Fundamental Group of Sn

Theorem 2.11. For n ≥ 2, the fundamental group π1(Sn, x0) = 0 is trivial.

Proof. The complement of a point in Sn is homeomorphic to Rn which is simply-connected,so any non-surjective loop in f : I → Sn is homotopic to the constant loop. Thus it sufficesto prove that every loop in Sn is homotopic to a non-surjective one.

Choose a point x 6= x0 in Sn and let B be a small ball around x0 with closure B. Givenany based loop f , the preimage f−1(B) is a disjoint union of open intervals (ai, bi) ⊂ I.Because f−1(x) is closed in I, it is compact, so it is contained in a finite collection of theintervals (ai, bi). For each of these, f restricts to a map f[ai,bi] : [ai, bi]→ B, sending ai, bi

13

to the boundary of B. Because B is simply connected and of dimension greater than one,we may replace f[ai,bi] by homotopic map f ′[ai,bi] : [ai, bi]→ B that misses x. Sticking thisfinite collection of homotopies together gives us a loop f ′ : I → Sn, which is homotopicto f and not surjective.

Corollary 2.12. R2 is not homeomorphic to Rn for n ≥ 3.

Proof. If there is a homeomorphism h : R2 → Rn, then by composing with translation wemay assume h(0) = 0. Then h should restrict to a homeomorphism R2 \ 0 to Rn \ 0. But

R2 \ 0 ∼= S1 × R

has fundamental group Z, while

Rn \ 0 ∼= Sn−1 × R

has trivial fundamental group if n ≥ 3, so they cannot be homeomorphic.

2.3.1 The Borsuk-Ulam Theorem

Theorem 2.13 (Borsuk-Ulam Theorem). For every continuous map f : S2 → R2,there exists a point x ∈ S2 such that f(x) = f(−x).

Proof. Suppose f : S2 → R2 contradicts the theorem. Then we may define the continuousmap g : S2 → S1 by the formula

g(x) =f(x)− f(−x)

|f(x)− f(−x)|.

Then g(−x) = −g(x). Consider the “equatorial loop” η(s) := (cos(2πs), sin(2πs), 0).Claim: The winding number of h := g η is odd.

Proof. The key property is that

h(s) = −h(s+ 1/2) (2)

for s ∈ [0, 1/2]. Choose a lift h : I → R with h(0) = 0, with respect to the mapp : R→ S1, p(s) = (cos(2πs), sin(2πs)). Then because h(1/2) = (−1, 0), it must be truethat h(−1/2) is a half integer, say q + 1/2. Observe that (2) implies that

p(h(s) + q + 1/2) = p(h(s+ 1/2))

for s ∈ [0, 1/2], so by uniqueness of lifting h(s) + q+ 1/2 = h(s+ 1/2) for s ∈ [0, 1/2] andin particular, h(1) = 2q + 1 proving the claim.

The claim implies in particular that g∗([h]) 6= 0 ∈ π1(S1). But this implies that[h] 6= 0 ∈ π1(S2), which contradicts the triviality of π1(S2).

Corollary 2.14. Given a covering of S2 by three closed sets, A1, A2, A3, then at least oneof the Ai contain a pair of anti-podal points.

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Proof. Consider S2 ⊂ R3 as a metric space in the usual way, and define a functiondi : S2 → R by di(x) = d(x,Ai), the distance from x to Ai. It is a standard result inanalysis that di is a continuous function and di(x) = 0 if and only if x ∈ Ai (this dependson Ai being closed).

Consider now the continuous map S2 → R2, x 7→ (d1(x), d2(x)). By the Borsuk-Ulam theorem, there exist an antipodal pair x,−x such that d1(x) = d1(−x) and d2(x) =d2(−x). If either of these is zero, then one of A1 or A2 contains antipodal points. Ifneither is zero, then x,−x do not lie in either A1 or A2, so they must both lie in A3.

The number three in Corollary 2.14 is the best possible. This can be seen by consider-ing the covering of the regular tetrahedron by its four faces, and projecting radially ontothe sphere.

3 General Concepts in Homotopy Theory

3.1 Homotopy and Homotopy Equivalence

Definition 19. Let X and Y be spaces and let Ai ⊂ X and Bi ⊂ Y subsets for i = 1, ..., n.A map or morphism of (n+ 1)− tuples

ϕ : (X,A1, A2, ..., An)→ (Y,B1, ..., Bn)

is a continuous map ϕ : X → Y satisfying ϕ(Ai) ⊂ Bi for all i.A homotopy of maps between (X,A1, ..., An) and (Y,B1, ..., Bn) is a family of maps

ϕt : (X,A1, ..., An)→ (Y,B1, ..., Bn) such that the associated map

Φ : X × I → Y

is continuous. The two maps ϕ0 and ϕ1 are called homotopic.

Example 12. A homotopy of paths in X with end points x0, x1 is the same thing as ahomotopy of maps from (I, 0, 1) to (X, x0, x1).

By the same reasoning as for Proposition 2.1, homotopy is an equivalence relation onthe set of maps of tuples.

We most often consider maps of spaces ϕt : X → Y or maps of pairs ϕt : (X,A) →(Y,B). In the special case that the subset is a point, we call (X, x0) = (X, x0) apointed space.

Proposition 3.1. Let ϕt : (X, x0)→ (Y, y0) be a homotopy of pointed spaces. The inducedmaps on fundamental groups (ϕ0)∗ and (ϕ1)∗ are equal.

Proof. Let f be loop representing [f ] ∈ π1(X, x0). Then ϕtf defines a homotopy betweenϕ0 f and ϕ1 f , so

ϕ0∗([f ]) = [ϕ0 f ] = [ϕ1 f ] = ϕ1∗([f ]).

15

Definition 20. A pair of tuples (X,A1, ..., An) and (Y,B1, ..., Bn) are called homotopyequivalent if there exist maps

ϕ : (X,A1, ..., An)→ (Y,B1, ..., Bn)

andψ : (Y,B1, ..., Bn)→ (X,A1, ..., An)

such that both ϕ ψ and ψ ϕ are homotopic to the identity map on (Y,B1, ..., Bn) and(X,A1, ..., An) respectively. The maps ϕ and ψ are called homotopy equivalences.

3.1.1 Homotopy invariance of π1

Corollary 3.2. Let ϕ : (X, x0) → (Y, y0) be a homotopy equivalence of pointed spaces.Then ϕ∗ : π1(X, x0) ∼= π1(Y, y0) is an isomorphism.

Proof. By definition, there exists a map of pairs ψ : (Y, y0)→ (X, x0) such that ϕ ψ andψ ϕ are homotopic to identity maps. It follows from Proposition 3.1 and functorialitythat ϕ∗ ψ∗ = id∗ is the identity and also that ψ∗ ϕ∗ is the identity, so ψ∗ and φ∗ areinverse isomorphisms.

In fact, we have a slightly stronger result that allows us to ignore basepoints.

Proposition 3.3. If ϕ : X → Y is a homotopy equivalence, then ϕ∗ : π1(X, x0) →π1(Y, ϕ(x0)) is an isomorphism for any x0 ∈ X.

We begin with a lemma.

Lemma 3.4. If ϕt : X → Y is a homotopy, and h is the path h(s) = ϕs(x0) where x0 ∈ Xis a basepoint, then ϕ0∗ = βh ϕ1∗. That is, we have commuting diagram

π1(Y, ϕ1(x0))

βh

π1(X, x0)

φ1∗77nnnnnnnnnnnn

φ0∗

''PPPPPPPPPPPP

π1(X,ϕ0(x0))

.

Proof. Define a family of paths ht in Y , by ht(s) = h(st). Thus h0(s) = ϕ(x0) and h1(s) =h(s). For a based loop f : (I, 0, 1) → (X, x0), we have a homotopy , ht · ϕt(f) · h−1

t ofloops in Y based at ϕ0(x0), which equals ϕ0(f) at t = 0 and h ·ϕ1(f) ·h−1 at t = 1. Thus

ϕ0∗[f ] = [ϕ0(f)] = [h · ϕ1(f) · h−1] = βh ϕ1∗[f ]

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Proof. Let ϕ : X → Y be a homotopy equivalence with homotopy inverse ψ. Considerthe diagram

π1(X, x0)ϕ∗→ π1(Y, ϕ(x0))

ψ∗→ π1(X,ψϕ(x0))ϕ∗→ π1(X,ϕψϕ(x0))

Since ψ ϕ is homotopic to IdX , the composition of the first two arrows

ψ∗ ϕ∗ = (ψ ϕ)∗ = βh

as maps from π1(X, x0) to π1(X,ψϕ(x0)). Applying Proposition 2.3, we deduce thatψ∗ ϕ∗ is an isomorphism, so ϕ∗ must be injective. Similarly, ψ∗ ϕ∗ is an isomorphism,so ψ∗ is injective, so ϕ must also be surjective.

3.1.2 Deformation Retractions

It will be convenient to have methods to easily recognize and construct homotopy equiv-alences. One important class is deformation retractions.

Definition 21. Let X be a space and A ⊂ X. A deformation retraction of X onto Ais a homotopy of pairs φt : (X,A)→ (X,A), such that

• φ0 = IdX and φ1(X) ⊆ A,

• φt(a) = a for all a ∈ A and t ∈ I.

We say that A is a deformation retract of X.

Observe that if φt is a deformation retraction, then φ1 is a retraction (if considered asmap into A).

Proposition 3.5. Let φt be a deformation retraction of X onto A. Then φ1 : X → A isa homotopy equivalence, with homotopy inverse the inclusion i : A → X.

Proof. In one direction, we have φ1 i = IdA, which is of course homotopic to IdA. In theother direction, i φ1 is just the same as φ1 (viewed as a map with codomain X) whichby definition is homotopic to φ0 = IdX .

Example 13. If X ⊂ Rn is convex, then it deformation retracts onto any point c ∈ Xby the deformation retraction φt(x) = (1− t)c+ tx.

Example 14. The space Rn \ 0 deformation retracts onto Sn−1 via the deformation

retraction φt(x) =(t+(1−t)|x||x|

)x

Definition 22. Let f : X → Y be a continuous map. The mapping cylinder is thespace

Mf := ((X × I)∐

Y )/ ∼

where ∼ is generated by (x, 1) ∼ f(y), for x ∈ X and y ∈ Y . Then Mf deformationretracts onto Y , via the homotopy

ht : Mf →Mf

by ht(x, s) = (x, (1− t)s+ t) and ht(y) = y for y ∈ Y .

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4 Van Kampen’s Theorem

4.1 Free Products of Groups

Let Gαα∈I be (disjoint) collection of groups, indexed by α ∈ I. A word in Gα isa finite sequence (g1, ..., gn) such that each gi ∈ Gα for some α (the empty sequence isallowed). A word is called reduced if both

1. No element gi is an identity element.

2. Adjacent elements gi, gi+1 lie in distinct Gα.

An arbitrary word (g1, ..., gn) can be simplified to a reduced word using two operations:

1. If some gi is identity element, then remove it.

2. If adjacent elements gi, gi+1 lie in the same Gα, then replace them with the singleentry gigi+1 ∈ Gα.

Since both the above operations shorten the length of the word, it is clear that this processmust terminate. It is however not immediately clear that it results in a unique reducedword.

Proposition 4.1. Let W denote the set of reduced words for Gα. Then the operation µ :W ×W → W , which sends (g1, ..., gm)× (h1, ..., hn) to a reduction of (g1, ..., gm, h1, ..., hn)is well-defined, and makes W into a group. We call this group the free product ofGα and denote it ∗αGα.

Proof. Postponing for a moment the question of whether µ is well-defined, observe thatthe empty word () acts as an identity element, and that every word (g1, ..., gn) has aninverse (g−1

n , ..., g−11 ).

It remains to prove µ is well-defined and associative (which amount to the same thing).For g ∈ Gα, define the permutation Lg : W → W by

Lg(h1, ..., hn) :=

(g, h1, ..., hn) if h1 6∈ Gα

(gh1, ..., hn) if h1 ∈ Gα

and then deleting the first entry if it is an identity element. It is easy to see that for eachGα, L defines an homomorphism Gα → P (W ) where P (W ) is the group of permutationsof W . Now extend to a map

L : W → P (W ), L(g1,...,gn) = Lg1 ... Lgn

Then L is injective because L(g1,...,gn) sends the empty word to (g1, ..., gn). Thus we canidentify W with its image P (W ). Since L sends µ to the product operation in P (W ), wededuce that µ is well-defined and associative, because the product operation in P (W ) iswell-defined and associative.

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Observe that Proposition 4.1 implies that any word (g1, ..., gn) simplifies to a uniquereduced word, namely the product g1g2....gn ∈ ∗αGα. From now on we usually denote aword (g1, ..., gn) simply by g1g2...gn whenever this ambiguity is harmless.

Free products satisfy the following universal property: Any collection of group homo-mophisms φα : Gα → H into a fixed group H determines a unique homomorphism

Φ : ∗αGα → H

that agrees with φα0 on each Gα0 ⊆ ∗αGα.

4.1.1 Free Groups

An important special case the free product is when all of the Gα = Z for all α ∈ I, inwhich case

FI := ∗α∈IZ

is called the free group on generators I. In the special case I = 1, ..., n is called thefree group of rank n and is denoted Fn := F1,...,n.

Proposition 4.2. For m and n positive integers, the free group Fm is isomorphic Fn ifand only if m = n.

Proof. Given any group G, we can form the abelianizaton

G/[G,G],

where [G,G] is the smallest normal subgroup of G containing all commutators, that iselements of the form [a, b] = aba−1b−1 for a, b ∈ G. We think of G/[G,G] as the “largestabelian group onto which G maps surjectively”. In the case of the free group of rank n,it is easy to see that

Fn/[Fn, Fn] = Z⊕ ...⊕ Z = Zn

is the free abelian group of rank n. Since any isomorphism Fm ∼= Fn would imply

Zm = Fm/[Fm, Fm] ∼= Fn/[Fn, Fn] = Zn

this can only be true if m = n.

4.1.2 Presentations of Groups

One way to construct groups is using generators and relations.

Definition 23. A group presentation is a pair< aαα∈I |Rββ∈J > where the aαα∈Iis a set symbols and the Rβ are words in the symbols aα, a−1

α α∈I . A presentation definesa group

G := FI/N

where FI is the free group on the symbols aαα∈I and N is the smallest normal subgroupcontaining the words Rβ. The symbols aα are called generators and the words Rβ arecalled relations.

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Every group admits a presentation. For example, given G we have a presentation withone generator for every element in G and with relations abc−1 for every triple a, b, c ∈ Gsuch that ab = c. Of course, this is a very inefficient presentation.

Example 15. If groups G and H have presentations < aα|Rβ > and < bγ|Sδ >,then the free product G ∗H has a presentation

< aα ∪ bγ|Rβ ∪ Sδ >

Example 16. The presentation < a, b | aba−1b−1 > defines a group isomorphic to Z2,because there are two generators and the only relation is that they commute with oneanother.

4.2 Van Kampen’s Theorem

Let X be a path-connected space and let A and B be two path connected open sets suchthat A∪B = X and A∩B is path connected. Then the inclusions induces homomorphismsof groups

π1(A)j1

$$IIIIIIIII

π1(A ∩B)

i188rrrrrrrrrr

i2

&&LLLLLLLLLLπ1(X)

π1(B)

j2::uuuuuuuuu

.

Theorem 4.3 (Van Kampen’s Theorem). Let X,A,B be as above. The homorphisms j1and j2 induce an isomorphism of groups

π1(X) ∼= (π1(A) ∗ π1(B))/N

where N is the smallest normal subgroup of π1(A)∗π1(B) containing all words of the formi1(ω)i−1

2 (ω) for ω ∈ π1(A ∩B).

Remark 3. In terms of presentations, Van Kampen’s Theorem say that a presentationof π1(X) can constructed from a presentation of π1(A) ∗ π1(B) by adding a relation foreach generator of π1(A ∩B).

We postpone the proof briefly to consider some applications.An important special case is the wedge product of spaces. Suppose that (X, x0) and

(Y, y0) are pointed spaces. Then the wedge product of (X, x0) and (Y, y0) is

X ∨ Y := (X∐

Y )/ ∼

where x0 ∼ y0.

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Proposition 4.4. Suppose that there exist open neighbourhoods x0 ∈ U ⊂ X and y0 ∈V ⊂ Y that deformation retract onto the basepoints. Then there is an isomorphism offundamental groups

π1(X ∨ Y ) ∼= π1(X) ∗ π1(Y ).

Proof. Defining open sets A = X∨V and B = U∨Y , then we have homotopy equivalences

A ∼= X, B ∼= Y, A ∩B = U ∨ V ∼= point.

Since homotopy equivalences determine isomorphisms on π1, π1(A ∩ B) = π1(point) istrivial, and by Van Kampen

π1(X ∨ Y ) ∼= π1(A) ∗ π1(B) = π1(X) ∗ π1(Y ).

Example 17. An important special case of Proposition 4.4 is a wedge of circles: π1(S1∨S1) = Z∗Z = F2. By induction, the wedge product of n circles ∨nS1 satisfies π1(∨nS1) ∼=Fn. We will see later that a similar statement holds for free groups of infinite rank.

Proof of Theorem 4.3. For notational simplicity, relabel A,B by A1, A2. Fix a basepointx0 ∈ A1 ∩ A2 ⊂ X.

For k = 1, 2 we are homomorphisms jk : π1(Ak) → π1(X) induced by inclusion. Thisinduces a map

J : π1(A1) ∗ π2(A2)→ π1(X)

by the universal property of free products. Our first task is to prove that this homomor-phism is surjective.

Let f : (I, 0, 1) → (X, x0) be a loop in X. The preimages f−1(A1) and f−1(A2) forman open cover of I, so by the Lebesgue number theorem, we can subdivide 0 = s0 <s1 < ... < sm = 1 such that f([si−1, si]) lands in one of A1 or A2 for all i = 1, ...,m. Bymerging intervals if necessary, we may assume that f(si) ∈ A1 ∩ A2 for all i. Choose foreach i a path gi : (I, 0, 1)→ (A1 ∩ A2, x0, f(si)) (this is always possible because A1 ∩ A2

is path-connected). Let fi := f |[si−1,si]. Then f is homotopic in X to the product

f ∼ (g0 · f1 · g−11 ) · ... · (gm−1 · fmg−1

m )

which is a product of loops each lying in A1 or A2. Thus J is surjective.It remains to prove that the kernel of J is equal to N . Certainly N ⊆ ker(J), because

for any loop f in A1∩A2, the element J(i1([f ])·i2([f ])−1) ∈ π1(X) is represented by f ·f−1

which is of course homotopic to the identity. The hard part is to show that N ⊇ ker(J).Let [f1]....[fr] be a word in ker(J) ⊆ π1(A1) ∗ π1(A2). Then f1, ..., fr be a sequence of

loops in either A1 or A2 such that the product f := f1 · ... ·fr is homotopic to the constantloop in X. Thus there exists a homotopy

H : I × I → X

21

with H(s, 0) = x0, H(s, 1) = f(s) and H(0, t) = H(1, t) = x0 for all s and t. Employingagain the Lebesgue number Lemma, we may choose a partition 0 = s0 < ... < sm = 1 and0 = t0 < ... < tn = 1 such that

H([si−1, si]× [tj−1, tj]) ⊆ Ak

for some k ∈ 1, 2 (we may assume that the parametrization of f = f1 · ... · fr is chosenso that each fl ends on a subset of the si ). We refer to the various closed subsets I × Idefined by the partition as rectangles, edges, and vertices.

For each vertex v = (si, tj), choose a path

gv : (I, 0, 1)→ (X, x0, H(v))

subject to the condition that the image of gv lies in the open set Ak assigned to allrectangles incident to v in I × I, and that gv is constant if H(v) = x0. Thus if e is anedge linking vertices v1 to v2, then the path he := gv1 ·H|e · g−1

v2is a loop contained in all

Ak assigned to a rectangle incident to e.To any path

γ : (I, 0, 1)→ (I × I, 0× I, 1× I)

defined by tracing a sequence of edges e1, e2, ..., eq, we have a homotopy

H γ ∼ he1 · ... · heq

where each hei is a loop in A1 or A2 or both. We will prove by induction that thecorresponding word(s)

[he1 ]...[heq ] ∈ π1(A1) ∗ π1(A2)

must lie in N .First of all, if he lies in A1∩A2 there is ambiguity in choosing [he]: it may be represented

by i1([he]) or i2([he]) where [he] ∈ π1(A1 ∩ A2). But since i1([he])i2([he])−1 ∈ N , this

ambiguity is not important for our question.Next observe that if γ′ is obtained by sliding γ over one rectangle, then replacing one

subword [hei ]...[hej ] by another sequence [he′i ]...[he′j′ ] which represents the same element in

π1(Ak) for the appropriate k, and thus γ and γ′ determine the same element of π1(A1) ∗π1(A2) (modulo the ambiguity described above). Finally, the path γ(s) = (s, 0) determinesthe trivial element in π1(A1) ∗ π1(A2), while the curve γ(s) = (s, 1) determines the word[f1]...[fs], which consequently must lie in N .

The following generalization of Van Kampen’s Theorem is often useful.

Theorem 4.5. If X is a union of path-connected open sets Aα each containing the base-point x0 ∈ X and if each intersection Aα ∩Aβ is path-connected, then the induced homo-morphism J : ∗απ1(Aα)→ π1(X) is surjective.

If in addition each intersection Aα ∩ Aβ ∩ Aγ is path-connected, then the kernel of Jis the normal subgroup N generated by all elements of the form iαβ(ω)iβα(w)−1, and soJ induces an isomorphism π1(X) ∼= ∗απ1(Aα)/N . Here iαβ : π1(Aα ∩ Aβ) → π1(Aβ) isinduced by subspace inclusion.

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The proof of Theorem 4.5 is similar to the proof of Theorem 4.3, but more finicky.

Example 18. Using Theorem 4.5, Example 17 generalizes to an isomorphism

π1(∨IS1) ∼= FI

for any set I.

Example 19. The Hawaiian Earing is a subspace of X ⊂ R2 defined as a unionX = ∪∞n=1Xn where Xn is the circle of radius 1/n centred at (1/n, 0). At first glance Xlooks like a countable wedge of circles, but we will use the fundamental group to showthat it is different.

For each n = 1, 2, ... define the retraction rn : X → Xn by sending all circles Xm withm 6= n to the origin. This determines a homomorphism

R : π1(X)→∞∏n=1

Z

by R = (rn)∞n=1. We claim that R is surjective.Let (kn)∞n=1 be a sequence of integers. Then we may construct a loop f : (I, 0, 1)→

(X,~0), which winds kn times around Xn in the interval [1 − 1n, 1 − 1

n+1]. By the past-

ing lemma, f is continuous on [0, 1 − 1n] for every n and f is continuous at 1 because

lims→1f(s) = ~0. Clearly R([f ]) = (kn)∞n=1, so R is surjective.This means in particular that π1(X) is uncountable, because

∏∞n=1 Z is uncountable.

However π1(∨Z+S1) = FZ+ is countable, so ∨Z+S

1 and X are not homeomorphic (or evenhomotopy equivalent).

4.3 The effect of attaching cells

Let Dn := x ∈ Rn||x| ≤ 1 denote the unit disk or closed n-cell with boundarySn−1 = ∂Dn := x ∈ Rn||x| = 1. Given a topological space X and a continuous mapf : Sn−1 → X, we may construct a new space

Y := (X qDn)/ ∼

where we quotient by the equivalence relation generated by p ∼ f(p) for all p ∈ Sn−1.We say that Y is obtained from X by attaching an n-cell. The map f is called theattaching map. More generally, if we have a collection of maps fα : Sn−1 → X, then weconstruct

Y = (X q (∐α

Dnα))/ ∼

where p ∼ fα(p) for all p ∈ Sn−1α and α.

Example 20. A wedge of circles ∨IS1 is constructed by attaching I many 1-cells onto apoint X = p by the only possible attaching map f : S0 → p.

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Example 21. The torus S1×S1 can be constructed by attaching a 2− cell onto a wedgeof two circles X = S1 ∨ S1. If we denote by a and b the loops defined by the two circlesin X, then the attaching map f : S1 → X is the loop a · b · a−1 · b−1.

Proposition 4.6. Let X be a path-connected space with base point x0 ∈ X, and let Y beobtained from X by attaching 2 − cells by a collection of attaching maps fα : S1 → X.For each α, choose a based loop f ′α := gα · fα · g−1

α where gα : (I, 0, 1)→ (X, x0, fα(p)) forsome p ∈ S1. Then the inclusion X → Y induces a surjective map

i∗ : π1(X) π1(Y )

with kernel N equal to the smallest normal subgroup containing [f ′α] for all α.

Proof. We begin by expanding Y to a slightly larger space Z that deformation retractsonto Y . For each α, we attach a strip Sα = I × I to Y , with the lower edge I × 0attached along gα, the right edge 1× I attached along an arc in D2

α, and all the left edges0 × I identified with each other. The top edges are left unidentified, and this allows adeformation of Z onto Y . For each α, choose a point yα in the interior of D2

α but disjointto the arc attaching the strip Sα. Define open sets A := Z − ∪αyα and B := Z − X.These form an open cover of Z such that

• A deformation retracts onto X

• B is contractible

• A ∩B is path-connected

It follows from Van Kampen’s Theorem that the map π1(X)→ π1(Y ) is surjective, withkernel generated as a normal subgroup by the image of π1(A ∩B)→ π1(A) ∼= π1(X).

It remains to show that π1(A ∩ B) is generated by loops homotopy equivalent to thef ′α. This follows from arguing that A ∩ B is homotopy equivalent to a wedge of circles∨αS1, with one circle for each cell D2

α.

Example 22. From the construction of the torus in Example 21, we deduce that

π1(S1 × S1) =< a, b | aba−1b−1 > .

More generally, the genus g surface Σg is constructed by gluing a 2-cell to a wedge of2g circles. If the loops defined by the circles are called a1, b1, ..., ag, bg, then the attachingmap sends S1 to the loop

∏gi=1[ai, bi], where [ai, bi] = aibia

−1i b−1

i is the commutator. Itfollows that

π1(Σg) =< a1, b1, ..., ag, bg |g∏i=1

[ai, bi] > .

The example above suggests the following result.

Proposition 4.7. Every group G is isomorphic to the fundamental group of some topo-logical spaces X.

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Proof. Every group G has a presentation < aα|Rβ >. Begin with a wedge of circles∨αS1, and then construct X by attach two cells D2

β using attaching maps

fβ : S1 → ∨αS1

determine by the words Rβ. By Proposition 4.6, π1(X) ∼= G.

Remark 4. Observe that in the proof of Proposition 4.7, the space X is determined by thepresentation of G; different presentations may result in different spaces. A simple exampleare the presentations < a | a > and < a|a, a > which determine spaces homeomorphicto D2 and S2 respectively. Both of these spaces are simply connected, but they are nothomotopy equivalent to each other. Later in the course, we will show how to assign toa group G a space BG that is uniquely determined up to homotopy equivalence andsatisfying π1(BG) = G.

An argument similar to 4.6, establishes the following.

Proposition 4.8. Let X be a path-connected space, and let Y be constructed by attachingn-cells to X, where n ≥ 3. Then the inclusion X → Y induces an isomorphism π1(X) ∼=π1(Y ).

Proof. Exercise on problem set 3.

5 Cell complexes

Wedges of circles ∨IS1 and the genus g surfaces Σg constructed in Example 22 are bothexamples of cell complexes (also called CW-complexes). Cell complexes are spacesthat are constructed inductively by attaching cells. For instance,

• A 0-dimensional complex X0 is a discrete set of points (i.e. a disconnected union of0-cells).

• A 1-dimensional cell complex X1 is a space constructed by attaching a collection of1-cells to X0.

• A 2-dimensional cell complex X2 is constructed by attaching 2-cells to X1.

• and so on ...

In general, a cell complex X may have cells in arbitrarily high dimensions, in which caseit is called ∞-dimensional. Each n-cell determines a characteristic map φα : Dn → X.A subset S ⊆ X is open/closed if and only if φ−1

α (S) ⊆ Dn is open/closed for all cells.A subspace A ⊆ X is called a subcomplex if it is a closed union of cells (that is,

of images of characteristic maps). Given a subcomplex A ⊂ X, the quotient space X/Adefined by identifying all points in A with each other, is naturally a cell complex called aquotient complex of X.

The subcomplex Xn ⊆ X consisting of all cells of dimension ≤ n is called the n-skeleton of X. If X is infinite dimensional, the topology on X satisfies S ⊂ X is open

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(resp. closed) if and only if S ∩Xn is open (resp. closed) in Xn for all n. In particular, amap f : X → Y is continuous if and only if the restrictions fn : Xn → Y are continuousfor all n. We say X has the direct limit topology with respect to Xn.

Lemma 5.1. Let X be a cell complex and C ⊂ X a compact subspace. Then C iscontained within finitely many cells of X.

Proof. Choose a sequence of points xi ∈ C lying in distinct cells. We will show that theset S := xi∞i=1 is closed in X.

First observe that

S ⊆ X is closed⇔ S ∩Xn is closed in X, ∀ n⇔ S ∩Xn is closed in Xn, ∀ n.

We use induction on n.Clearly S∩X0 is closed in X0 hence in X, because every subset of X0 is closed. Assume

by induction that S ∩Xn−1 is closed in X. Thus for any characteristic map

φα : Dk → X

the pre-image φ−1α (S∩Xn−1) is closed in Dk. For k ≤ n, the pre-image φ−1

α (S∩Xn) ⊆ Dk

equals φ−1α (S ∩Xn−1) plus at most one point, thus it is a union of two closed sets, hence

is closed in Dk. We deduce that S ∩Xn is closed in Xn hence also in X. By induction,this holds for all n so S is closed in X.

The same argument shows that every subset of S is also closed, so S has the discretetopology. But S is a closed subset of the compact set C, so it is compact. We concludethat S is finite.

The following result says that the fundamental group is only sensitive to the 2-skeletonof a cell-complex.

Corollary 5.2. If X is a cell complex with 2-skeleton X2, then the inclusion X2 → Xdetermines an isomorphism

π1(X2) ∼= π1(X)

Proof. For finite dimensional X, this is an immediate consequence of Proposition 4.6 andProposition 4.8.

If X is infinite, then we use Lemma 5.1. Given a loop f : S1 → X, the imagef(S1) ⊆ Xn for n sufficiently large, so [f ] ∈ π1(X) lies in the image of π1(Xn) = π1(X2),so

π1(X2)→ π1(X)

is surjective. Similarly, if f : S1 → X2 is contractible in X by F : S1 × I → X, then it iscontractible in some Xn, and [f ] = 0 in π1(X2) = π1(Xn) so

π1(X2)→ π1(X)

is injective.

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5.1 Homotopy extension property and subcomplexes

Definition 24. A pair of spaces (X,A) is said to possess the homotopy extensionproperty (or hep) if either of the following equivalent conditions hold for every space Y

• Given f : X → Y and a homotopy gt : A → Y such that g0 = f |A, there exists ahomotopy ft : X → Y such that ft|A = gt ∀t.

• Every continuous map φ : X × 0 ∪ A × I → Y extends to a continuous mapϕ : X × I → Y .

Our first application of hep is a new criterion for homotopy equivalence.

Proposition 5.3. Let (X,A) be a pair of spaces satisfying hep and suppose that A iscontractible. Then the quotient map q : X → X/A is a homotopy equivalence (where X/Ais defined by identifying all points in A with each other).

Proof. A is contractible if it is homotopy equivalent to a point. Thus there must exist ahomotopy gt : A → A with g0 = IdA and g1 is a constant map. By hep gt extends to ahomotopy of pairs

ft : (X,A)→ (X,A)

such that f0 = IdX and f1(A) is a point. Let ft : X/A → X/A be the induced map onquotients. Since f1(A) is a point, there exists a map h : X/A→ X such that f1 = h q.Thus we have a pair of commuting diagrams

Xft //

q

X

q

X/A

ft // X/A

Xf1 //

q

X

q

X/A

f1 //

h

;;wwwwwwwwwX/A

We claim that h and q are homotopy inverses. In particular, h q = f1 is homotopic tof0 = IdX through ft, and q h = f1 is homotopic to f0 = IdX/A through ft.

Our main examples of pairs possessing the homotopy extension property come fromthe following.

Theorem 5.4. Let X be a cell complex and let A ⊂ X be a subcomplex. Then (X,A)satisfies the homotopy extension property.

We begin with a lemma.

Lemma 5.5. The pair (X,A) satisfies the homotopy extension property if and only ifX × 0 ∪ A× I is a retract of X × I.

Proof. Suppose that (X,A) satisfy hep. Then setting Y = X × 0 ∪A× I and φ = IdYimplies the existence of a retraction

ϕ : X × I → X × 0 ∪ A× I.

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Conversely, if r : X×I → X×0∪A×I is a retraction, then any map φ : X×0∪A×I →Y extends to a homotopy

ϕ := φ r : X × I → Y.

Proof of Theorem 5.4. By Lemma 5.5, it suffices to produce a retraction X × I → X ×0 ∪ A× I. In fact, we will produce a deformation retraction.

To begin, observe that there exists a retraction

r : Dn × I → Dn × 0 ∪ Sn−1 × I.

This may be constructed geometrically by radial reflection from (0, 2) ∈ Dn×R. BecauseDn × I ⊂ Rn+1 is convex, r extends to a deformation retraction

rt := tr + (1− t)Id : Dn × I → Dn × I.

Applying this to n-cells produces a deformation retraction of

Xn × I onto Xn × 0 ∪ (Xn−1 ∪ A)× I.

Finally, define a deformation retraction

Rt : X × I → X × 0 ∪ A× I,

by performing the retraction of Xn × I onto Xn × 0 ∪ (Xn−1 ∪ A)× I during the timeinterval t ∈ [1/2n+1, 1/2n]. Observe that this deformation retraction is continuous onX × I because it is continuous on each Xn × I.

When considering 1-dimensional cell complexes, it is conventional to use alternativeterminology.

Definition 25. A graph is a 1-dimensional cell complex. The 0-cells are called verticesand the 1-cells are called edges.

Proposition 5.6. Every connected graph is homotopy equivalent to a wedge of circles.

Proof. Let X be a connected graph. We will show that X contains a contractible subgraphA that contains the vertices X0. Then X is homotopy equivalent to X/A which is a wedgeof circles because X/A has only one vertex.

We first define a nested sequence of subgraphs X(i) of X as follows. Let X(0) denotea single vertex in X. Given X(i) let X(i+ 1) be the closure of all edges with an endpointin X(i). Because X is connected, it is easy to see that

∞⋃i=0

X(i) = X.

Next, we construct inductively a nested sequence of contractible graphs A(i) ⊂ X(i)containing all the vertices in X(i). Begin with A(0) = X(0). Now assume inductivelythat A(i) has been constructed. For each vertex

v ∈ X(i+ 1) \X(i),

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choose a single closed edge ev in X(i+1) linking v with A(i) and define A(i+1) to be theunion of A(i) with these ev. Then A(i + 1) ⊂ X(i + 1) contains all vertices in X(i + 1)and it deformation retracts onto A(i) by retracting along the edges ev.

Finally, let A :=⋃∞i=0 A(i). This contains all the vertices of X and it deformation

retracts onto the point A(0) by performing the contraction A(i+ 1) onto A(i) during thetime interval [ 1

2i+1 ,12i

]

Corollary 5.7. Every connected cell complex X is homotopy equivalent to a quotient cellcomplex X/A with only one 0-cell.

Proof. If X is connected then X1 is a connected graph. If A ⊂ X1 is the subgraphconstructed in the proof of Proposition 5.6 then X/A has only one 0-cell.

Another nice application of hep concerns constructing spaces using attaching maps.Given two spaces X and Y , a subspace A ⊆ Y and a map f : A→ X, recall that we canconstruct a new space

X qf Y := (X q Y )/ ∼

where a ∼ f(a) for a ∈ A ⊆ Y . In this context, f is called an attaching map.

Proposition 5.8. If Y is a cell complex and A ⊆ Y is a sub-cell complex and ft : A→ Xis a homotopy of continuous maps. Then there is a homotopy equivalence

X qf0 Y ∼= X qf1 Y.

Proof. Let F : A × I → X be the homotopy F (a, t) = ft(a), and consider the spaceX qF (Y × I). Then the deformation retraction from Y × I onto (Y × 0) ∪ (A × I),determines (just by attaching) a deformation retraction

r : X qF (Y × I) X qF (Y × 0 ∪ A× I) ∼= X qf0 Y,

which is a homotopy equivalence by Proposition 3.5. Similarly X qF (Y × I) is homotopyequivalent to Xqf1Y . Finally, Xqf0Y ∼ Xqf1Y by transitivity of homotopy equivalence.

Example 23. The n-disk Dn has a cell complex structure with a 0-cell, a (n-1)-cell andone n-cell, such that the boundary ∂Dn = Sn−1 is a subcomplex. Thus given a space Xand two homotopic attaching maps

f0, f1 : Sn−1 → X,

the spaces X qf0 Dn ∼ X qf1 Dn are homotopy equivalent.A more particular example: If X is simply connected, then any space obtained by

attaching a 2-cell to X is homotopy equivalent to the wedge product X ∨ Sn.

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5.2 Cell complex propaganda

We present some results showing that many interesting spaces are either homeomorphicor homotopy equivalent to cell complexes. The material in this section will not be tested.

Definition 26. A real analytic function f : Rn → R is a infinitely differentiablefunction such that at every point p ∈ Rn, f equals its Taylor series at p on some positiveradius. A real analytic set X ⊂ Rn is the solution set finite collection of equationsf1(x) = ... = fn(x) = 0, for fi real analytic.

Example 24. Polynomial functions, exponential functions, trigonometric functions, etc.are real analytic.

Theorem 5.9 (Whitney?). Every real analytic set X ⊆ Rn is homeomorphic to a cellcomplex.

Let X and Y be two topological spaces. Let Cont(X, Y ) be the set of all continuousmaps from X to Y with the compact-open topology.

Example 25. The space LY = Cont(S1, Y ) is called the free loop space of Y . Thesubspace Ω(Y, y0) of loops based at y0 is called the based loop space of Y . It is easy toshow that π1(Y ) is equal to the set of path-components of Ω(Y, y0).

Theorem 5.10 (Milnor 1959). If X and Y are cell complexes and X is compact, thenCont(X, Y ) is homotopy equivalent to a cell complex.

Definition 27. A topological space X is called a (topological) n-manifold if it is Haus-dorff and if every point p is contained in an open neighbourhood p ∈ U ⊂ X that ishomeomorphic to Rn.

• Every open set in Rn is an n-manifold.

• The sphere Sn is an n-manifold.

• Surfaces of any genus are 2-manifolds.

• The product of an m-manifold and an n-manifold is an m+ n-manifold.

An example of a space that is locally Euclidean but is not a manifold is constructedby taking two copies of the real line R q R = R × a, b and forming the quotient by(t, a) ∼ (t, b) if t 6= 0. This space looks locally like R, but the points (0, a) and (0, b)cannot be separated by open sets.

Theorem 5.11. Every compact n-manifold is homotopy equivalent to a cell complex. Itremains an open question whether or not every compact n-manifold is homeomorphic toa cell complex.

There is a famous conjecture proven recently by Perelman concerning fundamentalgroups of 3-manifolds.

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Theorem 5.12 (Poincare conjecture). Let M be a compact, simply connected 3-manifold.Then M is homeomorphic to the 3-sphere S3.

The groups that arise as fundamental groups of compact 3-manifolds are rather special.This is no longer true for dimension greater than 3.

Proposition 5.13. Every finitely presented group G arises as the fundamental group ofa compact 4-manifold.

It is also true that there are many examples of simply connected 4-manifolds. Forexample S4 and S2×S2 are both simply connected. Thus the most obvious generalizationof the Poincare conjecture is not true in dimension higher than 3. However, there is aless obvious (and proven) generalization of the Poincare conjecture to n-manifolds ofdimension greater than 3. This generalization involves higher homotopy groups, whichwe will learn about next.

5.3 Higher homotopy groups and Whitehead’s Theorem

For any n ≥ 1, we can define the degree n homotopy group πn(X, x0) as follows.Let In = I × ... × I = [0, 1]n denote the n-cube. The boundary ∂In ⊂ In is the set

of n-tuples for which at least one entry equals 0 or 1 (by convention I0 is a point and∂I0 = ∅). For a space X with basepoint x0 ∈ X, define πn(X, x0) to equal the set ofhomotopy classes of maps

f : (In, ∂In)→ (X, x0).

Alternatively, because the quotient In/∂In ∼= Sn we may define πn(X, x0) to be homotopyclasses of maps f : (Sn, ∗)→ (X, x0) for some base point ∗ ∈ Sn.

Remark 5. If we adopt the conventions I0 = a point and ∂I0 = ∅, then we can definethe set π0(X, x0) = π0(X) as above (exercise: prove π0(X) may be identified with the setof path-components of X). However π0(X) is not a group in general.

For n ≥ 2 we define an operation on πn(X, x0) by [f ] + [g] = [f + g] where

(f + g)(s1, ..., sn) =

f(2s1, s2, ..., sn) if s1 ≤ 1/2

g(2s1 − 1, s2, ..., sn) if s1 ≥ 1/2.

By the same arguments that worked for n = 1, this operation is well-defined on homo-topy classes, the constant map acts as an identity, and we have inverses by f−1(s1, ..., sn) =f(1− s1, s2, ..., sn). So πn(X, x0) is a group for all n ≥ 1.

Unlike the fundamental group though, πn(X, x0) is an abelian group for n ≥ 2 (this iswhy we use + to denote the multiplication). This can be seen using a reparametrizationto produce a homotopy between f + g and g+ f (this is best shown with a picture). Theidea is to shrink the domains of f and g to small cubes and then use the extra dimensionsto move one around the other.

The homotopy groups πn(X, x0) possess many nice properties familiar from π1(X, x0)that are proved in similar fashion.

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• A product of spaces πn(X × Y, (x0, y0)) ∼= πn(X, x0)× πn(Y, y0).

• A path γ : (I, 0, 1) → (X, x0, x1) determines an isomorphism βγ : πn(X, x0) ∼=πn(X, x1). This isomorphism depends only on the homotopy class of βγ. Thus if Xis path connected, πn(X, x0) = πn(X) is independent of the basepoint.

• A continuous map φ : X → Y induces a homomorphism φ∗ : πn(X, x0)→ πn(Y, φ(x0)).This determines a functor from pointed topological spaces to abelian groups ( inparticular Id∗ = Id and (φ ψ)∗ = φ∗ ψ∗).

• Given a homotopy φt : X → Y , then the induced maps φ0∗ and φ1∗ agree as mapsπn(X) → πn(Y ). Consequently if two spaces X and Y are homotopy equivalent,then πn(X) ∼= πn(Y ) for all n.

The higher homotopy groups differ from π1 in one important respect: Van Kampen’sTheorem does not hold for πn if n ≥ 2. This makes the higher homotopy groups noto-riously difficult to calculate, even for relatively simple spaces. Consider for example thespheres Sk. One can show using an argument similar to ... that

πn(Sk) = 0, for n < k.

It is also true thatπn(Sn) = Z.

However, the groups πn(Sk) is not known in general for k > n (see Hatcher 4.1 for moreon this). Calculating the higher homotopy groups of spheres is one of the great unsolvedproblems in mathematics.

In view of the following result, the difficulty of calculating πn is should perhaps notbe surprising.

Theorem 5.14 (Whitehead’s Theorem). Let X and Y be path-connecteh topologicalspaces that are homotopy equivalent to cell-complexes. Suppose a map φ : X → Y in-duces isomorphisms πn(X) ∼= πn(Y ) for all n ≥ 1. Then φ is a homotopy equivalence.

This means that the homotopy groups contain a great deal of information about thehomotopy type of cell complexes. The proof of Whitehead’s Theorem is beyond the scopeof this course.

Corollary 5.15. Suppose X is homotopy equivalent to a connected cell-complex andπn(X) = 0 for all n ≥ 0. Then X is contractible.

Proof. The inclusion of a point into X is a homotopy equivalence by Whitehead’s Theo-rem.

Using homotopy groups we can state the correct generalization of the Poincare con-jecture to higher dimensions. The following result was proven by Smale in dimension ≥ 5and Freedman in dimension 4.

Theorem 5.16 (Poincare Conjecture). Let M be a connected, compact n-manifold suchthat πk(M) = 0 for k ≤ n/2. Then M is homeomorphic to Sn.

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So for example, we can now distinguish between S4 and S2 × S2 because π2(S4) = 0while

π2(S2 × S2) ∼= π2(S2)× π2(S2) ∼= Z× Z

so S4 and S2 × S2 are not homotopy equivalent.

6 Covering Spaces

Definition 28. Let X be a topological space. A covering space of X is a space Xand a map p : X → X satisfing the following condition: there exists an open cover Uαof X such that p−1(Uα) is homeomorphic to a disjoint union of open sets each mappinghomeomorphically to Uα by p (we do not require p−1(Uα) to be non-empty).

Example 26. The map p : R → S1, sending p(s) = (cos(2πs), sin(2πs)) is a coveringspace.

Example 27. The map ωn : S1 → S1, sending ωn(s) = (cos(2πns), sin(2πns)) is acovering space.

In fact, we will later prove that these are the only examples of (nonempty, connected)covering spaces of S1. Observe that the induced maps on fundamental groups p∗ : π1(R)→π1(S1) and ωn∗ : π1(S1)→ π1(S1) are all injective, and that every subgroup of π1(S1) ∼= Zis realized as the image of one of these maps. This is a special case of a general resultthat we will prove soon.

Example 28. Define a quotient space RP n := Sn/ ∼ with respect to the relation x ∼ −x.Then the quotient map Sn → RP n is a covering space. The n-manifold RP n is called thereal projective space of dimension n.

A key property about covering spaces is the following homotopy lifting property.

Proposition 6.1. Given a covering space p : X → X, a homotopy ft : Y → X and a“lift” f0 : Y → X such that

p f0 = f0.

Then there exists a unique homotopy ft : Y → X lifting ft and extending f0. Equivalently,denoting F (y, t) = ft(y) and F (y, t) = ft(y), we have a commutative diagram

Y × 0

f0 // X

p

Y × I F //

∃! F77nnnnnnnnnnnnnnX

Proof. The method of proof is identical to the proof of Lemma 2.6.

Consider the case that Y is point.

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0

f0(0)=x0 // X

p

I

F //

∃! F88ppppppppppppppX

Then by Proposition 6.1, given a path F : I → X, and a point point x0 ∈ p−1(F (0)),there is a unique path F : I → X such that F (0) = x0 and p F . This special case iscalled the path lifting property.

Proposition 6.2. Let p : X → X be a covering space and let f : Y → X be a continuousmap with Y path-connected. Suppose f admits lifts f1 and f2 such that f1(y0) = f2(y0),for some point point y0 ∈ Y . Then f1 = f2.

Proof. Let y ∈ Y . Because Y is path connected, there exists a path γ : (I, 0, 1) →(Y, y0, y). The composition fγ : (I, 0, 1) → (X, f(y0), f(y)) is a path in X, so by theuniqueness of path lifting we have

˜(fγ) = f1γ = f2γ

and in particular f1(y) = f1γ(1) = f2γ(1) = f2(y).

It is possible the weaken the hypothesis in Proposition 6.2 to allow Y to be connectedrather than path-connected (see Hatcher Prop 1.34).

Now consider the case that Y = I and that

ft : (I, 0, 1)→ (X, x0, x1)

is a homotopy with fixed endpoints. Then by the homotopy lifting property, any liftf0 : Y → X extends uniquely to lifts ft : I → X. Furthermore, the paths ft(0) andft(1) with parameter t are lifts of the constant loops ft(0) = x0 and ft(1) = x1, so byuniqueness of path lifting we know that ft(0) and ft(1) are also constant loops, so ft is infact a homotopy of paths with fixed endpoints.

Proposition 6.3. The map p∗ : π1(X, x0) → π1(X, x0) induced by a covering spacep : (X, x0) → (X, x0) is injective. The image of p∗ consists of homotopy classes of loopsin (X, x0) that lift to loops in (X, x0).

Proof. An element of the kernel of p∗ is represented by a path

f0 : (I, 0, 1)→ (X, x0)

such that the exists a homotopy

ft : (I, 0, 1)→ (X, x0)

with f0 = p f0 and f1 = x0 the constant loop. By the homotopy lifting property (andprevious discussion), f0 extends to a unique lift

ft : (I, 0, 1)→ (X, x0).

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Since f1 lifts the constant loop f1, we find that f1 is constant, so f0 represents the identityin π1(X, x0), so p∗ is injective.

The second part follows immediate: the image of p∗ is those loops that are the imageof loops in (X, x0) and this is the same as lifts of loops in (X, x0).

Given a covering space p : X → X, it is clear from the definition that the cardinalityof the fibres p−1(x) is locally constant with respect to X (because it is constant on theopen sets Uα). If X is connected, this implies that the cardinality of a fibre f−1(x) isindependent of x. We call this cardinality the number of sheets of the covering space.

Recall that if G is a group, then the index of subgroup H ⊆ G is the number of cosetsH\G := Hg|g ∈ G.

Proposition 6.4. Suppose p : (X, x0) → (X, x0) is a covering map with both X and Xare path-connected. Then the index of the subgroup p∗(π1(X)) ⊆ π1(X) is equal to thenumber of sheets of the cover.

Proof. Given a loop g in X based at x0, let g denote the lift to X with g(0) = x0. LettingH := p∗(π1(X)), we define a map

Φ : H\π1(X, x0)→ p−1(x0)

by sending [g] to g(1), where g is any representative of the coset [g]. We claim that Φ isa bijection.

To see that Φ is well-defined, first recall that based homotopies of paths lift to basedhomotopies of paths, so Φ is independent of the representative in π1(X, x0). Next, observeif h is a path representing an element of H, then by Proposition 6.3 h is a loop based atx0. Thus for any loop g

˜(h · g)(1) = (h · g)(1) = g(1),

so Φ is well defined.To see that Φ is surjective, recall that X is path connected, so for any point in

z ∈ p−1(x0) we can find a path g : (I, 0, 1)→ (X, x0, z). Then the composition g := p gis a based loop in (X, x0) and

Φ(Hg) = g(1) = z.

To prove injectivity, suppose that Φ(H[g1]) = Φ(H[g2]). Then g1 · g−12 is a loop based

at x0, so by Proposition 6.3, [g1][g2]−1 ∈ H and

H[g2] = H[g1][g2]−1[g2] = H[g1].

Example 29. The covers p : R→ S1 is countably sheeted and π1(R) = 0, so the index

[π1(S1) : p∗(π1(R))] = [Z : 0]

is countable. The cover ωn : S1 → S1 is n-sheeted, and the index

[π1(S1) : (ωn)∗(π1(S1))] = [Z : nZ] = n.

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These lifting properties can be extended beyond homotopies, arbitrary continuousmaps. We have the following lifting criterion for covering spaces.

Proposition 6.5. Let p : (X, x0) → (X, x0) is a covering space and let f : (Y, y0) →(X, x0) be a map, where Y is path-connected and locally path-connected. Then a liftf : (Y, y0)→ (X, x0) of f exists if and only if f∗(π1(Y, y0)) ⊆ p∗(π1(X, x0)).

Definition 29. A space X is locally path-connected if for each point x ∈ X andeach open nbhd x ⊂ U , there exists a smaller open nbhd x ∈ V ⊂ U such that V ispath-connected.

Example 30. Every cell complex is locally path-connected (Exercise?).

Example 31. An example of a space that is path-connected, but not locally path-connected, consider the set

X = (0 ∪ 1/n∞n=1)× I ∪ I × 0.

This space is not locally path-connected around (0, 1/2).

Proof of Proposition 6.5. One direction is clear. If f lifts to f , then by functoriality

f∗(π1(Y )) = p∗(f∗(π1(Y ))) ⊆ p∗(π1(X)).

We turn now to the converse. Suppose that f∗(π1(Y )) ⊆ p∗(π1(X)). Because Y ispath connected, then for point y ∈ Y , we can find a path γ : (I, 0, 1) → (Y, y0, y). Themap fγ = f γ defines a path in X from x0 to f(y). By hlp, there exists a unique lift

˜(fγ) : (I, 0)→ (X, x0).

Define f(y) := ˜(fγ)(1). To see that this f is well defined, suppose we start with a differentpath γ′ : (I, 0, 1)→ (Y, y0, y). Then the concatenation γ · (γ′)−1 is a based loop in (Y, y0)and represents an element [γ · (γ′)−1] ∈ π1(Y, y0). By assumption

f∗([γ · (γ′)−1]) = [fγ · (fγ′)−1] ∈ p∗(π1(X))

so by Proposition ... there is a lift of fγ · (fγ′)−1 to a loop in (X, x0). By uniqueness of

path lifting, this loop must be the concatenation of fγ with ˜(fγ′)−1

so

fγ(1) = ˜(fγ′)(1).

Next we must check that f is continuous. That is, we must show that for all y ∈ Y andopen nbhds f(y) ⊂ U ⊂ X, we must find an open nbhd y ∈ V ⊂ Y such that f(V ) ⊂ U .

By replacing U with something smaller if necessary, we may assume that p restrictsto a homeomorphism between U and U := p(U). By continuity

y ∈ f−1(U) ⊂ Y

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is an open neighbourhood, so there exists a path-connected open set V , with y ∈ V ⊆f−1(U). For any point y′ ∈ V , we may choose a path

γ : (I, 0, 1)→ (Y, y0, y′)

that first passes through y and then connects to y′ without leaving V . Thus the pathfγ first x0 to f(y) and then connects to f(y′) without leaving U . It then follows fromthe uniqueness of path lifting that the lift connects x0 to f(y) and then connects to f(y′)

without leaving U . Thus in particular, f(y′) = ˜(fγ)(1) ∈ U , so f(V ) ⊂ U .

6.1 The Classification of Covering Spaces

In this section, we will prove that under reasonable hypotheses on a (pointed) connecedspace (X, x0), for every subgroup H ⊂ π1(X, x0) there exists a connected covering spacep : (X, x0)→ (X, x0) such that p∗(π1(X, x0)) = H. Furthermore, any two covering spacescorresponding to the same subgroup H are “isomorphic” in a sense we will make precise.Thus (pointed) covering spaces over (X, x0) are in one to one correspondence with thesubgroups of π1(X, x0).

The main step is to construct a simply-connected covering space.

Proposition 6.6. Let X be a space that is path-connected, locally path-connected andsemi-locally simply-connected. Then there exists a path-connected and simply connectedcovering space p : X → X.

Definition 30. A space X is semi-locally simply connected if every point x ∈ X iscontained in some open neighbourhood U such that the induced map π1(U, x)→ π1(X, x)is zero.

To see why semi-locally simply connected is a necessary condition for Proposition ??,observe that if X → X is a covering space, then every x ∈ X is contained in a open seti : U → X that lifts to an open set i : U → X that maps homeomorphically to U underp. This induces a commutative diagram

π1(U)

∼=

i∗ // π1(X)

p∗

π1(U)

i∗ // π1(X)

so if i∗ is not zero, then p∗ i∗ is not zero and π1(X) 6= 0.

Example 32. Cell complexes are semi-locally simply connected (proof to come).

Example 33. The Hawaiian Earring is not semi-locally simply connected at the origin.

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Proof of Proposition 6.6. Choose a base point x0 ∈ X. As a set X is the set of homotopiesof based paths starting at x0:

X := [γ] | γ : (I, 0)→ (X, x0).

This makes sense because the path lifting property implies that if a simply connectedcovering space exists, there is a natural bijection with the set X above. The projectionmap p : X → X is defined p([γ]) → γ(1) and this map is surjective because X is pathconnected.

Next, we must construct a topology on X. As a preliminary step, define B to be thecollection of path connected open sets U ⊂ X such that π1(U)→ π1(X) is zero. BecauseX is locally path-connected and semi-locally path connected, B forms an open cover ofX. Also, any path-connected open subset V ⊂ U must also lie in B by functoriality (i.e.the map π1(V )→ π1(X) factors through the zero map π1(U)→ π1(X) so must be zero).It follows that B forms a basis of the topology of X (i.e every open set in X is a union ofa collection of sets in B).

Given a set U ∈ B and a homotopy class of paths [γ] in X from x0 to a point in U ,define

U[γ] := [γ · η] | η is a path in U , η(0) = γ(1) ⊆ X.

Observe that (as the notation suggests) U[γ] is independent of representative γ ∈ [γ]. We

will show that the sets U[γ] form the basis of a topology on X.Claim: If [γ′] ∈ Uγ then U[γ′] = U[γ]

Proof. By hypothesis, there exists a representative γ′ = γ · η for [γ′]. Thus the elemntesof U[γ′] have the form

[γ′ · µ] = [γ · η · µ]

where both η and µ lie in U , so U[γ′] ⊆ U[γ]. On the other hand, elements of U[γ] have theform

[γ · µ] = [γ · η · η−1 · µ] = [γ′ · η−1 · µ] ∈ U[γ′]

so U[γ] ⊆ U[γ′].

Let B denote the collection of all sets U[γ]. We will prove that B is a basis for a

topology on X. Clearly B covers X. Now consider an intersection of the form U[γ] ∩ V[γ′]

and let[γ′′] ∈ U[γ] ∩ V[γ′].

Choose W ∈ B such that γ(1) ∈ W ⊂ U ∩ V . Then

U[γ] ∩ V[γ′] = U[γ′′] ∩ V[γ′′] ⊃ W[γ′′]

so B is a basis that generates a topology on X.Claim: p : X → X is a covering space.

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Proof. Let U ∈ B, then

p−1(U) =⋃

γ(1)∈U

U[γ]

is a union of open sets (this implies p is injective). In fact this union is disjoint becauseif U[γ] ∩ Uγ′ is non-empty, then the previous claim implies they are equal.

It remains to show that the maps p : U[γ] → U are homeomorphisms.Surjectivity follows, because U is path-connected, for every point x ∈ U there exist a

path η in U from γ(1) to x so p([γ · η]) = (γ · η)(1) = x. Injectivity follows because if ηand η′ are two paths in U such that γ · η(1) = γ · η′(1) then

[γ · η] = [γ · η′ · (η′)−1 · η] = [γ · η′]because (η′)−1 · η is a loop in U hence is homotopic to the trivial loop in X.

Finally, p is a homeomorphism because for all open V ⊂ U , then the preimagep−1(V ) = V[γ] ⊂ U[γ] is open (where we use our flexibility to choose [γ] so that γ(1) ∈ V )and p(V[γ]) = V ⊂ U is open, so p is local homeomorphism.

Lastly, we must show that π1(X) is path-connected and simply connected. Choose abasepoint [x0] ∈ X to be the constant path at x0 ∈ X. Then any element [γ] ∈ X isconnected to x0 along the path [γt] where γt is the restriction of γ to the interval [0, t], soX is path connected.

To see that X is simply connected, recall from Proposition 6.3 that

π1(X, x0) ∼= p∗(π1(X, x0)) = [γ] ∈ π1(X, x) | γ is a loop.

As we’ve already seen γ = [γt] so γ is a loop if and only if γ(1) = [γ] = x0 = [x0]. I.e. γlifts to a loop only if γ is homotopic to the trivial loop. It follows that p∗(π1(X, x0)) isthe identity, so π1(X, x0) is trivial.

Example 34. Consider again the projective plane RP n = Sn/ ∼ where x ∼ −x. Thenthe quotient map q : Sn → RP n is a covering space. If n ≥ 2, then Sn is simply connectedso it is the universal cover of RP n. Furthermore, it is a 2-sheeted cover so we deduce thatif n ≥ 2, then π1(RP n) has order 2, so it must be the cyclic group of order two.

Proposition 6.7. Suppose X is path-connected, locally path-connected, and semilocallysimply-connected. Then for every subgroup H ⊆ π1(X, x0) there is a covering space p :XH → X such that p∗(π1(XH , x0)) = H for a suitably chosen basepoint x0 ∈ XH .

Proof. Let p : X → X be the universal cover of X constructed previously. For points[γ], [γ′] ∈ X, define the equivalence relation [γ] ∼ [γ′] if γ(1) = γ′(1) and [γ][γ′]−1 ∈ H.It is easily verified that ∼ is an equivalence relation because H is a group. In particular,symmetry follows from the fact that H contains inverses and transitivity follows the factthat H is closed under multiplication (prove this). We define XH to be the quotient ofX under this equivalence relation. The map p : XH → X defined by p([γ]) = γ(1) isstill well-defined because the equivalence relation only identifies paths with common endpoint. To see that p : XH → X is a covering map observe that

p−1(U) = p−1(U)/ ∼= (⋃

γ(1)∈U

U[γ])/ ∼

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and by a previous claim, open sets of the form U[γ] and U[γ′] are either disjoint or equal....

If we choose basepoint x0 to be the equivalence class of the constant path [x0], thenthe image p∗ : π1(XH , x0)→ π1(X, x0). This is because the lift of a loop γ in X based atx0 lifts to a path that starts at [x0] and ends at [γ], so this lifted path is a loop preciselyif [γ] = [γ][x0]−1 ∈ H.

Theorem 6.8. The subgroup of a free group is a free group.

Proof. Let FI denote the free group on generators I. Then by ... we know that FI isisomorphic to the fundamental group of a wedge of circles X := ∨I(S1):

FI = π1(X).

Let H ⊆ FI be a subgroup. Then by ... there exists a covering space XH . But any coveringspace of X must be a one dimensional cell-complex. Consequently, by ... π1(XH) ∼= H isa free group.

Example 35. Consider the wedge of two circles S1∨S1 with fundamental group F2. Thenwe can form a covering space with fundamental group FZ (as shown in class). Thus F2

has a subgroup isomorphic to the free group on a countable number of generators (indeedif F2 has generators a, b, then the generators of FZ are anba−n|n ∈ Z). By choosing asubset of these, we construct subgroups H ⊂ F2 isomorphic to Fn for any positive integern.

Definition 31. A morphism between two covering spaces p1 : X1 → X and p1 : X1 →X, is a continuous map f : X1 → X2 such that p1 = p2 f .

X1

f //

p1 A

AAAA

AAX2

p2~~

X

The morphism f is called an isomorphism of covering spaces if there exists a morphismg : X2 → X1 such that f g = IdX2 and g f = IdX1 .

If the covering spaces (Xi, x0) are pointed, then we revise the above definition torequire that f and g preserve the basepoint.

(X1, x1)f //

p1 %%KKKKKKKKK(X2, x2)

p2yysssssssss

(X, x0)

Theorem 6.9. Let X be a path-connected, locally path-connected, and semilocally simply-connected. Then there is a bijection between the set of isomorphism classes of pointedpath-connected covering spaces p : (X, x0) → (X, x0) and the set subgroups of π1(X, x0).If basepoints are ignored, then there is a bijection between isomorphism classes of path-connected covering spaces p : X → X and conjugacy classes of subgroups of π1(X, x0).

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Proof. We begin with the pointed case.Clearly if two pointed covering spaces are isomorphic then they determine the same

subgroup. To see the converse, suppose that we have two covering spaces ... such that

(p1)∗π1(X1, x1) = (p2)∗π1(X2, x2).

Then by ... we know there exists a unique lift

f : (X1, x1)→ (X2, x2)

such that p1 = p2 f (i.e. f is a morphism of pointed covering spaces). Similarly,there exists a unique map g : (X2, x2) → (X1, x1) such that p2 = p1 g. Consequentlyp1 = p2 f = p1 g f , so g f is a lift of p1 by p1. By uniqueness, we conclude thatg f = IdX1 and similarly that f g = IdX2 . Thus the theorem is proven for pointedcovering spaces.

It remains to show that given a path-connected covering space p : X → X, changingthe base-point x0 corresponds exactly to changing

H0 := p∗(π1(X, x0))

by conjugation. So let x1 be a different base-point with p(x0) = p(x1) = x0. Let γ be apath from x0 to x1. Then pγ is a based loop in X, representing an element g ∈ π1(X, x0).Then we have an inclusion

gH0g−1 ⊆ H1

because the elements of ...

Definition 32. An isomorphism of a covering space to itself is called a deck transfor-mation. The deck transformations of given covering space p : X → X form a groupunder composition G(X) called the group of deck transformations.

Proposition 6.10. Let p : (X, x0) → (X, x0) be a path-connected covering space of thepath-connected, locally path-connected space X and let H = p∗(π1(X, x0)) ⊂ π1(X, x0).Let N(H) denote the normalizer of H in π1(X, x0). Then

G(X) ∼= N(H)/H.

Proof. A deck transformation

Xf //

p@

@@@@

@@X

p~~

~~~~

~

X

is uniquely determined by f(x0) ∈ p−1(x0), so we have a natural bijection between G(X)and a subset of p−1(x0) = π1(X, x0)/H. Furthermore, given x1 ∈ p−1(x0), such an f existsif and only if

H = p∗(π1(X, x0)) ⊆ p∗(π1(X, x1)) = gHg−1

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where g ∈ π1(X, x0) represents a class of loops in (X, x0) that lift to paths from x0 tox1. Since H ⊂ gHg−1 iff g ∈ N(H), we have constructed a natural bijection G(X) ∼=N(H)/H. Proving that this is a group homomorphism is an exercise using the path liftingproperty.

In situations were H ⊂ π1(X, x0) is normal, we call the covering space normal or Ga-lois. Such covering spaces are distinguished by the fact that the group of deck transfor-mations acts transitively on the fibres p−1(x0) ( i.e. given any two points x, y ∈ p−1(x0),there is some deck transformation f ∈ G(X) such that f(x) = y ).

Example 36. Every covering space of S1 is normal, because π1(S1) is abelian. Thedeck transformation group for the universal cover p : R → S1 is π1(S1) = Z and forωn : S1 → S1 is Z/n.

Example 37. The covering spaces of S1∨S1 are often not normal. For example, considerexample (3) on page 57 of Hatcher. A little reasoning shows that this 3-sheeted coveringadmits no non-trivial deck transformations (done in class).

Deck transformations provide a different way to think about the classification result... If p : X → X is the universal covering space of X, then G(X) ∼= π1(X, x0) and wehave a homeomorphism

X/G(X) ∼= X

between X and the orbit space X/G(X). Similarly, if H ⊆ π1(X, x0) ∼= G(X), then weform the associated covering space as an orbit space

XH∼ //

X/H

X∼ // X/π1(X)

Example 38. A famous class of 3-manifolds are the lens spaces. Let S3 ⊂ C2 denotethe unit sphere and choose a pair of relatively prime integers p, q ∈ Z. Define an action bythe cyclic group Z/p on S3 under which the generator acts by (z1, z2) 7→ (e2πi/p, e2πiq/p).Then one shows that the orbit space

L(p, q) := S3/(Z/p)

is a 3-manifold, and that the quotient map S3 → L(p, q) is a normal covering spacewith deck transformation group Z/p. Thus π1(L(p, q)) ∼= Z/p. In 1919, J.W.Alexanderproved that the lens spaces L(5, 1) and L(5, 2) are not homeomorphic, or even homotopyequivalent, despite both being 3-manifolds with the same fundamental group. Observethat the special case L(2, 1) ∼= RP 3.

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