Introduction to Algebraic Topology

Home , Ball (mathematics), Induced homomorphism, Unit disk, Unit interval

Paul Arnaud Songhafouo Tsopméné 2 Contents

0 General Topology (a brief recall) 5 0.1 Deﬁnition and Example ...... 5 0.2 Subspace ...... 6 0.3 Continuous function ...... 6 0.4 Homeomorphism ...... 7 0.5 Compactness ...... 7 0.6 Connectedness ...... 7

1 Homotopy type 9 1.1 Deﬁnition and examples ...... 9 1.2 Contractible spaces ...... 11

2 The fundamental group 13

2.1 Deﬁnition of π1(X) ...... 13 2.1.1 Paths ...... 13 2.1.2 Homotopy of paths ...... 13 2.1.3 Loops and the deﬁnition of the fundamental group ...... 15

2.1.4 The dependence of π1(X, x) on the choice of the basepoint ...... 16 2.2 The fundamental group of the circle ...... 17

1 2.2.1 Four lemmas in order to prove that π1(S ) = Z ...... 17 1 2.2.2 Proving that π1(S ) = Z ...... 19 2.3 Induced homomorphism ...... 20 2.4 Applications of the fundamental group ...... 23 2.4.1 The Brouwer ﬁxed point theorem ...... 23 2.4.2 The fundamental theorem of algebra ...... 23 2.5 The Van Kampen theorem ...... 25 2.5.1 Free products of groups ...... 25

3 4 CONTENTS

2.5.2 Van Kampen’s theorem ...... 27

3 Homology 31 3.1 Chain complexes ...... 31 3.2 Simplicial homology ...... 33 3.2.1 ∆-complexes ...... 34 3.2.2 Deﬁnition of the simplicial homology and examples ...... 36 3.3 Singular homology ...... 39 3.3.1 Deﬁnition ...... 39 3.3.2 Exact sequences ...... 39 3.3.3 Relative homology groups ...... 39 3.4 Applications ...... 39

Bibliography 41 Chapter 0

General Topology (a brief recall)

In this chapter we recall brieﬂy the following concepts: Topological space, subsapce, continuous function, homeomorphism, compactness, and connectedness.

0.1 Deﬁnition and Example

Deﬁnition 0.1.1. A topological space or just a space is a couple (X, O(X)) where X is a set, and O(X) is a family of subsets of X satisfying the following three axioms:

(T1) The emptyset and the set X itself belong to O(X).

(T2) O(X) is closed under ﬁnite intersections. That is, for any ﬁnite family {Uj}j∈J of elements of O(X), ∩j∈J Uj ∈ O(X).

(T3) O(X) is closed under unions. That is, for any family {Ui}i∈I of elements of O(X), ∪i∈I Ui ∈ O(X).

Elements of O(X) are called open subsets of X, and the complement of elements of O(X) are called closed subsets of X. The family O(X) is called a topology on X.

Example 0.1.2. Let a ∈ Rn, and let r ≥ 0 be a real number. An open ball in Rn of center a and radius r is the subset B(a, r) deﬁned by

n B(a, r) = {x ∈ R : kx − ak < r}.

Here k.k stands for the standard Euclidean norm in Rn. Note that B(a, r) = ∅ when r = 0. Deﬁne O(Rn) as

n O(R ) = set of unions of open balls.

One can easily check that O(Rn) is a topology on Rn.

Throughout this course, Rn will be equipped with the topology O(Rn) we just deﬁned. We will often write X for (X, O(X)) when the topology is understood.

5 6 CHAPTER 0. GENERAL TOPOLOGY (A BRIEF RECALL) 0.2 Subspace

Deﬁnition 0.2.1. Let (X, O(X)) be a topological space, and A be a subset of X. We say that A is endowed with the subspace topology if the topology, O(A), on A is of the form:

O(A) = O(X) ∩ A = {U ⊆ A : there exists W ∈ O(X) such that U = W ∩ A} .

In that case A is called a subspace of X, and O(A) is called the induced topology.

In this course, all subsets of Rn will be endowed with the subspace topology. We will often deal with the following examples of subspaces.

Example 0.2.2. (i) X = Rn+1, and A = Sn is the n-dimensional sphere or just the n-sphere. Recall that ( n+1 ) n n+1 X 2 S = (x1, ··· , xn+1) ∈ R : xi = 1 . i=1

If n = 0 we have S0 = {−1, 1} ⊆ R. So S0 consists of only two elements. The 1-sphere S1 is called the circle, and the 2-sphere will be sometimes just called the sphere.

(ii) X = Rn, and A = Dn is the n-dimensional disk or just the n-disk. Recall that

( n ) n n X 2 D = (x1, ··· , xn) ∈ R : xi ≤ 1 . i=1

The space D0 is a single point, while D1 = [−1, 1] ⊆ R. The 2-disk D2 will sometimes just called the disk. Note that the boundary of Dn is Sn−1.

(iii) X = R, and A = [0, 1] is the unit interval. Here are some examples of open subsets of [0, 1]. For any a ∈ (0, 1), the interval [0, a) is an open subset of [0, 1] because [0, a) = (−1, a) ∩ [0, 1] and (−1, a) is an open subset of R. Similarly, (a, 1] and (a, b), where b ∈ (0, 1) is such that a < b, are open subsets of [0, 1].

0.3 Continuous function

Most of spaces we will encounter in this course are “metrizable”, that is, there is a notion of distance between two points. If one writes x ≈ y to saying that the distance between x and y is small enough (or equivalently to saying that x and y are very closed to each other), then roughly speaking a function f : X −→ Y between two topological spaces is continuous if for any x, x0 ∈ X, we have the implication

x ≈ x0 ⇒ f(x) ≈ f(x0).

Explicitly, we have the following deﬁnition.

Deﬁnition 0.3.1. A function f :(X, O(X)) −→ (Y, O(Y )) is continuous if for every U ∈ O(Y ), f −1(U) ∈ O(X).

The word “map” will mean continuous function. Or if we want to emphasize about the continuity, we will say continuous map. 0.4. HOMEOMORPHISM 7 0.4 Homeomorphism

Roughly speaking, two spaces X and Y are homeomorphic if one can carry X to Y by a continuous deformation and vise versa. By continuous deformation, we mean a deformation in which nearby points stay nearby during all the process. So when performing such a deformation, it is allowed to stretch or bend, but it is not allowed to glue or tear. To be more precise, we have the following.

Deﬁnition 0.4.1. We say that two topological spaces X and Y are homeomorphic, and we denote X =∼ Y , if there exists a bijection f : X −→ Y such that both f and f −1 are continuous.

π π Example 0.4.2. The open interval (0, 1) is homeomorphic to R. Indeed, let f : (0, 1) −→ (− 2 , 2 ) deﬁned π π π π by f(t) = (1 − t)(− 2 ) + t 2 , and let g :(− 2 , 2 ) −→ R deﬁned as g(x) = tan(x). One can easily check that ∼ π π f is a homeomorphism. So (0, 1) = (− 2 , 2 ). It is well known that the second map is a homeomorphism as π π ∼ well. So (− 2 , 2 ) = R.

The map f from Example 0.4.2 or more generally a map of the form (1 − t)a + tb will be extensively used in next section, and also in Chapter 2 to deﬁne most of homotopies.

0.5 Compactness

Intuitively, a space is said to be compact if it contains no “holes”. For instance, the circle without one point is not compact.

Deﬁnition 0.5.1. A topological space X is said to be compact if for any family {Ui}i∈I of open subsets of X such that X = ∪i∈I Ui, there is a ﬁnite subset J of I such that X = ∪j∈J Ui.

Such a family {Ui}i∈I is called an open cover of X.

Example 0.5.2. We admit the following theorem due to Heine-Borel. Recall that a subset of Rn is bounded if it is contained in one ball.

Theorem. A subspace of Rn is compact if and only if it is closed and bounded. As a consequence, we have the following.

(i) All closed and bounded intervals (that is, intervals of the form [a, b]) of R are compact. In particular, the interval [0, 1] is compact.

(ii) For all n ≥ 0, the n-sphere Sn is compact.

(iii) For all n ≥ 1, Rn is not compact because it is not bounded. However R0, which is a single point, is compact.

0.6 Connectedness

Roughly speaking, a space X is said to be connected if it can be “held by only one hand” or if it consists of only one piece. Explicitly, we have the following deﬁnition.

Deﬁnition 0.6.1. We say that a space X is connected if it cannot be represented as the union of two or more disjoint nonempty open subsets. 8 CHAPTER 0. GENERAL TOPOLOGY (A BRIEF RECALL)

Example 0.6.2. We admit the following well known topological results.

(i) For all n ≥ 0, Rn and Dn are connected. (ii) For all n ≥ 1, Sn is connected. But S0 = {−1, 1} is not connected since S0 = {−1} ∪ {1} where both {−1} and {1} are open subsets of S0. Recall that S0 ⊆ R is endowed with the subspace topology as we mentioned just after Deﬁnition 0.2.1. Chapter 1

Homotopy type

In Section 0.4, we introduced the notion of homeomorphism between two topological spaces X and Y . In this chapter we are going to introduce a weaker notion called homotopy equivalence. By “weaker”, we mean that two spaces can be homotopy equivalent without being homeomorphic. For instance, X = D2 and Y = {(0, 0)} are not homeomorphic since manifestly there is no bijection from X to Y . But they are homotopy equivalent as we shall see in Example 1.1.9 below. Contrary to the topological word (where two spaces are the same if they are homeomorphic), in the algebraic topology world two spaces are the same if they are homotopy equivalent. From now on I = [0, 1]. Before we introduce the notion of homotopy equivalence, we need to ﬁrst deﬁne the notion of homotopy between maps.

1.1 Deﬁnition and examples

Deﬁnition 1.1.1. Let f, g : X −→ Y be continuous maps. We say that f is homotopic to g, and we write f ∼ g, if there exists a continuous map H : X × I −→ Y such that H(x, 0) = f(x) and H(x, 1) = g(x) for all x ∈ X. The map H is called a homotopy from f to g.

Sometimes we will write Ht(x) for H(x, t). One should think t as the time running from 0 to 1. At the initial time t = 0 we have H0 = f, and at the ﬁnal time t = 1 we have H1 = g.

Example 1.1.2. The maps f, g : R −→ R defined as f(x) = ex and g(x) = x2 are homotopic. Indeed, one can easily check that the map H : R × I −→ R defined as H(x, t) = (1 − t)ex + tx2 is a homotopy from f to g. In fact, any continuous maps f and g from R to R are always homotopic (just define H as H(x, t) = (1 − t)f(x) + tg(x)).

Deﬁnition 1.1.3. A homotopy H is called linear if it is of the form H(x, t) = (1 − t)a(x) + tb(x).

Proposition 1.1.4. Let X and Y be two spaces. The relation ∼ is an equivalence relation in the set of maps from X to Y .

Proof. (i) Reﬂexivity. For any map f : X −→ Y , one has f ∼ f, the homotopy from f to f being the constant homotopy H : X × I −→ Y,H(x, t) = f(x).

9 10 CHAPTER 1. HOMOTOPY TYPE

(ii) Symmetry. Let f, g : X −→ Y be two maps such that f ∼ g. We want to show that g ∼ f. Let H be a homotopy from f to g, and deﬁne K : X × I −→ Y by K(x, t) = H(x, 1 − t). We claim that K is a homotopy from g to f. Indeed, ﬁrst of all, the map K is continuous since H is continuous. Next at the time t = 0, we have K(x, 0) = H(x, 1) = g(x), and at the time t = 1 we have K(x, 1) = H(x, 0) = f(x).

(iii) Transitivity. Let f, g, h: X −→ Y be three maps such that f ∼ g and g ∼ h. We want to prove that f ∼ h. Let H be a homotopy from f to g, and K be a homotopy from g to h. Deﬁne L: X × I −→ Y by 1 H(x, 2t) if 0 ≤ t ≤ 2 L(x, t) = 1 K(x, 2t − 1) if 2 ≤ t ≤ 1. 1 1 The map L is continuous because at the time t = 2 we have H(x, 2 2 ) = H(x, 1) = g(x), and at the same time we have K(x, 2t − 1) = K(x, 0) = g(x). Moreover L(x, 0) = H(x, 0) = f(x), and L(x, 1) = K(x, 1) = h(x) for all x.

Given maps f : X −→ Y and g : Y −→ Z, the composition g ◦ f will be sometimes just denoted gf.

Proposition 1.1.5. Let f0, f1 : X −→ Y and g0, g1 : Y −→ Z be continuous maps such that f0 ∼ f1 and g0 ∼ g1. Then g0f0 ∼ g1f1.

Proof. Let H : X × I −→ Y be a homotopy from f0 to f1, and K : Y × I −→ Z be a homotopy from g0 to g1. Then for any x ∈ X,H(x, 0) = f0(x) and H(x, 1) = f1(x). And for any y ∈ Y,K(y, 0) = g0(y) and K(y, 1) = g1(y). Deﬁne L: X × I −→ Z by L(x, t) = K(H(x, t), t). The map L is continuous since it is a composition of continuous maps. Furthermore, we have

L(x, 0) = K(H(x, 0), 0) = g0(H(x, 0)) = g0f0(x), and L(x, 1) = K(H(x, 1), 1) = g1(H(x, 1)) = g1f1(x),

which completes the proof.

The following is an immediate consequence of Proposition 1.1.5.

Corollary 1.1.6. (i) Let f0, f1 : X −→ Y and g : Y −→ Z be maps such that f0 ∼ f1. Then gf0 ∼ gf1.

(ii) Let f : X −→ Y and g0, g1 : Y −→ Z such that g0 ∼ g1. Then g0f ∼ g1f. Deﬁnition 1.1.7. A map f : X −→ Y is called a homotopy equivalence if there exists another map g : Y −→

X such that gf ∼ idX and fg ∼ idY .

Here idX : X −→ X stands for the identity map. Note that g is just a map (not necessarily the inverse of f) from Y to X. Now we are ready to deﬁne the notion of homotopy between topological spaces.

Deﬁnition 1.1.8. We say that two topological spaces X and Y have the same homotopy type or are homotopy equivalent, and we denote X ' Y , if there exists a homotopy equivalence X −→ Y .

Before we see some examples, recall that a subspace C ⊆ Rn is said to be convex if for any a, b ∈ C, for any t ∈ [0, 1], the point (1 − t)a + tb belongs to C. In convex subspaces, the linear homotopies are very useful as we will see in the following examples and in Chapter 2.

Example 1.1.9. (i) The space R2\{(0, 0)} has the same homotopy type as the circle S1. That is,

2 1 R \{(0, 0)}' S . 1.2. CONTRACTIBLE SPACES 11

(ii) The unit disk D2 is homotopy equivalent to the one-point space {(0, 0)}. That is,

D2 '{(0, 0)}.

Solution.

2 1 x 1 2 (i) Define f : R \{(0, 0)} −→ S by f(x) = kxk , and define g : S −→ R \{(0, 0)} by g(x) = x. Of course both f and g are continuous. Moreover, a simple calculation shows that fg(x) = x for all x ∈ S1. So fg = id, which implies in particular that fg ∼ id. To prove that gf ∼ id, define H : R2\{(0, 0)} × I −→ R2\{(0, 0)} by H(x, t) = (1 − t)gf(x) + tx. The map H is well defined because for any x 6= (0, 0), the segment [x, gf(x)] does not pass through the origin (0, 0). Clearly the map H is continuous. Moreover, at the time t = 0 we have H(x, 0) = gf(x), and at the time t = 1 we have H(x, 1) = x.

(ii) Let f : D2 −→ {(0, 0)} be the unique map from D2 to {(0, 0)}. And let g : {(0, 0)} −→ D2 deﬁned by g(0, 0) = (0, 0). Of course we have fg = id. As before, the homotopy from gf to id is again the linear homotopy deﬁned as H(x, t) = (1 − t)gf(x) + tx.

1.2 Contractible spaces

We will write pt for the one-point space. That is, the space that contains only one element.

Deﬁnition 1.2.1. A topological space X is said to be contractible if X ' pt.

Example 1.2.2. (i) The space R is contractible. (ii) The disk D2 is contractible as well.

Solution.

(i) Take pt = {0}, and deﬁne g : pt −→ R as g(0) = 0. Let f : R −→ pt be the unique map from R to pt. Certainly one has gf = id. Deﬁne H as H(x, t) = tx. Clearly this is a homotopy from fg to id. So R ' pt. (ii) The fact that D2 is contractible comes immediately from Example 1.1.9 (ii).

Example 1.2.3. The sphere S0 = {−1, 1} is not contractible. Indeed, assume that it is. Then there 0 0 exist continuous maps f : S −→ pt and g : pt −→ S such that gf ∼ idS0 and fg ∼ idpt. Since gf is 0 0 homotopic to idS0 , there exists a continuous map H : S × [0, 1] −→ S such that H(x, 0) = gf(x) and H(x, 1) = idS0 (x) = x. Depending on the value of g, we need to deal with two cases (set pt = {∗}).

1. Assume g(∗) = 1. Then for every x ∈ S0, one has gf(x) = 1. Deﬁne φ: [0, 1] −→ S0 as φ(t) = H(−1, t). Certainly φ is continuous since H is continuous. So φ([0, 1]) is connected because φ is continuous and [0, 1] is connected. Now let us compute φ([0, 1]). We have φ(0) = H(−1, 0) = gf(−1) = 1 and φ(1) = H(−1, 1) = −1. This implies that φ([0, 1]) = {−1, 1}, which is a contradiction since {−1, 1} is not connected.

2. The case g(∗) = −1 is handled in the same way.

To prove directly that a space X is not contractible (or more generally to prove that two topological spaces are not homotopy equivalent) is diﬃcult in general because we then need to prove that certain maps are not homotopic, and this is hard in most cases. A nice way to solve such problems is to assign an “algebraic 12 CHAPTER 1. HOMOTOPY TYPE invariant”, A(X), to X. By algebraic invariant, we mean any correspondence that assigns to any topological space X an algebraic object (such as a group, a ring, a vector space,...) A(X) such that A(X) is isomorphic to A(Y ) whenever X ' Y . The ﬁrst algebraic invariant we are going to deal with in this course is the fundamental group in Chapter 2. Using that invariant, we will show for example that the circle S1 is not contractible. Chapter 2

The fundamental group

2.1 Deﬁnition of π1(X)

The goal of this section is to properly deﬁne the fundamental group, π1(X), of a topological space X. That deﬁnition involves the notion of loop, which is a special path. So we begin with paths. As in Chapter 1, we reserve the letter I for the interval I = [0, 1].

2.1.1 Paths

Deﬁnition 2.1.1. A path in a topological space X is a continuous map α: [0, 1] −→ X.

One should think s ∈ I as the time. So any path α is running over within the time 1. The point α(0) is called the starting point or the initial point of α, and the point α(1) is called the ending point or the terminal point of α.

Deﬁnition 2.1.2. Let α, β : I −→ X be paths in X such that α(1) = β(0). One can then deﬁne a new path, denoted α· β, by 1 α(2s) if 0 ≤ s ≤ 2 α· β(s) = 1 β(2s − 1) if 2 ≤ s ≤ 1. The path α· β is called the composition of α and β.

A path is in particular a function. However, one has to notice that the composition of paths is quite different to the composition of functions (recall that the composition gf of two functions makes sense only if the codomain of f sit inside the domain of g). By Definition 2.1.2, to get α· β, we spend half of the time doing α (so we double the speed) and half of the time doing β. The definition of the fundamental group involves the notion of homotopy between paths that we now define.

2.1.2 Homotopy of paths

We say that two paths α, β have the same endpoints if α(0) = β(0), and α(1) = β(1). From now on, the homotopy between two paths makes sense only if they have the same endpoints.

13 14 CHAPTER 2. THE FUNDAMENTAL GROUP

Deﬁnition 2.1.3. Let α, β : I −→ X be paths with the same endpoints. We say that α is homotopic to β, and we denote α ∼ β, if there exists a continuous map H : I × I −→ X such that

(HP1) H(s, 0) = α(s), and H(s, 1) = β(s) for all s ∈ I, and

(HP2) for all t ∈ I, H(0, t) = α(0), and H(1, t) = α(1).

Such a map H is called a homotopy from α to β.

Sometimes we will write Ht(s) for H(s, t). That is, Ht(s) = H(s, t). This gives the map Ht : I −→ X, s 7→ Ht(s), which is a path in X for every t. The axiom (HP1) is then equivalent to H0 = α and H1 = β. And the second axiom says that during the homotopy (or continuous deformation) the endpoints remain ﬁxed.

Proposition 2.1.4. Let x, y ∈ X be two points in a space X. And let Pxy be the set of paths starting at x and ending at y. Then the relation ∼ we just deﬁned is an equivalence relation in Pxy.

Proof. The proof is the same as that of Proposition 1.1.4.

Notation. The equivalence class of a path α, with respect to the equivalence relation ∼, will be denoted [α]. Thus α ∼ β if and only if [α] = [β].

Deﬁnition 2.1.5. Let x be a point in a space X. The path cx : I −→ X deﬁned by cx(s) = x for all s is called the constant path at x.

Before we deﬁne π1X and prove that it is a group, we need four lemmas. Here is the ﬁrst one.

Lemma 2.1.6. Let x, y ∈ X, and α: I −→ X be a path such that α(0) = x and α(1) = y. Then α ∼ cx· α and α ∼ α· cy.

Proof. Let us ﬁrst prove cx· α ∼ α. By Deﬁnition 2.1.2, we have

1 x if 0 ≤ s ≤ 2 cx· α(s) = 1 (2.1.1) α(2s − 1) if 2 ≤ s ≤ 1.

We want to get a homotopy between α and cx· α. Deﬁne H : I × I −→ X by

( 1 x if 0 ≤ s ≤ 2 t H(s, t) = 2s−t 1 α 2−t if 2 t ≤ s ≤ 1.

1 2s−t The map H is continuous since at the time s = 2 t we have α 2−t = α(0) = x. To check that H is the required homotopy, we need to prove that H satisﬁes the axioms (HP1) and (HP2) from Deﬁnition 2.1.3.

2s−0 (HP1) At the time t = 0, H(s, 0) = α( 2−0 ) = α(s) and at the time t = 1, H(s, 1) agrees with (2.1.1).

2−t (HP2) By the deﬁnition of H, we have H(0, t) = x = α(0), and H(1, t) = α 2−t = α(1) = y.

Similarly, one proves that α ∼ α· cy. This ends the proof.

Deﬁnition 2.1.7. Let α: I −→ X be a path in a space X. The path α: I −→ X deﬁned by α(s) = α(1 − s) is called the inverse path of α.

Lemma 2.1.8. Let x, y ∈ X, and let α: I −→ X be a path such that α(0) = x and α(1) = y. Then α· α ∼ cx and α· α ∼ cy. 2.1. DEFINITION OF π1(X) 15 Proof. Exercise.

Lemma 2.1.9. The composition of paths is associative up to homotopy. That is, given three paths α, β, γ such that α(1) = β(0) and β(1) = γ(0), we have α· (β· γ) ∼ (α· β)· γ.

Proof. Exercise.

Because of Lemma 2.1.9, we have [α· (β· γ)] = [(α· β)· γ]. So we can remove parentheses and just write [α· β· γ]. The following lemma is the path version of Proposition 1.1.5. But their proofs are quite diﬀerent because the composition of paths is not deﬁned in the same way as the composition of functions.

Lemma 2.1.10. Let α0, α1, β0, β1 : I −→ X be four paths such that α0(0) = α1(0), α0(1) = α1(1) = β0(0) = β1(0), and β0(1) = β1(1). If α0 ∼ α1 and β0 ∼ β1, then α0· β0 ∼ α1· β1.

Proof. Let H : I × I −→ X be a homotopy from α0 to α1, and let Let K : I × I −→ X be a homotopy from β0 to β1. Deﬁne L: I × I −→ X by

1 H(2s, t) if 0 ≤ s ≤ 2 L(s, t) = 1 K(2s − 1, t) if 2 ≤ s ≤ 1. It is easy to check that the homotopy L does the work.

2.1.3 Loops and the deﬁnition of the fundamental group

Deﬁnition 2.1.11. Let X be a topological space. A loop in X is a path α: I −→ X such that α(0) = α(1).

The point x = α(0) = α(1) is called the basepoint of the loop α. We also say that α is a loop at x. Let

π1(X, x) denote the set of homotopy classes [α] of loops at x.

Proposition 2.1.12. The set π1(X, x) is a group with respect to the operation [α][β] = [α· β].

Proof. First of all, the group operation is well deﬁned: if α ∼ α0 and β ∼ β0, then α· β ∼ α0· β0 by Lemma 2.1.10. Therefore [α· β] = [α0· β0].

• The identity element in π1(X, x) is [cx]. Indeed, for any [α] we have [α][cx] = [α· cx] = [α] since α· cx ∼ α by Lemma 2.1.6. Similarly, one has [cx][α] = [α] by applying the same lemma.

• The group operation is associative since

[α] ([β][γ]) = [α] ([β· γ]) = [α· (β· γ)] = [(α· β)· γ] by Lemma 2.1.9 = ([α· β]) [γ] = ([α][β]) [γ].

• Every element [α] in π1(X, x) is invertible with [α] as its inverse. This follows from Lemma 2.1.8.

Now we are ready to deﬁne the fundamental group.

Deﬁnition 2.1.13. Let X be a topological space, and x ∈ X be a point in X. By Proposition 2.1.12, the

set π1(X, x) is a group called the fundamental group of X at the basepoint x. 16 CHAPTER 2. THE FUNDAMENTAL GROUP

Recall that the trivial group, denoted 0, is the group with only one element. Two groups G1 and G2 are the same if they are isomorphic. In that case, we will write G1 = G2.

Example 2.1.14. π1(R, 0) = 0.

Solution. This comes from the fact that any loop in R at 0 is homotopic to the constant loop c0. Indeed, let α: I −→ R be a loop at 0. Deﬁne H : I × I −→ R by H(s, t) = (1 − t)α(s) + tc0(s). It is clear that H is continuous as a product of continuous functions. Moreover, at the time t = 0 we have H(s, 0) = α(s), and at

the time t = 1 we have H(s, 1) = c0(s) for all s. We also have H(0, t) = α(0) = 0 and H(1, t) = α(1) = 0 for every t. So H satisﬁes conditions (HP1) and (HP2) from Deﬁnition 2.1.3. This implies that it is a homotopy between α to c0. Thus [α] = [c0] for any [α] ∈ π1(R, 0), which gives the desired result.

Proposition 2.1.15. Let X and Y be spaces, and let (x, y) ∈ X × Y . Then

∼ π1(X × Y, (x, y)) = π1(X, x) × π1(Y, y).

Proof. The main observation is the following. Let α: I −→ X × Y be a loop at (x, y) deﬁned as α(s) =

(α1(s), α2(s)). Then it is clear that α1 is a loop in X at x, and α2 is also a loop in Y at y. Conversely, given two loops β1 : I −→ X, and β2 : I −→ Y at x and y respectively, there exists a unique loop β : I −→ X × Y at (x, y) such that β(s) = (β1(s), β2(s)).

One can then deﬁne f : π1(X × Y, (x, y)) −→ π1(X, x) × π1(Y, y) by f[α] = ([α1], [α2]). One can also deﬁne g : π1(X, x) × π1(Y, y) −→ π1(X × Y, (x, y)) by g([β1], [β2]) = [β]. Clearly f and g are homomorphism of groups, and gf = id and fg = id.

2.1.4 The dependence of π1(X, x) on the choice of the basepoint

Let x1, x2 ∈ X be two points in a space X (x1 may be equal to x2). Assume that there is a path γ : I −→ X starting at x1 and ending at x2. Deﬁne

ϕγ : π1(X, x1) −→ π1(X, x2) by ϕγ [α] = [γ· α· γ].

The map ϕγ is well deﬁned because if α ∼ β, then γ· α· γ ∼ γ· β· γ by Lemma 2.1.10.

Proposition 2.1.16. The map ϕγ is an isomorphism of groups.

Proof. First of all, the map ϕγ is a homomorphism of groups since

ϕγ ([α][β]) = [γ· (α· β)· γ] = [γ· (α· γ· γ· β)· γ] = [(γ· α· γ)· (γ· β· γ)] = ϕγ ([α])ϕγ ([β]).

Now deﬁne

ψγ : π1(X, x2) −→ π1(X, x1) by ψγ [α] = [γ· α· γ].

One can easily check that ψγ is also a homomorphism of groups, and that ϕγ ψγ = id and ψγ ϕγ = id.

Recall that a group G is abelian if for any g1, g2 ∈ G, we have g1g2 = g2g1.

Proposition 2.1.17. If π1(X, x1) is abelian, then ϕγ is independent of the choice of the path class [γ]. That 0 0 is, if γ : I −→ X is another path starting at x1 and ending at x2 (γ need not be homotopic to γ), then ϕγ = ϕγ0 . 2.2. THE FUNDAMENTAL GROUP OF THE CIRCLE 17

0 Proof. Let γ as in the proposition, and let [α] ∈ π1(X, x1). We want to prove that ϕγ ([α]) = ϕγ0 ([α]), that 0 0 0 0 0 0 is, [γ· α· γ] = [γ · α· γ ]. Consider the loops γ · γ and γ· γ at x1. It is clear that (γ · γ)· (γ· γ ) ∼ cx1 , where 0 0 0 0 cx1 is the constant loop at x1. So [γ · γ][γ· γ ] = [cx1 ], which implies that [γ · γ][γ· γ ][α] = [α] because [cx1 ] is the identity element in π1(X, x1). Using now the fact that π1(X, x1) is abelian by hypothesis, the latter equality becomes [γ0· γ][α][γ· γ0] = [α], which is equivalent to γ0· (γ· α· γ)· γ0 ∼ α. This implies, by Lemma 2.1.10 and Lemma 2.1.6, that γ· α· γ ∼ γ0· α· γ0. Thus [γ· α· γ] = [γ0· α· γ0], and we get the desired result.

Deﬁnition 2.1.18. A space X is said to be path-connected if for any pair of points x, y ∈ X, there exists a path α: I −→ X such that α(0) = x and α(1) = y.

For example Rn is path-connected for all n ≥ 0: for any x, y ∈ Rn, the path α(t) = (1 − t)x + ty does the work. One can also see that the n-sphere Sn is path-connected when n ≥ 1. For example, if n = 1, given two points x, y ∈ S1 we select one of the two arcs connecting x to y to be our path. For n ≥ 2, we can use arcs again, but in a more subtle way.

Proposition 2.1.16 implies that if X is path-connected π1(X, x) is, up to isomorphism, independent of the choice of the basepoint. In that case, π1(X, x) will be just denoted π1(X). For instance we will often drop the basepoint when dealing with Rn or Sm, m ≥ 1, or with a product of those spaces.

Deﬁnition 2.1.19. A space X is said to be simply connected if it is path-connected and π1(X) = 0.

For instance, for all n ≥ 0, the space Rn is simply connected. We will see that Sn is simply connected when n ≥ 2. However the circle S1 fails to be simply connected as we shall see in the next section.

2.2 The fundamental group of the circle

1 The goal of this section is to prove Theorem 2.2.7, which states that π1(S ) = Z. To do this, we need a couple of lemmas.

1 2.2.1 Four lemmas in order to prove that π1(S ) = Z

By small connected neighborhood in S1 we mean a small closed arc in S1. Consider the map

1 2πis p: R −→ S deﬁned by p(s) = e = (cos(2πs), sin(2πs)) .

Note that the map p wraps once around the circle when restricting to any interval of the form [n, n+1), n ∈ Z. Also note that the restriction of p to [n, n + 1) is the same as the restriction to [m, m + 1) for any n, m.

Lemma 2.2.1. Let a, b ∈ [0, 1] with a < b, and let α:[a, b] −→ S1 be a continuous map that carries [a, b] to a small connected neighborhood. Let x0 ∈ R such that p(x0) = α(a). Then there exists a unique map αe:[a, b] −→ R such that αe(a) = x0 and pαe = α.

Proof. The key point of the proof is the following fact about the map p. Let x ∈ S1. Then there exists a 1 unique sx ∈ [0, 1) such that x = p(sx). This is by the deﬁnition of p, which wraps once around S if one restricts it to [0, 1). So the inverse image of x under p consists of points of the form sx + k, k ∈ Z. That is,

−1 p (x) = {sx + k, k ∈ Z.} 18 CHAPTER 2. THE FUNDAMENTAL GROUP

Now if U is a small connected neighborhood containing x, then one can easily see that

−1 −1 [ Fact: p (U) is of the form p (U) = Vk, (2.2.1) k∈Z

0 where each Vk is a closed interval containing sx + k and contained in [k, k + 1). So the Vks are pairwise disjoint and each of them is homeomorphic to U. The homeomorphism ϕk : Vk −→ U is nothing but the restriction of p on Vk. That is, ϕk = p|Vk. Set Vk = [yk, zk]. Without lost the generality, assume that the small closed arc U starts at α(a) and ends at α(b). This assumption implies that we must have p(yk) = α(a) 0 and p(zk) = α(b) for all k. Since by hypothesis x0 is such that p(x0) = α(a), it follows that x0 is one of yks. That is, there is an integer n, which is unique, such that x0 = yn.

−1 Deﬁne now αe:[a, b] −→ R by αe(s) = ϕn (α(s)). We have pαe = α because ϕn = p|Vn. We also easily have −1 αe(a) = ϕn (α(a)) = yn = x0. Such a map αe is unique by construction. 0 0 We prove the uniqueness. Suppose that there is another continuous map αe :[a, b] −→ R such that αe (a) = x0 0 0 and pαe = α. We want to prove that αe = αe . By the construction of αe, it follows that αe([a, b]) ⊆ Vn. For 0 0 the same n, we also have αe ([a, b]) ⊆ Vn because of the following three things. (i) αe ([a, b]) is connected 0 0 0 0 because αe is continuous and [a, b] is connected. (ii) αe ([a, b]) contains x0 since αe (a) = x0. (iii) pαe = α. 0 0 If αe(s0) 6= αe (s0) for some s0 ∈ [a, b], then ϕnαe(s0) 6= ϕnαe (s0) since ϕn is bijective (and then injective). 0 0 0 However, since pαe = pαe = α, it follows that ϕnαe(s0) = ϕnαe (s0). Therefore, having ϕnαe(s0) 6= ϕnαe (s0) is 0 a contradiction. So for all s ∈ [a, b] we have αe(s) = αe (s).

Because the map p: R −→ S1 satisﬁes (2.2.1), we say that R is a covering space of S1. This gives rise to a theory called the theory of covering spaces that we won’t be studied in this course. If the reader is interested in that theory, he can read [1, Pages 56–70].

Lemma 2.2.2. (lifting lemma) Let α: I −→ S1 be a path such that α(0) = 1. Then there exists a unique path αe: I −→ R such that αe(0) = 0 and pαe = α.

Proof. Let s ∈ I, and let Ns be a small open arc containing α(s). Since α is continuous, there exists an open neighborhood Is of s such that α(Is) ⊆ Ns. Consider then the family {Is}s∈I . It indeed forms an open cover of I. Therefore, by the compactness of I, there exists a0, a1, ··· , am ∈ I with a0 = 0, am = 1, ai < ai+1 such that m−2 ! [ I = [0, a1] ∪ [ai, ai+1] ∪ [am−1, 1]. i=1

Applying Lemma 2.2.1 to the restriction α|[0, a1], we get αe|[0, a1]. Applying the same lemma to α|[a1, a2], we get αe|[a1, a2], and so on. After a ﬁnite number of steps, we get the desired αe, which is unique since by Lemma 2.2.1 each αe|[ai, ai+1] is unique.

1 Definition 2.2.3. Let α: I −→ S be a path such that α(0) = 1. The path αe from Lemma 2.2.2 is called the lift of α. So αe must satisfies two conditions: αe(0) = 0 and pαe = α. 1 2πis Example 2.2.4. Let α: −→ S be the path defined as α(s) = e . We want to find αe. By definition, αe have to satisfy the equations αe(0) = 0 and pαe = α. That is, for all s ∈ I, one has

2πi(α(s)) 2πis e e = e and αe(0) = 0.

Solving these equations, we get αe(s) = s.

From now on, any lift will be along p. One can also lift the homotopies as it is said in the following result. 2.2. THE FUNDAMENTAL GROUP OF THE CIRCLE 19

1 Lemma 2.2.5. Let α, β : I −→ S be paths such that α(0) = β(0) = 1. Let αe and βe be the lifts of α and β respectively. Let H : I × I −→ S1 be a homotopy from α to β. Then there exists a unique continuous map He : I × I −→ R such that He(s, 0) = αe(s) for all s ∈ I, and pHe = H. Moreover He is a homotopy from αe to βe, which implies in particular that

αe(1) = βe(1). (2.2.2)

1 Proof. Let t ∈ I be any point in [0, 1]. Consider the path Ht : I −→ S defined by Ht(s) = H(s, t). By hypothesis we have H0 = α and H1 = β. Again by hypothesis H is a homotopy from α to β, and by the axiom (HP2) from Definition 2.1.3 we have H(0, t) = α(0) = 1. So Ht(0) = 1. Applying Lemma 2.2.2 to the path Ht, we obtain the existence of a unique path Het : I −→ R such that Het(0) = 0 and pHet = Ht. Define now He : I × I −→ R by He(s, t) = Het(s).

• The map He is continuous. Indeed, for (s, t) ∈ I × I pick a very small closed arc U in S1 such that H(s, t) ∈ U. As in the proof of Lemma 2.2.1, there exists a closed interval Vn that maps homeomor- −1 −1 phically to U under ϕn = p|Vn. The map He is then deﬁned by He(s, t) = ϕn (H(s, t)). Since ϕn and H are continuous, it follows that He is also continuous.

• Moreover, we have He0 = αe since He0 is the lift of H0 = α. We also have pHe = H since by construction Het is the lift of Ht for every t ∈ I.

Now we prove that He is a homotopy from αe to βe. Clearly He satisfies the axiom (HP1) from Definition 2.1.3 since He0 is the lift of H0 = α, and He1 is the lift of H1 = β. For the axiom (HP2), we easily have He(0, t) = Het(0) = 0 = αe(0), ∀t ∈ I, by the first property of the lifting. To see that He(1, t) = αe(1) for all 1 t, consider the path λ: I −→ S defined by λ(t) = H(1, t) = Ht(1). Clearly λ is the constant path at α(1) since H is a homotopy from α to β (and thus satisfies (HP2)). It is also clear that the lift of λ is the path λe: I −→ R defined by λe(t) = He(1, t) = Het(1). So pλe(t) = α(1) for all t. This implies that λe is the constant path as well. In particular we have λe(0) = λe(1), which is equivalent to αe(1) = βe(1). 1 Lemma 2.2.6. Let α, β : I −→ S be two loops at 1. Then the lifts αg· β, α,e and βe of α· β, α, and β respectively satisfy the equation αg· β(1) = αe(1) + βe(1).

Proof. Exercise.

1 2.2.2 Proving that π1(S ) = Z

Theorem 2.2.7. The fundamental group of the circle S1 is isomorphic to Z.

Proof. Since S1 is path-connected, the choice of the basepoint does not matter. To avoid some complications, 2πi0 1 take 1 = e to be our basepoint. The plan of the proof is to define two homomorphisms f : Z −→ π1(S ) 1 and g : π1(S ) −→ Z and prove that g is the inverse of f. 1 2πins Definition of f. For every n ∈ Z define a loop αn : I −→ S by αn(s) = e . That loop wraps n times around the circle. Define now f by

f(n) = [αn]

The map f is a homomorphism of groups since f(n + m) = [αn+m] = [αn· αm] = [αn][αm]. Indeed, one can 1 show that the composition αn· αm is homotopic to αn+m.

1 Here αn· αm is the composition of paths and not the standard product of two exponentials! 20 CHAPTER 2. THE FUNDAMENTAL GROUP

Definition of g. Define g by the formula g([α]) = αe(1), where αe is the lift of α as in Lemma 2.2.2. The map g is well defined because of the following two things. The first one is the fact that αe(1) is an integer. This is because p(αe(1)) = α(1) = 1. The second thing is the fact that αe(1) = βe(1) when α ∼ β. This immediately comes from (2.2.2) in Lemma 2.2.5. The map g is a homomorphism of groups since αβf (1) = αe(1) + βe(1) by Lemma 2.2.6. The inverse of f is g. There is an explicit lift of αn: the path αen : I −→ R defined as αen(s) = ns. Of course we have αen(0) = 0 and pαen = αn. We also have gf(n) = g[αn] = αen(1) = n for all n ∈ Z. So gf = id. To 2 1 end the proof, it suffices to show that g is injective . To do this, let [α], [β] ∈ π1(S ) such that g[α] = g[β]. Then αe(1) = βe(1), which implies that αe· βe is a loop at 0 in R. Since π1(R) = 0 by Example 2.1.14, it follows that [αe· βe] = [c0], which gives αe ∼ βe. Appying now Corollary 1.1.6 –(i) we get pαe ∼ pβe, that is, α ∼ β. Hence [α] = [β].

The following is a quick application of Theorem 2.2.7.

Example 2.2.8. The fundamental group of the torus T 2 is Z × Z. That is,

2 π1(T ) = Z × Z.

2 1 1 1 1 2 1 1 ∼ Solution. Recall that the torus, T = S × S , is the product of S with S . So π1(T ) = π1(S × S ) = 1 1 π1(S ) × π1(S ) by Proposition 2.1.15. The desired result follows now from Theorem 2.2.7.

2.3 Induced homomorphism

The aim of this section is to prove Theorem 2.3.7, which asserts that the fundamental group of a topological space X depends only on the homotopy type of X. In order words, the theorem says that if X is homotopy

equivalent to Y , then π1(X) is isomorphic to π1(Y ). We begin with the deﬁnition of the induced homomorphism. Let X and Y be two topological spaces, and let x and y be the basepoints of X and Y respectively. A pointed-map from X to Y is a continuous map f : X −→ Y such that f(x) = y. In order words, it is a continuous map that sends the basepoint to the

basepoint. Let f : X −→ Y be a pointed map. Deﬁne f∗ : π1(X, x) −→ π1(Y, y) by

f∗([α]) = [f ◦ α]. (2.3.1)

The map f∗ is well deﬁned because, by Corollary 1.1.6 –(i), if α ∼ β then fα ∼ fβ. Furthermore f∗ is a homomorphism of groups. Indeed, if α, β : I −→ X are two loops at x, by Deﬁnition 2.1.2, we have 1 α(2s) if 0 ≤ s ≤ 2 α· β(s) = 1 which implies that β(2s − 1) if 2 ≤ s ≤ 1,

1 f ◦ α(2s) if 0 ≤ s ≤ 2 (f ◦ (α· β))(s) = 1 = (f ◦ α)· (f ◦ β). f ◦ β(2s − 1) if 2 ≤ s ≤ 1,

Deﬁnition 2.3.1. The homomorphism f∗ : π1(X, x) −→ π1(Y, y) is called the induced homomorphism by f.

Proposition 2.3.2. (i) If X is a space with x as its basepoint, then (idX )∗ = idπ1(X,x).

2 In general, if f : X −→ Y and g : Y −→ X are two maps such that gf = idX and g is injective, then f is bijective. Indeed, 0 0 0 f is injective. This is because f(x) = f(x ) implies that g(f(x)) = g(f(x )), which is equivalent to x = x since gf = idX . The map f is also surjective since for every y ∈ Y , we have f(g(y)) = y. In fact, if f(g(y)) 6= y, then gfg(y) 6= g(y) because g is injective. We would then have g(y) 6= g(y), which is a contradiction. 2.3. INDUCED HOMOMORPHISM 21

f g (ii) If X / Y / Z are two pointed maps, then (gf)∗ = g∗f∗.

Proof. (i) By (2.3.1) we have (idX )∗[α] = [id ◦ α] = [α] for every [α] ∈ π1(X, x).

(ii) For every [α] ∈ π1(X, x) we have again by (2.3.1)

(gf)∗[α] = [(gf) ◦ α] = [g(f ◦ α)] = g∗([f ◦ α]) = g∗f∗([α]).

The following is an immediate consequence of Proposition 2.3.2.

Corollary 2.3.3. Let f : X −→ Y be a pointed map. If f is a homeomorphism then f∗ is an isomorphism.

Before we state the next proposition, we need to deﬁne the pointed version of the homotopy type. Let X and Y be two spaces, and let x0 ∈ X, y0 ∈ Y be the basepoints of X and Y respectively. Let f, g : X −→ Y be two pointed map. A continuous map H : X × I −→ Y is said to be a basepoint-preserving homotopy from

f to g if H is a homotopy in the sense of Deﬁnition 1.1.1, and H(x0, t) = y0 for all t. In that case we write 0 f ∼0 g. Note that by deﬁnition f ∼0 g implies f ∼ g. If there is another pointed map f : Y −→ X such 0 0 that f f ∼0 idX and ff ∼0 idY , we say that f is a base-point preserving homotopy equivalence. In that case we write X '0 Y meaning that X and Y have the same homotopy type in the pointed context.

Proposition 2.3.4. (i) If f ∼0 g then f∗ = g∗.

∼ (ii) If X '0 Y then π1(X, x0) = π1(Y, y0).

Proof. (i) Assume that f ∼0 g, and let [α] ∈ π1(X, x0). Then by Corollary 1.1.6 –(ii), we have fα ∼ gα, which is equivalent to [fα] = [gα]. So f∗([α]) = g∗([α]).

0 (ii) Suppose that X '0 Y . Then there exists two pointed maps f : X −→ Y and f : Y −→ X such that 0 0 0 f f ∼0 idX and ff ∼0 idY . By the ﬁrst part of this proposition, we deduce that (f f)∗ = (idX )∗ and 0 that (ff )∗ = (idY )∗. Applying now Proposition 2.3.2, we deduce that f∗ : π1(X, x0) −→ π1(Y, y0) is 0 ∼ an isomorphism (its inverse is indeed f∗). So π1(X, x0) = π1(Y, y0).

Having to pay so much attention to basepoints when dealing with the fundamental group is something nuisance. For homotopy equivalences one does not have to be quite so careful since the conditions on the basepoints can actually be dropped, as shown Proposition 2.3.6 below. Let f, g : X −→ Y be two continuous maps (not necessarily pointed), and let H : X ×I −→ Y be a homotopy

from f to g (that is, f ∼ g). For every x ∈ X, deﬁne a path γx : I −→ Y by γx(t) = H(x, t). Clearly the

starting point of γx is f(x), and its ending point is g(x). Deﬁne also ϕγx : π1(Y, f(x)) −→ π1(Y, g(x)) by the formula

ϕγx ([α]) = [γx· α· γx].

By Proposition 2.1.16, ϕγx is an isomorphism of groups. 22 CHAPTER 2. THE FUNDAMENTAL GROUP

Lemma 2.3.5. For any x ∈ X, the diagram

π1(Y, g(x)) 5 O g∗

ϕ π1(X, x) γx

f∗ ) π1(Y, f(x))

commutes. That is, ϕγx ◦ f∗ = g∗.

Proof. Exercise.

Proposition 2.3.6. If f : X −→ Y is a homotopy equivalence in the sense of Deﬁnition 1.1.7, then for any

x ∈ X the induced homomorphism f∗ : π1(X, x) −→ π1(Y, f(x)) is an isomorphism.

Proof. Exercise.

As an immediate and nice consequence of Proposition 2.3.6, we have the following, which is the main result of the section. It is also a stronger version of Proposition 2.3.4 –(ii) because, here, we don’t care about the basepoint.

Theorem 2.3.7. Let X and Y be two topological spaces not necessarily pointed. Assume that X and Y are homotopy equivalent. Then their fundamental groups are isomorphic. That is, ∼ X ' Y ⇒ π1(X) = π1(Y ).

∼ The converse is not true: one can have π1(X) = π1(Y ) without having X ' Y . For instance, by Exam- 2 ∼ 2 ple 2.1.14, π1(R) = 0. As we will see later in Example 2.5.6, π1(S ) = 0. So π1(R) = π1(S ). But, manifestly, R is not homotopy equivalent to the sphere S2. Here are some quickly applications of Theorem 2.3.7. Example 2.3.8. The fundamental group of the disk D2 is trivial. Indeed, by Example 1.1.9 –(ii) we have 2 2 ∼ D ' pt where pt = {(0, 0)}. Applying now Theorem 2.3.7 we get π1(D ) = π1(pt). Since the fundamental 2 group of the one point space is trivial, it follows that π1(D ) = 0. More generally, if X is contractible 2 then π1(X) = 0 by Theorem 2.3.7. Another example is π1(R \{(0, 0)}), which is isomorphic to Z by Exam- ple 1.1.9 –(i), Theorem 2.3.7, and Theorem 2.2.7.

We close this section with the notion of retraction for spaces. Recall the inclusion map i: A,→ X, which is deﬁned by i(x) = x if A is a subspace of X.

Deﬁnition 2.3.9. Let X be a topological space, and A be a subspace of X. We say that A is a retract of X if there is a continuous map r : X −→ A such that r ◦ i = idA. If in addition i ◦ r ∼ idX , we say that A is a deformation retract of X.

For instance, by Example 1.1.9, the circle S1 is a deformation retract of R2\{(0, 0)}. Again by the same example, the one-point space {(0, 0)} is a deformation retract of D2. However, as we will see in Section 2.4, S1 is not a retract of D2. The following result immediately follows from Proposition 2.3.6.

Proposition 2.3.10. Let i: A,→ X be the inclusion map. If A is a deformation retract of X then the induced homomorphism i∗ is an isomorphism. 2.4. APPLICATIONS OF THE FUNDAMENTAL GROUP 23 2.4 Applications of the fundamental group

In this section, we will see two applications (among many others) of the fundamental group. The first one is topological, and regards the Brouwer fixed point theorem, which says that any continuous map from D2 to itself has a fixed point. The second application is more algebraic, and regards the fundamental theorem of algebra, which says that any non constant polynomial with complex coefficients has at least one root. Another interesting application is the Borsuk-Ulam theorem (see [1, Theorem 1.10]), which asserts that any continuous map f from the sphere S2 to the plane R2 admits a pair of antipodal points x and −x in S2 such that f(x) = f(−x). So at any time, there are two antipodal points on the earth with the same temperature and the same barometric pressure!

2.4.1 The Brouwer ﬁxed point theorem

The goal here is to prove Theorem 2.4.2 below. Recall the unit disk D2 = {(x, y) ∈ R2 : x2 + y2 ≤ 1}. Its boundary is the circle S1. Proposition 2.4.1. Let i: S1 ,→ D2 be the inclusion map. There is no continuous map r : D2 −→ S1 such that r ◦ i = idS1 .

Proof. Assume that such a map r exists, and consider the following diagram of homomorphisms of groups

1 i∗ 2 r∗ 1 π1(S ) / π1(D ) / π1(S ) .

1 1 The assumption r ◦ i = idS and Proposition 2.3.2 imply that r∗ ◦ i∗ = idπ1(S ). But this is not possible 2 1 since π1(D ) = 0 by Example 2.3.8, and π1(S ) = Z by Theorem 2.2.7.

Before we state the theorem, recall that a map f : X −→ X admits a fixed point if there is some x0 such that f(x0) = x0. Theorem 2.4.2 (Brouwer fixed point theorem). Any continuous map f : D2 −→ D2 has a fixed point.

Proof. Let f : D2 −→ D2 be a continuous map. Assume that for every x ∈ D2, f(x) 6= x. This allows 2 1 1 us to deﬁne r : D −→ S by r(x) = Rx ∩ S , where Rx is the ray that starts at f(x) and passes through x. Obviously r ◦ i(x) = x for all x ∈ S1. Moreover r is continuous because so is f. This contradicts Proposition 2.4.1.

2.4.2 The fundamental theorem of algebra

Let us begin with the notion of degree of a map. Roughly speaking, the degree of a continuous map f : S1 −→ S1 is a number that represents the number of times that f(S1) wraps around the codomain S1. Explicitly, it is deﬁned as follows. First consider the map q : I −→ S1 deﬁned by q(s) = e2πis. Consider now the composition fq = f ◦ q : I −→ S1. Clearly fq is a loop at f(1) (note that f(1) is not necessarily equal to 1 1 1, the basepoint we used in the previous sections). So [fq] ∈ π1(S , f(1)). To carry f(1) to 1, let γ : I −→ S 1 be a path starting at f(1) and ending at 1. By Proposition 2.1.17, the element [γ· fq· γ] ∈ π1(S , 1) is independent of the choice of [γ]. Set

αf = γ· fq· γ,

and consider the lift αff : I −→ R of αf , which is unique by Lemma 2.2.2. Notice that αff (1) is an integer 1 because p ◦ αff (1) = αf (1) by Lemma 2.2.2, and αf (1) = 1 (recall that p: R −→ S was deﬁned by p(s) = e2πis). 24 CHAPTER 2. THE FUNDAMENTAL GROUP

Deﬁnition 2.4.3. The degree of a map f : S1 −→ S1 is an integer, denoted deg(f), and deﬁned as the image of 1 ∈ [0, 1] under αff . That is, deg(f) = αff (1). Example 2.4.4. (i) If f : S1 −→ S1 is constant, then deg(f) = 0.

1 1 n (ii) Let n ∈ Z. The degree of the map fn : S −→ S deﬁned by fn(x) = x is n.

Solution.

(i) Assume that f : S1 −→ S1 is a constant map. Then fq : I −→ S1 is the constant path at f(1). That

is, fq = cf(1). Therefore, for any path γ as before, we have γ· fq· γ ∼ γγ ∼ c1. This implies that αf ∼ c1, and that deg(f) = αff (1) = ce1(1) by (2.2.2) from Lemma 2.2.5. Since the lifting is unique by Lemma 2.2.2, it follows that ce1 : I −→ R is the constant map at 0. Thus ce1(1) = 0, which implies that deg(f) = 0.

(ii) We want to show that deg(fn) = n. Notice ﬁrst that fn(1) = 1 by the deﬁnition of fn. So we can take γ to be the constant path at 1. This implies that αf = γ· fnq· γ ∼ fnq, and therefore we have 2πins deg(fn) = αff (1) = fgnq(1) by (2.2.2) from Lemma 2.2.5. Since fnq(s) = e for all s, by the uniqueness of the lifting, it follows that fgnq(s) = ns (we have used the lift s 7→ ns in the proof of Theorem 2.2.7). So fgnq(1) = n, which implies that deg(fn) = n.

Proposition 2.4.5. If two maps f, g : S1 −→ S1 are homotopic then deg(f) = deg(g).

Proof. Assume that f ∼ g. Then, by Corollary 1.1.6 –(ii), it follows that fq ∼ gq. This implies that for any 1 path γ in S such that γ(0) = f(1) and γ(1) = 1, we have γ· fq· γ ∼ γ· gq· γ. That is, αf ∼ αg. Applying now (2.2.2) from Lemma 2.2.5, we get αff (1) = αfg(1), meaning that deg(f) = deg(g).

We are now ready to state and prove the main result of this subsection.

n n−1 Theorem 2.4.6 (The fundamental theorem of algebra). Let f(x) = x + a1x + ··· + an−1x + an be a polynomial with complex coeﬃcients ai, where n > 0. Then there is a complex number x such that f(x) = 0. Therefore there are n such complex numbers (counted with multiplicities).

Proof. If there is x ∈ S1 such that f(x) = 0 then the proof is over. Assume that f(x) 6= 0 for all x ∈ S1. 1 1 f(x) This allows us to define fb: S −→ S by fb(x) = |f(x)| where | z | stands for the modulus of z. We proceed to calculate deg(fb). Suppose that f(x) 6= 0 inside the unit disk D2. This allows us to define H : S1 × I −→ S1 f(tx) f(0) by H(x, t) = |f(tx)| . Then H is a homotopy from the constant map at |f(0)| to fb, and we conclude that deg(fb) = 0 by Proposition 2.4.5 and Example 2.4.4 –(i). Suppose next that f(x) 6= 0 for all x such that | x |≥ 1. This allows us to define K : S1 × I −→ S1 by

L(x, t) x K(x, t) = , where L(x, t) = tnf( ) = xn + t(a xn−1 + ta xn−2 + ··· + tn−1a ). | L(x, t) | t 1 2 n

Then K is a homotopy from fn to fb, and we conclude that deg(fb) = n by Proposition 2.4.5 and Exam- ple 2.4.4 –(ii). Since by hypothesis n > 0, it is not possible to get at the same time deg(fb) = 0 inside the disk D2, and deg(fb) = n outside. So we must then have f(x) = 0 for some x either inside or outside D2. 2.5. THE VAN KAMPEN THEOREM 25 2.5 The Van Kampen theorem

The Van Kampen theorem gives a method for computing the fundamental groups of spaces than can be decomposed into simpler spaces whose fundamental groups are already known. The theorem involves the notion of free product of groups.

2.5.1 Free products of groups

(a) Disjoint union of sets. The disjoint union of two sets X and Y , denoted X ` Y , is a set deﬁned by a X Y = (X × {0}) ∪ (Y × {1}) .

For example, if X = {2, 3} and Y = {2, 4}, then X ` Y = {(2, 0), (3, 0), (2, 1), (4, 1)}. One can see in this example the diﬀerence between the disjoint union and the classical union of X and Y (X ∪ Y = {1, 2, 3}). (b) Words on an alphabet. Let A be a set, viewed as an alphabet. A word on A is a ﬁnite sequence

a1a2 ··· ak, k ≥ 0, where each letter ai belongs to A. The integer k is called the length of the word. A word of length k = 0 is called the empty word. ` (c) Free product of two groups. Let G1 and G2 be two groups. Let A = G1 G2. Deﬁne G1 ?G2 to be the set of words a1a2 ··· ak, k ≥ 0, on the alphabet A such that adjacent letters ai and ai+1 belong to diﬀerent groups. Words satisfying these conditions are called reduced. One can endow G1 ?G2 with a natural group structure:

• The group operation or the multiplication is the juxtaposition of words. For instance,

(a1a2 ··· ak)(b1b2 ··· bm) = a1a2 ··· akb1b2 ··· bm.

Note that the word a1a2 ··· akb1b2 ··· bm may not be reduced! The point is we can always carry any word to a unique reduced one (see Example 2.5.2 below) by using the following two rules. Rule1: if two 0 0 adjacent letters, say a and a , belong to the same group G1 (or G2), we replace the word aa , which is 0 a word with two letters, by the word a·G1 a (which is a word with only one letter. Here “·G1 ” stands for the group operation in G1). Rule2: if the identity element appears somewhere in a word, we take it out. For example, if a ∈ G1 and e2 is the identity element of G2, then we replace the word ae2 by the word a.

• The identity element is the empty word.

−1 −1 −1 • The inverse of a1a2 ··· ak is the word ak ··· a2 a1 .

Remark 2.5.1. The free product of two groups G1 and G2 is not abelian in most of cases. This is because the word ab, a ∈ G1, b ∈ G2, is not the same as the word ba. For instance Z ? Z is not abelian.

Example 2.5.2. Take G1 = G2 = Z. In this example we just want to ﬁgure out the group multiplication in ` Z ? Z. Since the alphabet is the disjoint union A = Z Z, we need to distinguish the generators. Let a be the generator of the ﬁrst copie of Z, and b be the one of the second copie. That is, G1 =< a > and G2 =< b >.

(i) (a2bab−3)(a4b) = a2bab−3a4b.

(ii) (a2bab−3)(b5a) = a2bab−3b5a = a2bab2a.

(iii) (a2bab−3)(b3a−1b−2) = a2bab−3b3a−1b−2 = a2baa−1b−2 = a2bb−2 = a2b−1. 26 CHAPTER 2. THE FUNDAMENTAL GROUP

(d) A natural fact about G1 ?G2. Let G be another group, and let i1 : G −→ G1 and i2 : G −→ G2 be group homomorphisms. Consider the following diagram.

i2 G / G2 (2.5.1)

i1 G1

−1 Let NG be the normal subgroup of G1?G2 generated by words of the form i1(g)(i2(g)) , g ∈ G. Consider the quotient group G1 ?G2/NG. There are canonical homomorphisms j1 : G1 −→ G1 ?G2/NG, x 7→ j1(x) = xNG, and j2 : G2 −→ G1 ?G2/NG, y 7→ j2(y) = yNG. So the original diagram becomes the commutative square

i2 G / G2 (2.5.2)

i1 j2

j1 G1 / G1 ?G2/NG.

Here is the fact.

Proposition 2.5.3. Given another group A, and homomorphisms f1 : G1 −→ A and f2 : G2 −→ A such that f1i1 = f2i2, there exists a unique homomorphism Φ: G1 ?G2/NG −→ A such that Φj1 = f1 and Φj2 = f2.

Proof. For an element a1a2 ··· akNG of G1 ?G2/NG, we will assume without lost the generality that a1 ∈ G1, a2 ∈ G2, ··· , ak ∈ G2. Deﬁne Φ by

Φ(a1a2 ··· akNG) = f1(a1)f2(a2) ··· f2(ak).

We need to check four things.

−1 −1 • Φ is well deﬁned. Indeed, let b1b2 ··· bm ∈ G1 ?G2 such that a1a2 ··· ak(b1b2 ··· bm) = i1(g)(i2(g)) for some g ∈ G. A simple computation gives

−1 −1 −1 f1(i1(g))f2 (i2(g)) = f1(i1(g)) (f2(i2(g))) = f2(i2(g)) (f2(i2(g))) .

−1 The last equality comes from the hypothesis f1i1 = f2i2. So f1(i1(g))f2 (i2(g)) = e where e is the

identity element of A. Therefore, f1(a1)f2(a2) ··· f2(ak) = f1(b1)f2(b2) ··· f2(bm), which implies that Φ(a1a2 ··· akNG) = Φ(b1b2 ··· bmNG).

• Φ is a homomorphism of groups since

Φ((a1 ··· akNG)(b1 ··· bmNG)) = Φ(a1 ··· akb1 ··· bmNG) = (f1(a1) ··· f2(ak))(f1(b1) ··· f2(bm)) = Φ(a1 ··· akNG)Φ(b1 ··· bmNG).

• For any g1 ∈ G1, we have Φj1(g1) = Φ(g1NG) = f1(g1) by the deﬁnition of Φ. So Φj1 = f1. Similarly, we have Φj2 = f2.

0 0 • Uniqueness of Φ. Let Φ : G1 ?G2/NG −→ A be another homomorphism such that Φ j1 = f1 and 0 0 Φ j2 = f2. We want to prove that Φ = Φ . To do this, let a1a2 ··· akNG ∈ G1 ?G2/NG. By the deﬁnition of the group operation in G1 ?G2, and by the deﬁnition of j1 and j2, we have

a1a2 ··· akNG = (a1NG)(a2NG) ··· (akNG) = j1(a1)j2(a2) ··· j2(ak). 2.5. THE VAN KAMPEN THEOREM 27

Therefore, because Φ0 is a homomorphism of groups, we have

0 0 0 0 Φ (a1a2 ··· akNG) = Φ j1(a1)Φ j2(a2) ··· Φ j2(ak) = f1(a1)f2(a2) ··· f2(ak) = Φ(a1a2 ··· akNG).

0 0 The second equality comes from the assumption Φ j1 = f1 and Φ j2 = f2, and the last equality from the deﬁnition of Φ.

In a more sophisticated language, the group G1 ?G2/NG is called the pushout or the colimit of the diagram (2.5.1).

Remark 2.5.4. (i) If G = 0 then NG is trivial, and therefore G1 ?G2/NG = G1 ?G2. In that case, we have Φ: G1 ?G2 −→ A.

(ii) If G1 = G2 = 0 then G1 ?G2 = 0, and therefore Φ: 0 −→ A is the trivial homomorphism. We have defined the free product of two groups. One can easily extend this definition to the free product of a finite number of groups G1,G2, ··· ,Gn, and get a generalization of Φ (see for example [1, Pages 40–42]).

2.5.2 Van Kampen’s theorem

There are many versions of the Van Kampen theorem. We will use only the simplest one in this course (for a more general statement, we refer the reader to [1, Theorem 1.20]).

Let X be a topological space. Assume that X = U1 ∪ U2 where U1 and U2 are path-connected open subsets of X such that U1 ∩ U2 is also path-connected. Choose the basepoint x0 inside U1 ∩ U2. Consider the inclu- sions maps i1 : U1 ∩ U2 ,→ U1, i2 : U1 ∩ U2 ,→ U2, f1 : U1 ,→ X, and f2 : U2 ,→ X. Applying the fundamental i1 i2 group to U1 o ? _ U1 ∩ U2 / U2 we get the diagram

i2∗ π1(U1 ∩ U2) / π1(U2)

i1∗ π1(U1), which has the same form as (2.5.1). Set G = π1(U1 ∩ U2),A = π1(X), and consider the induced homomorphisms f1∗ : π1(U1) −→ π1(X) and f2∗ : π1(U2) −→ π1(X). Clearly we have f1∗i1∗ = f2∗i2∗. Applying now Proposition 2.5.3, there exists a unique homomorphism

Φ: π1(U1) ? π1(U2)/NG −→ π1(X) (2.5.3) such that Φj1 = f1∗ and Φj2 = f2∗ where j1, j2 are as in (2.5.2). We are now ready to state the main theorem of the section.

Theorem 2.5.5 (The Van Kampen theorem). Assume that we have the underlined above things. Then the homomorphism Φ from (2.5.3) is an isomorphism.

Proof. See [1, Pages 44–46].

n Example 2.5.6. π1(S ) = 0 for all n ≥ 2. 28 CHAPTER 2. THE FUNDAMENTAL GROUP

n n n Solution. Write S = U1 ∪ U2 where U1 = S \{(0, ··· , 0, 1)} and U2 = S \{(0, ··· , 0, −1)}. It is clear ∼ n ∼ that U1,U2, and U1 ∩ U2 are path-connected. Moreover π1(U1) = π1(U2) = 0 since U1 = R = U2 by the n stereographic projection, and since π1(R ) = 0. So by Theorem 2.5.5, we have n ∼ π1(S ) = π1(U1) ? π1(U2)/NG = 0 ? 0/NG = 0. Proposition 2.5.7. For all n 6= 2, R2 is not homeomorphic to Rn.

Proof. If n = 0 then the result immediately follows since R0 is the one point space. For n = 1, the result follows because removing one point in R provides two connected components, while removing one point in R2 does not change the number of connected components of R2 (which is actually one). Now assume that n ≥ 3 and suppose that there is a homeomorphism f : R2 −→ Rn. Then the map g : R2\{0} −→ Rn\{f(0)} deﬁned by g(x) = f(x) is still a homeomorphism. Therefore, by Corollary 2.3.3, the induced homomorphism 2 n g∗ : π1(R \{0}) −→ π1 (R \{f(0)}) is an isomorphism of groups. But this is not possible since 2 1 n n−1 π1(R \{0}) = π1(S ) = Z and π1(R \{f(0)}) = π1(S ) = 0.

For the ﬁrst π1 we have used Example 1.1.9 − (i), Theorem 2.3.7 and Theorem 2.2.7. And for the second one, we have used Exercise2 − (a) (from Assignment 1), Theorem 2.3.7 and Example 2.5.6.

Remark 2.5.8. The fundamental group is not abelian in general as we can see in Example 2.5.9 below.

Let X and Y be two spaces, and let x0 ∈ X, y0 ∈ Y . The wedge of X and Y is the space denoted X ∨ Y , and deﬁned by a X ∨ Y = X Y/(x0 ∼ y0). ` That is, X ∨ Y is obtained from the disjoint union X Y by identifying x0 with y0. An illustration of that is given by Figure 2.1.

Figure 2.1: An example of the wedge

1 Example 2.5.9. Let S be the circle endowed with its usual basepoint x0 = y0 = 1. Then the fundamental group of S1 ∨ S1 is the free product Z ? Z. That is, 1 1 π1(S ∨ S ) = Z ? Z.

Solution. The space we are dealing with is shown in Figure 2.2.

Figure 2.2: The wedge of two copies of the circle

1 1 Write S ∨ S = U1 ∪ U2 where U1 and U2 are as in Figure 2.3.

1 1 Clearly U1,U2, and U1 ∩ U2 are path-connected. Moreover we have U1 ' S ,U2 ' S , and U1 ∩ U2 ' pt. So π1(U1) = π1(U2) = Z by Theorem 2.3.7 and Theorem 2.2.7, and π1(U1 ∩U2) = 0 since U1 ∩U2 is contractible. 1 1 ∼ Thus, by the Van Kampen theorem, and by Remark 2.5.4 –(i), it follows that π1(S ∨ S ) = Z ? Z. 2.5. THE VAN KAMPEN THEOREM 29

Figure 2.3: A decomposition of S1 ∨ S1 30 CHAPTER 2. THE FUNDAMENTAL GROUP Chapter 3

Homology

3.1 Chain complexes

Deﬁnition 3.1.1. A chain complex is a collection {An}n∈Z, denoted A∗, of abelian groups equipped with group homomorphisms ∂n : An −→ An−1 such that

∂n ◦ ∂n+1 = 0 for all n ∈ Z. (3.1.1)

Example 3.1.2. Consider the collection {An}n∈Z deﬁned as  0 if n < 0  An = Z if n ∈ {0, 1, 2}  0 if n > 2.

Deﬁne ∂2 : Z −→ Z and ∂1 : Z −→ Z as ∂2(n) = 2n and ∂1(n) = 0. The other maps ∂n are deﬁned to be 0.

Clearly one has ∂n+1∂n = 0 for all n. So {An}n∈Z is a chain complex.

A chain complex A∗ will be writing in the following way:

∂n+1 ∂n · · · −→ An+1 / An / An−1 −→ · · · (3.1.2)

The elements of An are called n-chains and the homomorphisms ∂n are called the boundary maps or dif- ferentials. The groups An may be endowed with extra structure; for example, they may be vector spaces or modules over a ﬁxed ring R. The diﬀerentials must preserve the extra structure if it exists; for example, they must be linear maps or homomorphisms of R-modules.

Notation 3.1.3. If An happens to be 0 for all n < 0, the sequence (3.1.2) will be denoted

∂n+1 ∂n ∂n−1 ∂2 ∂1 ··· / An / An−1 / ··· / A1 / A0.

In that case A∗ will be just denoted {An}n≥0.

Deﬁnition 3.1.4. Let A∗ be a chain complex. An element a of An is called

(i) an n-cycle (or just a cycle) if ∂n(a) = 0;

(ii) an n-boundary (or just a boundary) if there exists b ∈ An+1 such that ∂n+1(b) = a.

31 32 CHAPTER 3. HOMOLOGY

Example 3.1.5. Consider the chain complex A∗ from Example 3.1.2.

• Since ∂0 : Z −→ 0 is the zero map, for every n ∈ Z, we have ∂0(n) = 0. So every n ∈ Z = A0 is a 0-cycle. For a similar reason, every element in A1 = Z is a 1-cycle. But in A2 = Z, the only 2-cycle is 0 since the kernel of ∂2 is the trivial group {0}.

• For m ∈ A2, m 6= 0, we have ∂2(m) = 2m 6= 0, and this implies that m is not an 2-cycle.

• One can see that every element of A1 of the form 2n is a 1-boundary since 2n = ∂2(n) by the deﬁnition of ∂2. One can also see that every odd integer in A1 is not a boundary.

The kernel of ∂n is exactly the collection of all n-cycles, while the image of ∂n+1 is the collection of all n-boundaries.

Equation (3.1.1) implies that Im∂n+1 is a subgroup of Ker∂n. Indeed, if a ∈ Im∂n+1 then there exists b ∈ An+1 such that a = ∂n+1(b). Composing this latter equality by ∂n, we get ∂n(a) = ∂n ◦ ∂n+1(b). Since ∂n ◦ ∂n+1 = 0 by (3.1.1), it follows that ∂n(a) = 0, which implies that a ∈ Ker∂n. So Im∂n+1 ⊆ Ker∂n. Furthermore, Im∂n+1 is a normal subgroup of Ker∂n since An is abelian and therefore Ker∂n is abelian as 1 well . One can then consider the quotient group Ker∂n/Im∂n+1, which is denoted Hn(A). That is,

Hn(A) := Ker∂n/Im∂n+1. (3.1.3)

By the deﬁnition of a quotient group, an element of Hn(A) is the class [a] of an n-cycle a with respect to the equivalence relation

0 0 a ∼ a if and only if there exists b ∈ An+1 such that a − a = ∂n+1(b).

In other words, a lies in the same class as a0 if and only if the diﬀerence a − a0 is a boundary. In particular, if a is an n-boundary, there exists b ∈ An+1 such that a = ∂n+1(b). This is equivalent to a − 0 = ∂n+1(b), which implies that [a] = [0]. So any boundary element in An is zero in Hn(A). The non-zero elements of Hn(A) are precisely the n-cycles which are not the boundary of something.

Deﬁnition 3.1.6. The group Hn(A) is called the nth homology group of A∗, and the collection {Hn(A)}n∈Z is called the homology of A∗, and it is denoted H∗(A).

As in the preceding chapter, the trivial group {0} will be just denoted 0. If two groups G and G0 are isomorphic, we will sometimes just write G = G0 instead of G =∼ G0. Given a group G, one can easily see that the quotient G/{0} of G by the trivial group is isomorphic to G, that is, G/{0} = G.

Example 3.1.7. Consider the chain complex A∗ from Example 3.1.2. We want to compute the homology groups Hn(A), n ∈ Z.

• If n < 0, then An = An−1 = 0 by the deﬁnition of An. This implies that Ker(∂n) = 0 and Im(∂n+1) = 0. Therefore Hn(A) = 0.

• If n > 2, a similar reasoning shows that Hn(A) = 0.

∂1 ∂0 • Assume n = 0 and consider the sequence A1 −→ A0 −→ A−1. Since A1 = A0 = Z,A−1 = 0 and ∂1 = 0 = ∂0 by deﬁnition, it follows that Im∂1 = 0 and Ker∂0 = Z. This implies that H0(A) = Ker∂0/ Im∂1 = Z.

1A classical fact in group theory says that any subgroup of an abelian group is normal. 3.2. SIMPLICIAL HOMOLOGY 33

∂2 ∂1 • Assume n = 1 and consider the sequence A2 −→ A1 −→ A0. Since ∂1 = 0, it follows that Ker∂1 = A1 = Z. Since ∂2 is the multiplication by 2, it follows that Im∂2 = {∂2(n): n ∈ Z} = {2n : n ∈ Z} = 2Z. So

H1(A) = Ker∂1/Im∂2 = Z/2Z = Z2.

∂3 ∂2 • Assume n = 2 and consider the sequence A3 −→ A2 −→ A1. Clearly Ker∂2 = 0 and Im∂3 = 0. This implies H2(A) = 0.

Conclusion:  0 if n < 0   if n = 0 H (A) = Z n if n = 1  Z2  0 if n ≥ 2.

There is a notion of “map” between chain complexes.

Deﬁnition 3.1.8. Let A∗ and B∗ be chain complexes. A chain map from f : A∗ −→ B∗ is a collection

{fn : An −→ Bn}n∈Z of group homomorphisms such that

∂nfn = fn−1∂n for all n. (3.1.4)

This amounts to saying that the following diagram commutes.

∂n+1 ∂n ··· / An+1 / An / An−1 / ···

fn+1 fn fn−1 ··· / Bn+1 / Bn / Bn−1 / ··· ∂n+1 ∂n

Proposition 3.1.9. Let f : A∗ −→ B∗ be a chain map.

(i) If a is an n-cycle, then so is fn(a).

(ii) If a is an n-boundary, then so is fn(a).

Proof. Exercise.

Proposition 3.1.9 implies that the map Hn(f): Hn(A) −→ Hn(B) that sends [a] to [fn(a)] is well deﬁned.

Deﬁnition 3.1.10. The collection {Hn(f)}n≥0 : H∗(A) −→ H∗(B), denoted f∗, is called the induced map in homology.

3.2 Simplicial homology

To define the simplicial homology of a topological space, we need to define first the notion of “∆-complex”. 34 CHAPTER 3. HOMOLOGY 3.2.1 ∆-complexes

We begin with the deﬁnition of a simplex.

m Definition 3.2.1. Consider a Euclidean space R , and let n ≤ m. Let v0, ··· , vn be a sequence of points m of R such that the difference vectors v1 − v0, ··· , vn − v0 are linearly independent. Define [v0, ··· , vn] as

( n n ) X X [v0, ··· , vn] = tivi | ti = 1 and ti ≥ 0 for all i . (3.2.1) i=0 i=0

This set is called the n-simplex generated by v0, ··· , vn. The coeﬃcients ti are called the barycentric coor- P dinates of the point i tivi in [v0, ··· , vn].

For purposes of homology it will be important to keep track of the order of the vertices of a simplex. From now on, “n-simplex” will really mean “n-simplex with an ordering of its vertices”. Unless stated otherwise,

we will always choose the natural order, that is, v0 ≤ · · · ≤ vn.

The points v0, ··· , vn are called the vertices of [v0, ··· , vn].A 0-simplex is just a point v0, while a 1-simplex is a line segment with endpoints v0 and v1.A 2-simplex is a triangle with vertices v0, v1, and v2, while a 3-simplex is a tetrahedron, and so on (see Figure 3.1).

Figure 3.1: Example of simplices

An equivalent deﬁnition of n-simplex is the following. A subset X of Rm is said to be convex if for any a, b ∈ X, the set {(1 − t)a + tb|t ∈ [0, 1]} is contained in X. One can see that the smallest convex subset in m 2 R containing v0, ··· , vn is exactly [v0, ··· , vn] . n+1 Example 3.2.2. In R , consider the points e0, ··· , en deﬁned as ei = (ei0, ··· , eij, ··· , ein) where 1 if j = i e = ij 0 otherwise.

For instance, if n = 2, one has e0 = (1, 0, 0), e1 = (0, 1, 0), and e2 = (0, 0, 1). The n-simplex generated by n e0, ··· , en is called the standard n-simplex and it is denoted ∆ . It follows immediately from the deﬁnition that

( n ) n n+1 X ∆ = (t0, ··· , tn) ∈ R | ti = 1 and ti ≥ 0 for all i . (3.2.2) i=0 See Figure 3.2 for the standard 2-simplex.

Remark 3.2.3. There is a canonical linear homeomorphism from the standard n-simplex ∆n to any other P n-simplex [v0, ··· , vn], preserving the order of vertices, namely, (t0, ··· , tn) 7→ i tivi. So n ∼ ∆ = [v0, ··· , vn].

2 m By “smallest” we mean the intersection of all convex subsets of R containing v0, ··· , vn. 3.2. SIMPLICIAL HOMOLOGY 35

Figure 3.2: The standard 2-simplex

Deﬁnition 3.2.4. Let [v0, ··· , vn] be an n-simplex, and let 0 ≤ k ≤ n. The (n − 1)-simplex generated by v0, ··· , vk−1, vk+1, ··· , vn is called a face of [v0, ··· , vn], and it is denoted [v0, ··· , vbk, ··· , vn].

So [v0, ··· , vn] has exactly n + 1 faces. In other words, a face of [v0, ··· , vn] is obtained by deleting one of the n + 1 vertices vk of [v0, ··· , vn]. We adopt the following convention: The vertices of a face, or any subsimplex generated by a subset of the vertices, will always be ordered according to their order in the larger simplex.

n The face of ∆ generated by e0, ··· , ek−1, ek+1, ··· , en is

 n n   X X  [e0, ··· , ebk, ··· , en] = tiei | ti = 1 and ti ≥ 0 for all i 6= k . i=0,i6=k i=0,i6=k 

k n For the sake of simplicity, [e0, ··· , ek, ··· , en] will be often just denoted ∂ ∆ . As in Remark 3.2.3, there is b ∼ a canonical linear homeomorphism φ: ∆n−1 −→= ∂k∆n preserving the order of vertices, namely,

φ(t0, ··· , tn−1) = t0e0 + ··· tk−1ek−1 + tkek+1 + ··· tn−1en. (3.2.3)

Deﬁnition 3.2.5. (i) The union of all faces of ∆n is called the boundary of ∆n, and it is denoted ∂∆n. So n [ ∂∆n = ∂k∆n. k=0

(ii) The interior of ∆n, denoted ∆˚n, is deﬁned to be ∆n\∂∆n. That is,

∆˚n = ∆n\∂∆n.

Now we are ready to deﬁne a ∆-complex.

Deﬁnition 3.2.6. A ∆-complex is a pair (X, ∆) where X is a topological space, and ∆ is a collection of n continuous maps σα : ∆ −→ X, with n depending on the index α, satisfying the following three conditions.

n (i) For every σα ∈ ∆, the restriction σα|∆˚ is injective, and each point of X is in the image of exactly n one such restriction σα|∆˚ .

n k n (ii) For any element σα : ∆ −→ X of ∆, each restriction σα|∂ ∆ is such that the composition

k n n−1 σα|∂ ∆ ◦ φ: ∆ −→ X

is one of the maps σβ from ∆. Here φ is the canonical linear homeomorphism deﬁned above. 36 CHAPTER 3. HOMOLOGY

−1 n (iii) A set A ⊂ X is open if and only if σα (A) is open in ∆ for each σα ∈ ∆.

Among other things, this last condition rules out trivialities like regarding all the points of X as individual vertices. If the structure is understood, we will just write X for (X, ∆).

1 0 Example 3.2.7. Let X be the unit circle S . Consider the following collection ∆ = {σ0, σ1} where σ0 : ∆ = 1 1 1 2πit0 {1} −→ S is defined as σ0(1) = (1, 0), and σ1 : ∆ −→ S is defined as σ1(t0, t1) = e . We want to show 1 that the pair (S , ∆) is a ∆-complex. First of all, the conditions t0 + t1 = 1 and t0, t1 ≥ 0 in the definition 1 of ∆ imply that t0, t1 ∈ [0, 1]. So the map σ1 wraps the circle exactly one time. Now let us check conditions (i), (ii), and (iii).

0 (i) We need to check (i) for every element of ∆. Let us begin with σ0. Since the boundary of ∆ is 0 0 0 empty, it follows that ∆˚ = ∆ . Therefore the restriction σ0|∆˚ is clearly injective. For σ1, a simple ˚1 2 computation gives ∆ = {(t0, t1) ∈ R |t0 + t1 = 1 and t0, t1 ∈ (0, 1)}. This implies that the restriction 1 2πis 1 σ1|∆˚ is injective (one can check that from the definition of σ1). Now let x = e ∈ S . If s = 0 then 0 x = (1, 0), which belongs to the image of σ0|∆˚ . If s ∈ (0, 1), then x definitely belongs to the image of 1 σ1|∆˚ . So (i) is satisfied.

k 1 (ii) This easily follows from the deﬁnition of σ1, ∂ ∆ , and φ. (iii) Exercise.

The idea behind a ∆-complex is as follows. Roughly speaking, a ∆-complex is a space X constructed by gluing together simplices. However, there are rules for what kinds of “gluing ” you are allowed to use.

(a) You have to start with all the 0-simplices, then glue in all the 1-simplices, then glue in all the 2-simplices, and so on. You are not allowed to drop this ordering condition.

(b) If Y is the ∆-complex you have built so far and f : ∂∆n −→ Y is any continuous map, you can build a ∆-complex X = Y ` ∆n/ ∼. , where “∼” is the equivalence relation that identiﬁes any x ∈ ∂∆n with f(x) = y. The highly restriction when adding a new simplex is the following. For each (n − 1)-simplex F = ∂k∆n which is a face of ∂∆n, the restriction of f to F must be equal to the inclusion of one of the (n − 1)-simplices you already have. That is, f maps the n vertices of F (with their canonical ordering) to the n vertices of some (n − 1)-simplex you have already added to your complex (with the same ordering on the vertices).

3.2.2 Deﬁnition of the simplicial homology and examples

n Let (X, ∆) be a ∆-complex. An element σα : ∆ −→ X of ∆ will be called an n-simplex of X.

Deﬁnition 3.2.8. Deﬁne ∆n(X) to be the free abelian group generated by all n-simplices of X.

P An element of ∆n(X) is a formal sum α nασα, where nα ∈ Z. We want to turn out the collection {∆n(X)}n≥0 to a chain complex (see Definition 3.1.1). So we need to define maps ∂n : ∆n(X) −→ ∆n−1(X) such that ∂n∂n+1 = 0 for all n. Define ∂n as

n X k ∂n(σα) = (−1) (σα|[e0, ··· , ebk, ··· , en]) ◦ φ, k=0 where φ is the homeomorphism from (3.2.3).

Proposition 3.2.9. For all n, one has ∂n∂n+1 = 0. 3.2. SIMPLICIAL HOMOLOGY 37

Proof. Exercise.

Proposition 3.2.9 tells us that ∂n is a diﬀerential. So the collection ∆∗(X) = {∆n(X)}n≥0 is a chain complex (see Deﬁnition 3.1.1 and Notation 3.1.3).

∆ Deﬁnition 3.2.10. The simplicial homology of X, denoted H∗ (X), is deﬁned to be the homology of the chain complex ∆∗(X). That is, ∆ Hn (X) := Ker∂n/Im∂n+1. ∆ The group Hn (X) is called the nth simplicial homology group of X. ∆ 1 Example 3.2.11. In this example, we would like to compute the simplicial homology H∗ (S ) of the circle. Consider the ∆-complex structure, ∆ = {σ0, σ1}, from Example 3.2.7. In that structure we have one 0- simplex, σ0, and one 1-simplex, σ1. So ∆0(X) is the free abelian group generated by σ0, that is, ∆0(X) = Z. Similarly ∆1(X) = Z. Since there is no n-simplex in ∆ with n∈ / {0, 1}, it follows that ∆n(X) = 0 for all n∈ / {0, 1}. So the chain complex ∆∗(X) looks like

∂3 ∂2 ∂1 ∂0 ∂−1 ··· / 0 / Z / Z / 0 / ···

∆ • Let us compute H0 (X). First of all, we have Ker(∂0) = Z. A simple computation shows that ∂1(σ1) = ∆ 1 0. So Im∂1 = 0, and therefore H0 (S ) = Z. ∆ • Now we compute H1 (X). Since ∂1(σ1) = 0, it follows that Ker∂1 = Z. Certainly Im∂2 = 0. So ∆ 1 H1 (S ) = Z. ∆ 1 • For the other values of n, we have Hn (S ) = 0.

Hence, if n = 0, 1 H∆(S1) = Z n 0 otherwise. n Remark 3.2.12. For the sake of simplicity, we will often identify an n-simplex σα : ∆ −→ X with its n n image σα(∆ ) (or with σα(∆˚ )). So instead of looking at σα as a map, we will look at it as a subspace of X. That point of view has the advantage that it makes computations easier. And the inconvenience that it makes much harder the veriﬁcation of (i), (ii), and (iii) from Deﬁnition 3.2.6. Since we are interested in computations, the ∆-complex structure should be always understood from the picture.

For example, let us consider again the circle S1. Deﬁne v ⊆ S1 as v = {(1, 0)} and e = S1. One should think e as a 1-simplex, staring at v and ending at v, that wraps the circle one time (see Figure 3.3).

Figure 3.3: A ∆-complex structure on S1

Assume that e is oriented in the counterclockwise. One can then replace the collection {σ0, σ1} before by {v, e}. So our ∆-complex structure on S1 consists of one 0-simplex, v, and one 1-simplex, e. This implies 1 1 that ∆0(S ) is the free abelian group generated by v that we denote hvi, and ∆1(S ) = hei. Since the the 1 1 free abelian group generated by one element is Z, it follows that ∆0(S ) = Z and ∆1(S ) = Z. Since there 1 is no n-simplex with n∈ / {0, 1}, it follows that ∆n(S ) = 0 for all n∈ / {0, 1}. Compare to what we did in Example 3.2.11, the diﬀerential is much easier to compute here. For instance, ∂1(e) = v − v = 0. 38 CHAPTER 3. HOMOLOGY

Example 3.2.13. Here we compute the simplicial homology of the torus T , as shown Figure 3.4.

Figure 3.4: Torus T

• How many 0-simplices do we have? Only one: v. So ∆0(T ) = hvi = Z. Since ∂0 : ∆0(T ) −→ 0 is the zero morphism, it follows that Ker∂0 = Z.

3 • How many 1-simplices do we have? Exactly 3: a, b, and c. So ∆1(T ) = ha, b, ci = Z . The boundary of a is ∂1(a) = v − v = 0. Similarly, ∂1(b) = ∂1(c) = 0. This implies that Im∂1 = 0 and Ker∂1 = ha, b, ci.

2 • How many 2-simplices do we have? Exactly 2: U and L. So ∆2(T ) = hU, Li = Z . The boundary of U is ∂2(U) = a + b − c = ∂2(L). This implies that Im∂2 = ha + b − ci and Ker∂2 = hU − Li = Z.

• How many n-simplices, n ≥ 3, do we have? Zero. So ∆n(T ) = 0 for n ≥ 3.

∆ ∆ Ker∂0 Now we compute H (T ). For n = 0, H (T ) = = since Ker∂0 = and Im∂1 = 0. For n = 1, we ∗ 0 Im∂1 Z Z have ∆ Ker∂1 ha, b, ci ha, b, a + b − ci ∼ ∼ H1 (T ) = = = = ha, bi = Z × Z. Im∂2 ha + b − ci ha + b − ci

∆ Ker∂2 ∼ ∆ For n = 2, we have H (T ) = = since Ker∂2 = and Im∂3 = 0. Certainly, H (T ) = 0 for n ≥ 3. 2 Im∂3 Z Z n Hence the homology of the torus is  Z if n ∈ {0, 2} ∆  Hn (T ) = Z × Z if n = 1  0 if n ≥ 3.

Notice that the orientation of simplices is very important in the calculations, especially when we want to compute the boundary or the diﬀerential of something.

Example 3.2.14. We would like to compute the simplicial homology of the real projective plane RP2, as shown Figure 3.5

2 • Since there are two 0-simplices, v and w, it follows that ∆0(X) = hv, wi = Z . Since ∂0 is the zero morphism, Ker∂0 = hv, wi.

3 • ∆1(X) = ha, b, ci = Z , ∂1(a) = w − v = ∂1(b), and ∂1(c) = 0. This implies that Im∂1 = hw − vi and Ker∂1 = ha − b, ci. 3.3. SINGULAR HOMOLOGY 39

Figure 3.5: Real projective plane RP2

2 • ∆2(X) = hU, Li = Z , ∂2(U) = a − b − c, and ∂2(L) = a − b + c. This implies that Ker∂2 = 0 and Im∂2 = ha − b − c, a − b + ci.

• Since there is no n-simplices when n ≥ 3, it follows that ∆n(X) = 0.

∆ Now we compute Hn (X). For n = 0, one has

∆ Ker∂0 hv, wi hv, w − vi ∼ H0 (X) = = = = hvi = Z. Im∂1 hw − vi hw − vi For n = 1,

∆ Ker∂1 ha − b, ci ha − b − c, ci ∼ hci Z ∼ H1 (X) ======Z2. Im∂2 ha − b − c, a − b + ci ha − b − c, 2ci 2hci 2Z

∆ Ker∂2 ∆ For n = 2, H (X) = = 0 since Ker∂2 = 0. Clearly, H (X) = 0 for n ≥ 3. So 2 Im∂3 n  Z if n = 0 ∆ 2  Hn (RP ) = Z2 if n = 1  0 if n ≥ 2.

3.3 Singular homology Work in progress...

3.3.1 Deﬁnition

3.3.2 Exact sequences

3.3.3 Relative homology groups

3.4 Applications 40 CHAPTER 3. HOMOLOGY Bibliography

[1] Allen Hatcher, Algebraic Topology, (2002) Cambridge University Press, xii+544 pp.