Introduction to Algrbraic Topology

Introduction to Algrbraic Topology

Ingo Waschkies

September 27, 2010

Abstract Lecture notes of a course given at the Université de Nice - Sophia Antipolis in fall 2010

Contents

1 Category Theory 2 1.1 Categories and Functors ...... 2 1.2 Cartesian squares ...... 5

2 The fundamental groupoid 9 2.1 Connected spaces ...... 9 2.2 Construction of the fundamental groupoid ...... 11 2.3 Homotopy invariance ...... 14

3 Coverings 16 3.1 Fiber bundles ...... 16 3.2 Coverings ...... 19 3.3 Group actions and coverings ...... 22 3.4 Universal coverings and fundamental groupoid ...... 24

1 1 Category Theory

1.1 Categories and Functors

Definition 1.1. A category C is given by the following data: (i) a class of objects Ob C,

(ii) for any two objects X, Y ∈ Ob C a set HomC (X, Y) whose elements are called the morphisms (or arrows) from X to Y (iii) and for any three objects X, Y, Z ∈ Ob C a map

◦XYZ : HomC (X, Y) × HomC (Y, Z) − HomC (X, Z);(f, g) 7 g ◦XYZ f

which is called composition map (the index XYZ→will be immediately omitted in the→ notations). such that the following two axioms are satisfied:

(C1) For any object X ∈ Ob C there exists a morphism idX ∈ HomC (X, X) such that for any object Y

and any morphisms f ∈ HomC (X, Y) we have f ◦ idX = f and idY ◦ f = f. (C2) The composition of morphisms is associative, i.e. for any objects X, Y, Z, W ∈ Ob C and any three

morphisms f ∈ HomC (X, Y), g ∈ HomC (Y, Z) h ∈ HomC (Z, W) we have

h ◦ (g ◦ f) = (h ◦ g) ◦ f.

Let C be a category and X, Y ∈ Ob C two objects. Instead of f ∈ HomC (X, Y) we will usually write f : X Y, the object X is called the source of the morphism f and Y is called the target of f. An endomorphism is a morphism which has the same source and target. A morphism f : X Y is→ called an isomorphism if there exixts a morphism g : Y X such that g ◦ f = idX −1 and f◦g = idY . If such a morphism g exists it is unique and called the inverse morphism f of f. → →

Let C be a category and X ∈ Ob C be an object. Then HomC (X, X) is a set endowed with a binary operation ◦ (the composition of morphisms) which is associative and has a unit element idX. Such a structure is called a monoid in basic algebra. It is not necessarily a group since mor- phisms need not be invertible. A category with only a single object ∗ is completely determined by the monoid HomC (∗, ∗), hence categories with one object are the same as monoids. In this sense a category is a generalization of the algebraic notion of monoid.

We have already mentioned that in order to have a group structure on HomC (X, X) we need to impose that every endomorphism of X is invertible. So similarly to the monoid situation we can state that a category with a single object such that every morphism is an isomorphism is the same as a group. This leads to the following generalization of the algebraic notion of group:

Definition 1.2. A groupoid is a category such that every morphism in G is an isomorphism. We will see soon that in some sense a groupoid is actually just a collection of groups, a collection that need not be a set. It will be convenient to introduce the opposite category C◦ of a category C. The category C◦ has the same objects as C but source and target of morphisms are formally exchanged, i.e. a morphism f : X Y in C◦ is a morphism f : Y X in C. More precisely:

→ →2 Definition 1.3. Let C be a category. Its opposite category C◦ is given by the following data: (i) Ob C◦ = Ob C, ◦ (ii) for any two objects X, Y ∈ Ob C we set HomC◦ (X, Y) = HomC (Y, X) (iii) and for any three objects X, Y, Z ∈ Ob C◦ the map

op ◦ : HomC◦ (X, Y) × HomC◦ (Y, Z) − HomC◦ (X, Z) is given by g ◦op f = f ◦ g (note that f, g are morphism f : Y X, g : Z Y in C so the → composition is well defined). Note that if we have a monoid (i.e. a category with just a single object),→ then the→ opposite cate- gory is given by the monoid with the opposite structure.

Definition 1.4. Let C, D be two categories. A functor F : C D is given by the following data (i) for each object X ∈ Ob C an object F(X) ∈ Ob D, → (ii) for any two objects X, Y ∈ Ob C a map (for which we will always omit the index XY )

FXY : HomC (X, Y) − HomD (F(X),F(Y)) such that the following axioms hold →

(F1) for any X ∈ Ob C we have F(idX) = idF(X) (F2) for any X, Y, Z ∈ Ob C and any morphisms f : X Y, g : Y Z we have

F(g) ◦ F(f) = F(g ◦ f) → →

The axioms (F1) and (F2) are often called the functoriality axioms, and a functor is often just referred to as a correspondance F : Ob C Ob D ; X 7 F(X) which is functorial (or natural) in X. This always means that there are maps as in (ii) which satisfy (F1) and (F2). In practical situations, correspondances→F : Ob C Ob→D ; X 7 F(X) will sometimes be natural in X but will invert the direction of arrows (and therefore are not functors in the sense of the above definition). The perhaps easiest example→ is the duality→ for vector spaces: a linear ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ map f : V W defines a linear map f : W V with idV = idV and (g ◦ f) = f ◦ g . This is what is called a contravariant functor. → → Definition 1.5. A contravariant functor F : C D is a functor F : C◦ D. This means, that it is given by the following data (i) for each object X ∈ Ob C an object F(X) ∈ Ob→D, → (ii) for any two objects X, Y ∈ Ob C a map

F : HomC (X, Y) − HomD (F(Y),F(X)) such that the following axioms hold →

(F’1) for any X ∈ Ob C we have F(idX) = idF(X) (F’2) for any X, Y, Z ∈ Ob C and any morphisms f : X Y, g : Y Z we have

F(f) ◦ F(g) = F(g ◦ f) → → 3 If we want to emphasize that a functor is not contravariant we sometimes say that it is covari- ant. The category Set whose objects are the sets and whose morphisms are maps between sets plays a special role in category theory. Every category C is automatically equipped with a collection of Hom-functors indexed by the objects of C

HX = HomC (X, · ): C − Set ; Y 7 HomC (X, Y) 0 On morphisms HX is defined as follows: for any morphism f : Y Y we define HX(f) as the composition with f, more precisely → → 0 → HX(f): HomC (X, Y) − HomC (X, Y ); g 7 HX(f)(g) = f ◦ g It is easy to check the functor axioms. Obviously we can also put the X in the second variable and get the so-called Yoneda-functors→ →

YX = HomC ( · ,X): C − Set ; Y 7 HomC (Y, X)

On morphisms YX is again defined as the composition, but beware this time we get a con- travariant functor. → → Given two functors F : C D and G : D E we can define the composition G ◦ F : C E by (G ◦ F)(X) = G(F(X)) for any object X ∈ Ob C and (G ◦ F)(f) = G(F(f)) for any morphism f from C. It is easily checked→ that G ◦ F satisfies→ the functor axioms F1 and F2. We may also easily→ see that the identity maps on objects and morphisms are functorial and define the identity functor IdC : C C, and we can form the category Cat of all categories (which leads to some set-theoretical problems as the collection of functors between two categories is not a set). Therefore we get→ the notion of an isomorphism between categories as a functor F : C D such that there exists a functor G : D C with G ◦ F = IdC and F ◦ G = IdD. This means just that F is an isomorphism of categories if and only if it is bijective on objects and morphisms.→ It turns out that this notion is of little importance→ because isomorphisms between categories arise only very rarely. This is because many constructions that can be formalized in category theory are only unique up to isomorphism, hence we encounter more frequently the situation that we have functors F : C D. G : D C such that for any object X ∈ Ob C the object GF(X) is not equal but isomorphic to X. A good example is the dual functor defined on the category of finite-dimensional vector→ spaces: → f f ∗ ∗ : Vect (k) − Vect (k); V 7 V = Homk (V, k) It is well-known that V V∗∗ is an isomorphism, but it is not an equality, so ∗ is not an iso- morphism of categories, it is what→ is called an equivalence→ of categories. To make a precise definition we need the notion→ of morphisms between functors:

Definition 1.6. Let F, F0 : C D be functors. A morphism of functors φ : F G is a collection of morphisms φX indexed by all objects X ∈ Ob C → 0 → φX : F(X) F (X) 0 such that for every morphism f : X Y in C we get F (f) ◦ φX = φX 0 ◦ F(f). This is visualized by the commutative diagram → F(f) → F(X) / F(Y)

φX φY   F0(X) / F0(Y) F 0(f) 4 In the literature morphisms of functors are sometimes called natural transformations. Mor- phisms of functors can be composed in an obvious way, and we may define for two fixed categories C, D the category of functors from C to D. In particular we have the notion of an iso- morphism of functors. We will need the following caracterisation of isomorphisms of functors:

Lemma 1.1. Let F, G : C D be functors and φ : F G a morphism of functors. Then φ is an isomorphism if and only if φX : F(X) G(X) is an isomorphism for all X ∈ Ob D. → → Proof. The condition is clearly necessary,→ let us prove that it is sufficent. Suppose that φX is an isomorphism and let us denote by ψX : G(X) F(X) its inverse mor- phism in D. We need to check that the morphisms ψX define a morphism of functors ψ : G F. Let f : X Y be a morphism in C, then φY ◦ F(f) = G(f) ◦ φX by composition→ with ψY on the left we get F(f) = ψY ◦ φY ◦ F(f) = ψY ◦ G(f) ◦ φX, and by composition with ψX on the right→ we get → F(f) ◦ ψX = ψY ◦ G(f) ◦ φX ◦ ψX = ψY ◦ G(f)

Hence the morphisms ψX define a morphism of functors ψ : G F such that by construction φ ◦ ψ = IdG and ψ ◦ φ = IdF. Definition 1.7. A functor F : C D is called → faithful if for every two objects X, Y the map Hom (X, Y) − Hom (F(X),F(Y)) is injective. → C C full if for every two objects X, Y the map Hom (X, Y) − Hom (F(X),F(Y)) is surjective. C → C fully faithful if for every two objects X, Y the map Hom (X, Y) − Hom (F(X),F(Y)) is bijective. →C C essentially surjective if for every Y ∈ Ob D there exists an object X ∈ Ob C and an isomorphism F(X) ' Y → conservative if any morphism f : X Y is an isomorphism if and only if F(f) is an isomorphism an equivalence of categories if there exists a functor G : D C such that F◦G ' Id and G◦F ' Id. → We will make use of the following theorem: → Theorem 1.1. A functor F : C D is an equivalence of categories if and only if it is fully faithful and essentially surjective. → 1.2 Cartesian squares

Definition 1.8. Let C be a category. A commutative diagram

p2 Z / X2 (i.e. f1 ◦ p1 = f2 ◦ p2)

p1 f2   X1 / Y f1 is called a Cartesian square if the following universal property is satisfied: Every two morphisms gi : W Xi (i = 1, 2) such that f1 ◦ g1 = f2 ◦ g2 factor uniquely through Z, i.e. there exists a unique continous map map g : W Z such that pi ◦ g = gi (i = 1, 2). In other → →

5 words we can complete the following commutative diagram of solid arrows by a unique dashed arrow:

W g g 2

p2 # Z / X g1 2

p1 f2    X1 / Y f1

It is useful to state the uniqueness of the factorisation seperately:

Lemma 1.2. Consider a Cartesian square

p2 Z / X2 (i.e. f1 ◦ p1 = f2 ◦ p2)

p1 f2   X1 / Y f1 and let g, h : W Z be two morphisms such that pi ◦ g = pi ◦ h for i = 1, 2 then g = h.

Proof. Set gi = p→i ◦ g(= pi ◦ h) for i = 1, 2, then we have f1 ◦ g1 = f1 ◦ p1 ◦ g = f2 ◦ p2 ◦ g hence the gi factor uniquely through Z, but g and h both define factorization. Hence g = h.

We shall now show that the universal property implies that in a Cartesian square the mor- phisms f1, f2 determine (Z, p1, p2) up to unique isomorphism.

Lemma 1.3. If p p 0 2 0 2 Z / X2 Z / X2 p 0 1 f2 p1 f2     X1 / Y X1 / Y f1 f1

∼ 0 0 are both Cartesian squares, then there exists a unique isomorphism φ : Z Z such that pi ◦ φ = pi

Proof. The universal property of both Cartesian squares assure us that→ there exist unique mor- 0 0 0 0 phisms φ : Z Z such that pi ◦ φ = pi and ψ : Z Z such that pi ◦ φ = pi. We only need to show that φ is an isomorphism with inverse ψ. 0 0 0 We get that pi→◦ φ ◦ ψ = pi ◦ ψ = pi = Id ◦ pi for i→= 1, 2, hence by unicity φ ◦ ψ = Id and similarly ψ ◦ φ = Id.

Definition 1.9. Let f1 : X1 X and f2 : X2 X be two morphisms. The Cartesian product (defined up to unique isomorphism) of (f1, f2) is a triplet (Z, p1 : Z X1, p2 : Z X2) such that → → p2 Z / X2 → →

p1 f2   X1 / Y f1 is a Cartesian square. The Cartesian product will be denoted by X1 × X2. X 6 Lemma 1.4. Let f1 : X1 X, f2 : X2 X be maps between sets. The Cartesian product in Set is given by X→1 × X2 = (x1→, x2) ∈ X1 × X2 | f1(x1) = f2(x2) X and by the restrictions of the natural projection pi : X1 × X2 Xi to X1 × X2. X

Proof. Given two maps gi : W Xi we can define a→ map g : W X1 × X2 by g(w) = (g1(w), g2(w)). It is the unique map that satisfies pi ◦ g = gi. The condition f1 ◦ g1 = f2 ◦ g2 means that for any w ∈ W we have→ g(w) ∈ X1 × X2, hence we have constructed→ a unique map X g : W X1 × X2 such that pi ◦ g = gi. X

This allows→ us to define Cartesian products in any category starting with the definition Set. Given a commutative diagram p2 Z / X2

p1 f2   X1 / Y f1 we get for any object W ∈ Ob C a well-defined map

HomC (W, Z) − HomC (W, X1) × HomC ◦(W, X2) g 7 (p1 ◦ g, p2 ◦ g) HomC (W,Y) Clearly the diagram is→ a Cartesian square if and only if this map is a→ bijection for any object W ∈ Ob C. Definition 1.10. We say that a category C has (or possesses) Cartesian products if for any two mor- phisms fi : Xi X (i = 1, 2) of C the Cartesian product exists. Let C, D be categories which have Cartesian products. A functor F : C D commutes with Cartesian products if it sends→ Cartesian squares to Cartesian squares. Lemma 1.5. Cartesian products exist in T op, and the forgetful functor→ T op − Set commutes to Cartesian products. → Proof. Let f1 : X1 X et f2 : X2 X be continous maps. We endow the set X1 × X2 with X the subspace topology induced by X1 × X2. Therefore the projections pi : X1 × X2 Xi are → → X automatically continous. Now let us prove the universal property. For any two contious maps gi : W Xi such that f1 ◦g1 = f2 ◦g2 we know that there exists a unique map g : W →X1 ×X2 X such that pi ◦ g = gi. We only have to check that g is continous, which is obvious since the composition→ g : W X1 × X2 X1 × X2 is continous. → X

Similarly we can show→ that Ab→and Vect(k) have Cartesian products and their forgetful func- tors commute to Cartesian products. Cartesian squares are sometimes called pullback diagrams in the following sens. If

p2 Z / X2

p1 f2   X1 / Y f1 is a Cartesian square then the morphism p1 is called the pullback of f2 along f1. The pullback of a morphism inherits many properties, here we will only give a very obvious one: 7 Lemma 1.6. Let p2 Z / X2

p1 f2   X1 / Y f1 be a Cartesian square. If f2 is an isomorphism then so is p1.

Proof. The universal property of the Cartesian diagram implies that there is a unique morphism −1 q : X1 Z such that p1 ◦ q = idX1 since the pair (idX1 , f2 ◦ f1) factors through the Cartesian product:

→ X1 −1 f ◦f1 q 2

p2 # / id Z X2 X1 p1 f2    X1 / Y f1

−1 −1 Moreover p1 ◦ q ◦ p1 = p1 and p2 ◦ q ◦ p1 = f2 ◦ f1 ◦ p1 = f2 ◦ f2 ◦ p2 = p2, hence by unicity q ◦ p1 = idZ, and p1 is an isomorphism.

We shall need the following proposition:

Proposition 1.1. Consider a commutative diagram

p 0 p 0 2 2 Z / Z / X2

0 p1 p1 f2

 0   X / X1 / Y f f1 and let us assume that the right square is Cartesian. Then

0 p2 Z0 / Z

0 p1 p1   X0 / X f 1 is Cartesian if and only if p ◦p 0 0 2 2 Z / X2

0 p1 f2   X0 / Y f1◦f is Cartesian.

8 Proof. Consider the chain of maps

0 0 HomC (W, Z ) − HomC (W, X ) × HomC (W, Z) HomC (W,X1) ∼ 0 −→ HomC (W, X ) × HomC (W, X1) × HomC (W, X2) HomC (W,X1) HomC (W,Y) ∼ 0 −→ HomC (W, X ) × HomC (W, X2) HomC (W,Y) → 0 0 here the first maps is given by g 7 (p1 ◦ g, p2 ◦ g), the second map is given by (g1, g2) 7 (g1, p1 ◦ g2, p2 ◦ g2) and the third is given by (g1, g2, g3) 7 (g1, g2). Note that the composition 0 0 of the three maps is given by g 7 →(p1 ◦ g, p2 ◦ p2 ◦ g). The third map is clearly a bijection→ with inverse (h1, h2) 7 (h1, f ◦ h1, h2), the second map→ is a bijection because the right square is Cartesian. Hence the first map is→ a bijection if and only if the composition is a bijection. → 2 The fundamental groupoid

2.1 Connected spaces

Definition 2.1. A topological space X is called connected if whenever we have a cover X = U ∪ V by two open disjoint subsets then U = ∅ or V = ∅. A subset S ⊂ X is called connected if S is connected with the subspace topology induced by X. We immediately see that a topological space is connected if and only if for every cover X = U∪V by two open disjoint subsets then U = X or V = X, or alternatively if and only if the only subsets of X that are open and closed are ∅ and X. A subset S ⊂ X is connected if for any open covering S ⊂ U ∩ V with open subsets U, V ⊂ X such that U ∩ V ∩ S = ∅ we have U ∩ S = ∅ or V ∩ S = ∅ (or equivalently S ⊂ U or S ⊂ V).

Proposition 2.1. Let S ⊂ X be a connected subspace. Then any subset W ⊂ X such that S ⊂ W ⊂ S is connected.

Proof. Let W ⊂ U ∪ V be an open cover such that U ∩ V ∩ W = ∅. Since S ⊂ W we get that U ∩ V ∩ S = ∅. Hence U ∩ S = ∅ or V ∩ S = ∅. By relabeling we may assume that U ∩ S = ∅. Now let x ∈ W be a point. Since x ∈ S we get that every open neighborhood of x has a non- empty intersection with S, so U cannot be a neighborhood of x. Hence x 6∈ U for all x ∈ W and W ∩ U = ∅. Therefore W is connected.

Proposition 2.2. Let {Si}i∈I be a family of connected subsets Si ⊂ X such that Si ∩ Sj 6= ∅ for all S i, j ∈ I. Then S = i∈I Si is connected.

Proof. Let S ⊂ U ∩ V be an open cover of S such that U ∩ V ∩ S = ∅. Since Si ⊂ S we get U ∩ V ∩ Si = ∅ and Si ⊂ S ⊂ U ∪ V for any i ∈ I, and since Si is connected we deduce that U∩Si = ∅ or V ∩Si = ∅. Assume that we have U∩Si = ∅ and V ∩Sj = ∅ and let x ∈ Si ∩Sj then x 6∈ I and x 6∈ V which is impossible, hence we either have U ∩ Si = ∅ for all i ∈ I or V ∩ Si = ∅ S S S S for all i ∈ I, so finally i∈I Si ∩ U = i∈I(Si ∪ U) = ∅ or i∈I Si ∩ V = i∈I(Si ∪ V) = ∅

Let x ∈ X be a point, then the proposition assures that the union of all connected subsets S ⊂ X containing x is again connected, hence there exists a biggest connected subset of X containing

9 x. This subset is called the connected component of x, and we will denote it by [x]0. By defini- tion we have that if S 3 x is a connected subset then S ⊂ [x]0. Since [x]0 is a connected subset containing x we get that [x]0 = [x]0 and connected components are closed.

Lemma 2.1. Let x, y ∈ X. We have [x]0 = [y]0 if and only if there exists a connected subset S ⊂ X containing x and y.

Proof. If [x0] = [y]0 then [x]0 is a connected subset containing x and y so the condition is clearly necessary. Let us assume that there exists a connected subset S ⊂ X containing x and y, then S ⊂ [x]0 as a connected set containing x. In particular we get y ∈ [x]0 which implies [x]0 ⊂ [y]0 as a connected subset containing y. Similarly we see that [y]0 ⊂ [x]0.

We have seen that the connected components [x]0 and [y]0 are either equal or disjoint. By chos- F ing a point xi in each connected component, we get that X = [xi]0 is the disjoint union of its connected components. But this decomposition is purely set-theoretical in general, it does not respect topologies since connected components need not be open (unless of course there are only finitely many). This is not particularly intuitive and the course will deal with spaces such that this decomposition is topological - therefore we introduce locally connected spaces.

Definition 2.2. A topological space is called locally connected if for every x ∈ X and any neighborhood V 3 x of x there exists connected neighborhood W such that x ∈ W ⊂ V. Lemma 2.2. Let X be a locally connected topological space. Let x ∈ X be a point and V 3 x a neighbor- hood. Then there exists an open connected subset U ⊂ X such that x ∈ U ⊂ V.

V Proof. We may assume that V is open. Let [x]0 be the connected component of V containing x. V V Let z ∈ [x]0 , there exists a connected neighborhood S of z such that S ⊂ V, therefore S ⊂ [z]0 = V V [x]0 . Hence [x]0 is an open and connected neighborhood of x contained in V. Proposition 2.3. Let X be a locally connected topological space. The connected components of X are open and X is homeomorphic to the disjoint union of its connected components.

Proof. Let z ∈ [x]0, there exists an open connected neighborhood S of z, therefore S ⊂ [z]0 = [x]0. Hence [x]0 is open, and the topological decomposition follows. Proposition 2.4. Let f : X Y be a continous map and suppose that X is connected. Then f(X) is connected. → −1 −1 Proof. Let f(X) ⊂ U ∪ V be an open cover such that U ∩ V ∩ f(X) = ∅. Then f (U)(= f (U ∩ f(X))) and f−1(V) are open subsets of X such that X = f−1(f(X)) = f−1(U∪V) = f−1(U)∪f−1(V) −1 −1 −1 −1 and therefore f (U)∩f (V) = f (U∩V) = f (U∩V ∩f(X)) = ∅. Since X is connected we get −1 −1 that f (U) = ∅ or f (V) = ∅ which means precisely that U ∩ f(X) = ∅ or V ∩ f(X) = ∅. Lemma 2.3. Let a, b ∈ R with a 6 b. The interval [a, b] is connected.

Proof. Let [a, b] = U t V be the disjoint union of two open subsets. We may assume that a ∈ U. Since U is open there exists a small interval [a, t) ⊂ U, and we can consider s = sup{t ∈ S [a, b] | [a, t) ⊂ U}. We have [a, s) = t

10 Definition 2.3. A path on X from x0 ∈ X to x1 ∈ X is a continous map γ : I X such that γ(0) = x0 and γ(1) = x1. A topological space X is called pathwise connected or arcwise connected if for→ any two points x0, x1 ∈ X there exists a path from x0 to x1 Lemma 2.4. A pathwise connected space X is connected.

Proof. Let X = U t V be a covering of X by two disjoint open subsets. If neither U nor V are −1 −1 empty then chose x0 ∈ U, x1 ∈ V and a path γ from x0 to x1. Then I = γ (U) t γ (V) is −1 −1 a covering of I by open disjoint subsets and therefore γ (U) = ∅ or γ (V) = ∅. Both are impossible since γ(0) ∈ U and γ(1) ∈ V. Hence X is connected. n Example 2.1. A subset X ⊂ R is called convex is for any two points x, y ∈ X the straight line connecting x and y is included in X, i.e.

tx + (1 − t)y | t ∈ [0, 1] ⊂ X

Clearly X is arcwise connected and therefore also connected.

2.2 Construction of the fundamental groupoid

In this section X will denote a topological space.

Definition 2.4. Recall that a path on X from x0 ∈ X to x1 ∈ X is a continous map γ : I X such that γ(0) = x0 and γ(1) = x1. We set → Ω(X, x0, x1) = {γ : I X continous | γ(0) = x0 and γ(1) = x1}

X x x x Ω(X, x ) A loop on with base point 0 is a path→ from 0 to 0. We will usually write 0 instead of Ω(X, x0, x0).

We will mostly write γ : x0 x1 instead of γ ∈ Ω(x0, x1). Note that a loop can be seen in a 1 2πti unique way as a map of pointed spacesγ ˜ :(S , 1) (X, x0) by setting γ(t) = γ˜ (e ), hence 0 1 ∼ we have a natural idetification C ((S , 1), (X, x0)) − Ω(X, x0). → The constant path at x0 ∈ X will be denoted by cx . It is defined by cx (t) = x0 for all t ∈ [0, 1]. 0 → 0 For each path γ : x0 x1 we define its inverse path γinv : I X by γinv(t) = γ(1 − t). Then γinv is a path from x1 to x0. → Let γ : x0 x1 and δ : x1 x2. Then γ(1) = δ(0) and we get a well-defined map

1 γ(2t) t ∈ [0, 2 ] γ ⊥ δ : I X t 7 1 δ(2t − 1) t ∈ [ 2 , 1] → → Clearly γ ⊥ δ| 1 and γ ⊥ δ| 1 are continous, hence γ ⊥ δ is continous and defines a path [0, 2 ] [ 2 ,1] from x0 to x2. This path is called the composition of γ and δ.

Note that if ρ : x2 x3 is another path in X then we pratically never have (γ ⊥ δ) ⊥ ρ = γ ⊥ (δ ⊥ ρ), i.e. the composition of paths is not associative in general. However, the paths (γ ⊥ δ) ⊥ ρ and γ ⊥ (δ ⊥ ρ) are very similar: they are homotopic.

Definition 2.5. Let γ0, γ1 : x0 x1 be two paths on X. We say that γ0 is homotopic to γ1 if there exists a continous map h : I × I − X (called a homotopy from γ0 to γ1) such that 11 → h(s, 0) = γ0(s) for all s ∈ [0, 1]

h(s, 1) = γ1(s) for all s ∈ [0, 1]

h(0, t) = x0 for all t ∈ [0, 1]

h(1, t) = x1 for all t ∈ [0, 1]

If γ0 and γ1 are homotopic by h we write h : γ0 γ1 or simply γ0 ∼ γ1. We also write γt = ht : I X for the path s 7 ht(s) = h(s, t). → We may think of a homotopy as a continous family of paths γt that deform γ0 into γ1. →

Lemma 2.5. The homotopy relation defines an equivalence relation on Ω(X, x0, x1).

Proof. Clearly we get γ ∼ γ by the constant deformation ht = γ for all t ∈ [0, 1], i.e. by the continous map h : I × I X ; h(s, t) = γ(s).

Consider γ ∼ δ. Then there exists a homotopy h : γ δ. Then ht deforms γ into δ, hence h1−t deforms δ into γ. More precisely, define→

h˜ : I × I − X (s, t) 7 h(s, 1 − t).

h˜ h˜ : δ γ Clearly is continous and hence defines→ a homotopy→ . Consider γ ∼ δ and δ ∼ ρ. We get homotopies h : γ δ and k : δ ρ. Define

1 h(s, 2t) t ∈ [0, 2 ] l : I × I − X ; l(s, t) = 1 k(s, 2t − 1) t ∈ [ 2 , 1] Clearly l is well defined since→ h(s, 1) = δ(s) = k(s, 0), and it is continous. It is then easily checked that l : γ δ. Definition 2.6. Let ∼ be the equivalence relation defined by the homotopy of paths. We set for all x0, x1 ∈ X π1(X, x0, x1) = Ω(X, x0, x1)/ ∼ π1(X, x0) = π1(X, x0, x0)

If γ : x0 x1 is a path its equivalence class will be denoted by [γ]. Proposition 2.5. We have the following basic properties of homotopies of paths:

(i) Let γ0, γ1 : x0 x1 and δ0, δ1 : x1 x2 be homotopic paths. Then their compositions are homotopic. More precisely if h : γ0 γ1 and k : δ0 7 δ1 are homotopies, then we get a homotopy l : γ0 ⊥ δ0 γ1 ⊥ δ1 such that lt = ht ⊥ kt. → (ii) Let γ : x0 x1 be a path, then γ ⊥ cx1 ∼ γ ∼ cx0 ⊥ γ.

(iii) Let γ : x0 x1. Then γ ⊥ γinv ∼ cx0 and γinv ⊥ γ ∼ cx1

(iv) Let γ : x0 x1, δ : x1 x2 and ρ : x2 x3 be paths, then (γ ⊥ δ) ⊥ ρ ∼ γ ⊥ (δ ⊥ ρ)

Proof. (i) Define

1 h(2s, t) s ∈ [0, 2 ] l : I × I − X ; l(s, t) = 1 k(2s − 1, t) s ∈ [ 2 , 1] → 12 Then l is well-defined since h(1, t) = x1 = k(0, t), and it is continous. An easy verification shows that we get a homotopy l : γ0 ⊥ δ0 γ1 ⊥ δ1. (ii) Set 2 1 γ( 1+t s) s ∈ [0, 2 (1 + t)] h : I × I − X ; h(s, t) = 1 x1 s ∈ [ 2 (1 + t), 1]

The function h is well defined→ since h(1, t) = x1 and clearly continous (since its restrictions to a finite covering of closed subsets are continous). We have h(s, 0) = γ ⊥ cx1 , h(s, 1) = γ, h(0, t) = γ(0) = x0 and h(1, t) = x1 hence we have a homotopy h : γ ⊥ cx1 γ. Similarly, let 2s−1+t 1 2s−1+t us note that 1+t is the affine bijection [ 2 (1 − t), 1] ' [0, 1] that sends 1+t to 0 and 1 to 1, hence we may set

1 x0 s ∈ [0, 2 (1 − t)] k : I × I − X ; h(s, t) = 2s−1+t 1 γ( 1+t ) s ∈ [ 2 (1 − t), 1] → It is easily checked that k defines a homotopy k : cx0 ⊥ γ γ. (iii) Let us set

1 x0 s ∈ [0, 2 t] γ( 2s−t ) s ∈ [ 1 t, 1 ] h : I × I − X ; h(s, t) =  1−t 2 2  2s−1 1 1 γinv( ) s ∈ [ , 1 − t]  1−t 2 2 x s ∈ [1 − 1 t, 1] → 0 2   It is easy to check that h is a homotopy h : γ ⊥ γinv cx0 . Another (arguably easier) possibility is to consider the homotopy.

1 γ((1 − t)2s) s ∈ [0. 2 ] k : I × I − X ; k(s, t) = 1 γ((1 − t)(2 − 2s)) s ∈ [ 2 , 1] →

The homotopy γinv ⊥ γ ∼ cx1 follows since by definition γ = (γinv)inv. (iv) A homotopy h :(γ ⊥ δ) ⊥ ρ γ ⊥ (δ ⊥ ρ) is given by

4s 1 γ( 1+t ) s ∈ [0, 4 (1 + t)] h : I × I − X ; h(s, t) = 1 1 δ(4s − 1 − t) s ∈ [ 4 (1 + t), 4 (2 + t)]  4s−2−t 1 ρ( 2−t ) s ∈ [ 4 (2 + t), 1] → We leave the verification to the reader.  Corollary 2.1. (i) The composition of paths induces well-defined maps

π1(X, x0, x1) × π1(X, x1, x2) − π1(X, x0, x2) ; ([γ], [δ]) 7 [γ][δ] = [γ ⊥ δ]

(ii) We may define a groupoid Π1(X) by the→ following data →

its objects are the points of X, i.e. Ob Π1(X) = X

for any two points x0, x1 ∈ X the morphisms are given by the homotopy classes of paths, i.e. Hom (x0, x1) = π1(X, x0, x1) Π1(X)

composition maps in Π1(X) are given by composition of paths as in (i) (formally we get the slight notational complication δ ◦ γ = [γ][δ])

13 −1 (iii) Composition of paths induces a group structure on π1(X, x0) with unit element [cx0 ] and [γ] = [γinv] Lemma 2.6. Let x0, x1 ∈ X be two points of X. Then any path γ : x0 x1 defines by an isomorphism of groups ∼ −1 π1(X, x0) − π1(X, x1)[δ] 7 [γ] [δ][γ]

Hence if x0 and x1 are in the same path-connected→ component→ of X, then π1(X, x0) and π1(X, x1) are isomorphic. Hence if X is path-connected we can talk about ’the fundamental group of X’ and denote it by π1(X). Obviously the isomorphism of the lemma depends only on the ho- motopy class of γ, but if π1(X, x0, x1) is non-trivial then it is not unique. We say that π1(X) is well-defined up to non-canonical isomorphism. The language of categories allows us to formu- late precisely what we mean: we will see that if X is arcwise connected then Π1(X) is equivalent to a group (i.e. a groupoid with a single object).

Definition 2.7. Let X be a topological space. The groupoid Π1(X) is called the fundamental groupoid of X.

Let f : X Y be a continous function. Then any path γ : x0 x1 defines a path f ◦ γ : f(x0) f(x1). Hence we get a map → Ωx0,x1 (f): Ω(X, x0, x1) − Ω(Y, f(x0), f(x0)) ; γ 7 f ◦ γ

Lemma 2.7. If γ0, γ1 are homotopic paths, then Ωx ,x (f)(γ0) ∼ Ωx ,x (f)(γ1) → 0 1 0 1 →

Proof. If h : γ0 γ1 is a homotopy then it is easily checked that f ◦ h is a homotopy from f ◦ γ0 to f ◦ γ1.

Hence we get well-defined maps

(π1)x0,x1 (f): π1(X, x0, x1) − π1(Y, f(x0), f(x0)) ; [γ] 7 [f ◦ γ]

Lemma 2.8. We have (π1)x,x(idX) = idπ (X,x) and (π1)x ,x (f)(δ)◦(π1)x ,x (f)(γ) = (π1)x ,x (f)([γ][δ]) 1 → 1 2 0 1 → 0 2 Proof. This follows easily from f ◦ (γ ⊥ δ) = f ◦ γ ⊥ f ◦ δ.

Hence we get a functor Π1(f): Π1(X) − Π1(Y)

which is simply defined by f : X Y on the level of objects and by π1(f) on the level of morphisms. →

It is further easily checked that Π1(Id→X) = IdΠ1(X) and if f : X Y and g : Y Z are continous maps then we have Π1(g ◦ f) = Π1(g) ◦ Π1(f). Hence the fundamental groupoid is functorial, and we get a functor→ →

Π1 : T op − Gr

→ 2.3 Homotopy invariance

Definition 2.8. Let A ⊂ X be a subspace. Two maps f0, f1 : X Y are homotopic relatively to A if f0|A = f1|A and there exists a continous map h : X × I − Y →

such that → 14 h(x, 0) = f0(x) for all x ∈ X

h(x, 1) = f1(x) for all x ∈ X

h(a, t) = f0(a) = f1(a) for all t ∈ [0, 1] and all a ∈ A

We write f0 ∼A f1 if f0 and f1 are homotopic relatively to A. We also often use the notation

ft = ht : X × Y ; x 7 h(x, t)

If A is the empty set, then we simply say that f and f are homotopic and we will write f ∼ f (although 0 1 → 0 1 this is not really compatible with the notation used for paths). Lemma 2.9. Homotopy relatively to A defines an equivalence relation on the set of continous functions from X to Y.

Proof. Very easy. Definition 2.9. A continous map f : X Y is a homotopy equivalence if there exists a continous map g : Y X such that g ◦ f ∼ idX and f ◦ g ∼ idY. An important special case of homotopy→ equivalences is given by deformation retracts. → Definition 2.10. Let i : Z X be the inclusion of a subspace. We say that Z (or i) is a deformation retract if there exists a continous map p : X Z such that p ◦ i = idZ and i ◦ p ∼ idX. If we further have i ◦ p ∼Z id→X we say that Z is a strong deformation retract. A space X is called contractible if there is a point→ x ∈ X that is a (strong) deformation retract of X. Lemma 2.10. Let f0, f1 : X Y be homotopic continous functions and h : X × I Y a homotopy from f0 to f1. Let γ : x0 x1 be a path on X. Define the paths hxi : I Y →; hxi (t) = h(xi, t) for i = 0, 1. → Then

→ hx0,inv ⊥ f0 ◦ γ ⊥ hx1 ∼ f1 ◦ γ.

Proof. Define

1 hx0 (1 − 3s) s ∈ [0, 3 t] k : I × I − Y ; k(s, t) = 3s−t 1 1 h1−t(γ( 3−2t )) s ∈ [ 3 t, 1 − 3 t]  1 hx1 (3s − 2) s ∈ [1 − 3 t, 1] → One easily checes that the map k is well-defined and continous. It defines a homotopy from f1 ◦ γ to 1 hx0,inv(3t) t ∈ [0, 3 ] δ(t) = 1 2 f0(γ(3t − 1)) t ∈ [ 3 , 3 ]  2 hx1 (3t − 2) t ∈ [ 3 , 1] δ ∼ h ⊥ f ◦ γ ⊥ h It is then an easy exercise to construct a homotopy x0,inv 0 x1

Corollary 2.2. Let h be a homotopy from f0 to f1, then h defines morphisms [hx] ∈ π1(f0(x), f1(x)). We get an isomorphism of functors

Π1(h): Π1(f0) − Π1(f1); Π1(h)x = [hx]

In particular we get → Proposition 2.6. Let f : X Y be a homotopy equivalence. Then Π1(f): Π1(X) Π1(Y) is an equivalence of categories. In particular, if X and Y are arcwise connected, then π1(X, x0) and π1(Y, y0) are isomorphic groups for any→ choice of points x0 ∈ X and y0 ∈ Y. → 15 Proof. Let g : Y X be a homotopy inverse of f and h : g ◦ f ∼ idX, k : f ◦ g ∼ idY homotopies. ∼ Then we get isomorphisms Π1(h): Π1(g) ◦ Π1(f) − Π1(idX) = IdΠ1(X) and Π1(k): Π1(f) ◦ ∼ → Π1(g) − Π1(idY) = IdΠ1(Y). → → 3 Coverings

3.1 Fiber bundles

Let f : Y X be a continous map and x ∈ X a point. The subset f−1(x) ⊂ Y is called the fiber of f at x. It can be visualized by the Cartesian square → f−1(x) / Y

f   {x} / X

A continous map can intuitively also be understood as the collection of its fibers that move ’continously’ inside Y. From this point of view the simplest continous function is a function whose fibers are all homeomorphic (or at least locally homeomorphic), and this gives rise to the notion of a fiber bundle.

Definition 3.1. A fiber bundle on X is a continous map p : Y X such that for every point x ∈ X there exists an open neighborhood U ⊂ X, a topological space F and a homeomorphism → ϕ : p−1(U) −∼ U × F

p ◦ ϕ = p p : U × F U such that U where U is the→ natural projection. This is visualized by the commutative diagram −1 ϕ p (U) → ∼ / U × F FF z FF zz p FF zzp FF zz U F |zz # U The space X is sometimes called the base of the fiber bundle, Y is called the total space, and the map ϕ is called a (local) trivialization of p on U. A fiber bundle p : Y X is called trivial if there exists a global homeomorphism φ : Y X × F such that pX ◦ φ = p. Note that if p : Y X is a fiber bundle then a local trivialization→ ϕ on U ⊂ X gives an identi- ∼ → fication ϕ : p−1(x) − {x} × F ' F for all x ∈ U, hence all fibers above U are homeomorphic, but note that in general→ the homeomorphism between fibers depend on the trivialization that we chose. →

Definition 3.2. Let p1 : Y1 X, p2 : Y2 X be fiber bundles. A morphism of fiber bundles is a continous map f : Y1 Y2 such that p2 ◦ f = p1: → → f → Y1 / Y2 @@ ~ @@ ~~ p1 @@ ~~p2 @ ~~ X

16 The composition of two morphisms of fiber bundles is defined in an obvious way and we get the category of fiber bundles on X. Note that a morphism f : Y1 Y2 is an isomorphism if and only if f is a homeomorphism (the inverse map will be automatically a morphism of fiber bundles). →

−1 ∼ −1 Let Uα,Uβ be two open subsets with trivializations ϕα : p (Uα) Uα×F and ϕβ : p (Uβ) ∼ Uβ × G. We get the homeomorphism →

∼ −1 ∼ → ϕαβ :(Uα ∩ Uβ) × F − p (Uα ∩ Uβ) − (Uα ∩ Uβ) × G

Clearly for any x ∈ U ∩ U we get the induced homeomorphism ϕ | : F − G. Hence α β → → α,β {x}×F we get a map ϕαβ : Uα ∩ Uβ − {homeomorphisms F − G} →

Lemma 3.1. Let p : Y X be a fiber bundle with discrete fibers and Uα,Uβ be two open subsets with trivializations. Then ϕαβ is a locally constant→ function. → Let p : Y X be a fiber→ bundle and Z ⊂ X a subspace. Then we can define the restriction of p −1 to Z as p|p−1(Z) : p (Z) Z or in other words by the Cartesian diagram → pY → p−1(Z) ' Z × Y / Y X

p|p−1(Z) p   Z / X f

Since this notation is cumbersome we shall often simply write p|Z : Y|Z Z, especially if Z is an open subset of X. 0 ∗ ∗ ∗ 0 This situation generalizes to any continous map f : X X. We define f→Y and f p : f Y X by the Cartesian square pY f∗Y / Y → →

f∗p p   X0 / X f Hence we may identify up to unique isomorphism f∗Y = X0 × Y and f∗p is the natural projec- X 0 ∗ ∗ 0 tion onto X , pY the projection onto Y. We call f p : f Y X the pullback of p.

0 Lemma 3.2. Let p : Y X be a fiber bundle and f : X →X be a continous map. Then the natural projection f∗p = f∗Y X0 is a fiber bundle on X0. → → Proof. First let us assume→ that p : X × F X is a trivial bundle. Then the map X0 × F − X0 × (X × F);(x0, λ) 7 (x0, (f(x0), λ)) X → is bijective, and it is easily checked→ that it is a homeomorphism.→ Therefore f∗p is a trivial bundle. Let x0 ∈ X0 be a point and chose an open subset U 3 f(x0) on X with a trivialization h : Y 0 −1 0 −1 0 0 −1 U × F. Then U = f (U) is an open neighborhood of x and pX 0 (U ) = X × p (U) X → 0 0 0 ∗ −1 0 −1 Lemma 3.3. Let x ∈ X . Then pY : X × Y Y induces a homeomorphism f p (x ) p (f(x)) Y S Proposition 3.1. Patching Lemma for maps of fiber bundles Let X = i∈I Ui be an open covering of a topological space X, p : Y X, q : Z X be→ fiber bundles on X. Let us write Uij = Ui →∩ Uj. 17 → → f, g : Y Z f| −1 = g| −1 f = g (1) Let be two maps of fiber bundles such that p (Ui) p (Ui). Then

(2) Let fi : Y|U Z|U be maps of fiber bundles on Ui such that fi|U = fj|U . Then there exists a →i i ij ij unique map of fiber bundles f : Y Z such that f|Ui = fi. → (3) Let f : Y Z be a map of fiber bundles. Then f is an isomorphism if and only if f|Ui is an isomorphism for all i ∈ I. → → S −1 S −1 Proof. (1) is obvious since Y = x∈X p (x) = i∈I p (Ui). −1 (2) Let us define f : Y Z by setting f(y) = fi(y) if y ∈ p (Ui). This is a well-defined map −1 −1 since whenever y ∈ p (Ui) and y ∈ p (Uj) we have fi(y) = fj(y) by assumption. Clearly f| −1 = f f Y p (Ui) i, hence is→ continuous on an open cover of , hence continuous. By construction it is a map of fiber bundles. Its unicity is clear by (1). g = f|−1 g | g | f | = f | (3) Let i Ui . Then i Uij and j Uij are then inverse of the same map i Uij j Uij and therefore they coincide. By (2) they patch to a morphism of fiber bundles g : Z = raY such that g|Ui = gi. We have (g ◦ f)|Ui = g|Ui ◦ f|Ui = id, hence by (1) we get g ◦ f = id. Similarly one proves f ◦ g = id, hence f is an isomorphism. Proposition 3.2. Patching Lemma for fiber bundles S Let X = i∈I Ui be an open covering of a topological space X, pi : Yi Ui fiber bundles on Ui. Let us ∼ write Uij = Ui ∩ Uj and Uijk = Ui ∩ Uj ∩ Uk, and let ϕij : Yj|(Uij) Yi|(Uij) be isomorphisms of

fiber bundles on Uij (i.e. pi|Uij ◦ ϕij = pj|Uij ) such that → → (i) ϕii = idYi −1 (ii) ϕji = ϕij

(iii) ϕij ◦ ϕjk = ϕik on Uijk (i.e. ϕij|p−1(U ) ◦ ϕjk|p−1(U ) = ϕik|p−1(U )) j ijk k ijk k ijk

−1 ∼ Then there exists a fiber bundle p : Y X and isomorphisms of fiber bundles ϕi : p (Ui) Yi (i.e. 0 0 pi ◦ ϕi = p). Moreover if p : Y X is another fiber bundle with isomorphisms of fiber bundles 0 0 −1 ∼ 0 ϕi :(p ) (Ui) Yi, then there exists→ a unique isomorphism of fiber bundles α : Y Y →such that 0 ϕi ◦ α = ϕi. → → → ˜ F ˜ Proof. Let us set Y = i∈I Yi. For yi, yj ∈ Y where we assume yi ∈ Yi, yj ∈ Yj. We put −1 −1 yi ∼ yj if there exists x ∈ Uij such that yi ∈ pi (x), yj ∈ pj (x) and ϕij(yj) = yi. Conditions (i), (ii), (iii) imply that we have defined an equivalence relation on Y˜ and we may set Y = Y/˜ ∼. By the universal property of the disjoint union there is a natural continous mapp ˜ : Y˜ X such thatp ˜(yi) = pi(yi) for all yi ∈ Yi. If yi ∼ yj thenp ˜(yi) = pi(yi) = pi(ϕij(yj)) = pj(yj) hence p˜ factors through Y and defines a continous map p : Y X such that p(yi) = pi(y→i) where yi is the equivalence class of yi ∈ Yi. Note that Y comes equipped with continous open maps σi : Yi Y˜ Y. By what we just saw we get that p ◦ σ→i = pi, hence we get a map of fiber bundles on Ui. Clearly σi is injective since σi(yi1 ) = σi(yi2 ) means that yi1 = ϕii(yi2 ) = yi2 . −1 −1 Now let→y ∈ →Y|Ui = p (Ui), say y ∈ Yj. Then y ∈ pj (Uij and by setting yi = ϕij(y) we get −1 that y = yi = σi(yi). We have shown that σi : Yi p (Ui) is bijective and open, hence a −1 homeomorphism and therefore an isomorphism of fiber bundles on Ui. By taking ϕi = σi we have shown the first part of the proposition. → p0 : Y 0 X ϕ0 : Y 0 ∼ Y Now let is another fiber bundle with isomorphisms of fiber bundles i Ui i, −1 0 0 ∼ then we have isomorphisms ϕi ◦ ϕi : Y |Ui − Y|Ui . On Uij we get → → ϕ−1 ◦ ϕ0| = (ϕ ◦ ϕ )−1 ◦ ϕ ◦ ϕ0 = ϕ−1◦ = varphi0 i i Uij ij j → ij j j j By the patching lemma for morphisms we get α. 18 3.2 Coverings

Definition 3.3. A continous map f : X Y is called a local homeomorphism if for every point x ∈ X there exists an open neighborhood W 3 x such that f|W : W f(W) is a homeomorphism. In particular we see immediately that→f : X Y is a local homeomorphism if and only if there S → exists an open covering X = i∈I such that f|Ui : Ui f(Ui) is a homeomorphism. → Definition 3.4. Let X be a topological space. A covering→ space (or simply a covering) of X is a fiber bundle p : Y X which is a local homeomorphism. A morphism of coverings is a morphism of the underlying fiber bundles. By the corresponding→ property of local homeomorphisms we get the following obvious lemma:

S Lemma 3.4. Let p : Y X be a fiber bundle, X = i∈I Ui a covering such that p|Ui : Y|Ui Ui is a covering for all i ∈ I, then p is a covering. In particular the patching lemmas for fiber bundles hold for coverings. → →

Proof. The first statement is obvious, the patching lemma for morphisms of coverings is also obvious since they are just bundle maps of the underlying bundles. The patching lemma for coverings holds since we may patch the coverings as a fiber bundle, since the fiber bundle will be locally isomorphic to the given coverings, it is a covering.

Obvious examples of covering maps are projections p : X × F X where F is a space with the discrete topology. We will see that locally all covering maps are obtained in this way. Indeed, ∼ let p : Y X be a covering space of X and consider a trivialization→ ϕ : Y|U U × F. Let y ∈ Y and W 3 y a neighborhood such that p|W is a homeomorphism, set (x, z) = ϕ(y) ∈ U × F. By definition→ of the product topology, there exists an open neighborhoods Ux ⊂→U of X and Vz ⊂ F −1 −1 of Z such that Ux × Vz ⊂ ϕ(W). Then p|ϕ (Ux×Vz) is a homeomorphism, but p ◦ ϕ = pU which means in particular that the projection p : Ux × Vz Ux is bijective. In particular that implies that V = {z} hence z is open. Therefore we have shown → Proposition 3.3. Let p : Y X be a fiber bundle on X. Then p is a covering space if and only if there ∼ exist local trivializations ϕ : Y|U U × F such that F is a discrete space. → ∼ If U is an open subset of X such that there exist local trivializations ϕ : Y|U U × F such F is a → discrete space, we say that U is evenly covered. The subspace p−1(U) is the topological disjoint union of spaces Si that are all homeomorphic to U, these subspaces are called→ the sheets of the covering space. Note that sheets over U are by definition open subsets of Y and that for any −1 x ∈ U the set p (x) intersects with each sheet Si in a single point.

Definition 3.5. Let p : Y X be a covering space, and f :(Z, z0) (X, x0) be a continous map. A −1 lift of f to p with respect to y0 ∈ p (x0) is a continous map f˜ :(Z, z0) (Y, y0) such that p ◦ f˜. → → (Y, y0) → f˜ u: uu p uu uu uu  (Z, z ) / (X, x ) 0 f 0

Lemma 3.5. Let p : Y X be a covering space. Let Z be a connected space, then for a fixed y0 ∈ −1 p (x0) any map f :(Z, z0) (X, x0) possesses at most one lift to p with respect to y0. → → 19 Proof. Let g1, g2 be two lifts and consider W = {z ∈ Z | g1(z) = g2(z)}. We will show that W is open and closed. Let z ∈ W and chose a neighborhood U 3 g1(z) in Y which is mapped −1 homeomorphically onto p(U) 3 f(z). Then for any w ∈ U we get g1(w) = p|U (p(g1(w))) = −1 −1 p|U (f(w)) = p|U (p(g2(w))) = g2(w) Hence U ⊂ W and W is open. Not let z 6∈ W i.e. g1(z) 6= g2(z), let U 3 p(g1(z)) = f(z) = p(g2(z)) be an open neighborhood such that p is −1 evenly covered on U. Then g1(z) and g2(z) are both in the fiber p (f(z)) hence they are on two −1 −1 different sheets S1 3 g1(z) and S2 3 g2(z). Since sheets are open we get that g1 (S1) ∩ g2 (S2) is an open neighborhood of z such that is mapped into sheet S1 by g1 and into sheet S2 by g2. −1 −1 Since S1 ∩ S2 = ∅ we get that g1(w) 6= g2(w) for every w ∈ g1 (S1) ∩ g2 (S2). This shows that the complementary of W is open, hence W is closed. Finally W is non-empty since z0 ∈ W. Hence W = Z and g1 = g2.

Lemma 3.6. Let p :(Y, y0) (X, x0) be a covering map of pointed spaces. Let f :(Z, z0) (X, x0) be a continous map such that f∗p is trivial. Then there exists a lift of f to p. → → Proof. We have a Cartesian square f^ Z × F / Y

pZ p   Z / X f

∼ −1 Note that f˜ : {z0} × F − p (x0) so there exists a unique λ ∈ F such that f^(z0, λ) = y0. Define iλ : Z Z × F by iλ(z) = (z, λ). Since F is discrete iλ is a homeomorphism onto the subspace ^ Z × {λ}. we get p ◦ f ◦ i→λ = f ◦ pZ ◦ iλ = f ◦ id = f and f˜ ◦ iλ(z0) = y0, hence f˜ ◦ iλ is a lift of f to p. →

Lemma 3.7. Let X be a locally connected topological space, F1,F2 discrete spaces and h : X × F1 S X × F2 a map of trivial covers. Then there exists an open covering X = i∈I Ui and a maps g : F1 F2 such that h|Ui×F1 = idX × g. → → Proof. Assume that X is connected, then for any λ ∈ F1 the set h(X×{λ}) is a connected subspace of X×F2 and therefore must be contained in X×{µ} for some µ ∈ F2. Since h is a map of bundles we get that h(x, λ) = (x, µ) for all x ∈ X. This defines a function g : F1 F2 such that h = id×g. Now if X is not connected that we can get the result by writing X as the disjoint union of its connected components which are all open because X is locally connected.→ Proposition 3.4. Let p : Y I be a cover of I = [0, 1]. Then p is a trivial bundle.

−1 ∼ Proof. Let us assume that→ we have 0 < t1 < t2 < t3 with trivializations ϕ0 : p ([0, t2) − −1 ∼ [0, t2) × F0 and ϕ1 : p ((t1, t3)) − (t1, t3) × F1. The intersection [0, t2) ∩ (t1, t3) = (t1, t2) is connected, and therefore the homeomorphism →

→ ∼ ϕ | −1 ◦ ϕ | −1 :(t , t ) × F − (t , t ) × F 1 p (t1,t2) 0 p (t1,t2) 1 2 0 1 2 1

∼ −1 is given as id × ϕ01 by some bijective map ϕ01 : F0 F1.→ Note that if y is in p (t1, t2) = −1 −1 p [0, t2) ∩ p (t1, t3) then we have (id(t1,t2) × ϕ01)(ϕ0(y)) = ϕ1(y) so that we may define → 1 (id[0,t ) × ϕ )(ϕ (y)) if y ∈ p ([0, t ) ψ : p−1([0, t )) − [0, t ) × F y 7 2 01 0 2 3 3 1 −1 ϕ1(y) if y ∈ p (t1, t3) → →

20 Clearly ψ is continous since its restriction to [0, t2) and (t1, t3) are continous. The map ψ is bijective with the inverse

−1 −1 −1 −1 ϕ0 (t, ϕ01 )(λ)) if t ∈ [0, t2) ψ :[0, t3) × F1 − p ([0, t3)) (t, λ) 7 −1 ϕ1 (y) if t ∈ (t1, t3) which is continious on [0, t→2) and (t1, t3), hence on→[0, t3). Together we see that ψ defines a −1 ∼ trivialization of p on [0, t3). Note that if t3 = 1 and ϕ1 : p ((t1, 1] − (t1, 1] × F1 is a trivialization, then the same construction holds and we have shown that p is globally trivial in that case. → Let p : Y [0, 1] be a cover. For any t ∈ (0, 1) we can find 0 < αt < t < βt < 1 and −1 ∼ trivializations ϕt : p (αt, βt) − (αt, βt) × Ft as well as trivializations on [0, β0) and (α1, 1]. Since I is compact→ we can cover I with finitely many of those intervals. By further removing any interval contained in the union of→ the others, we get a finite chain of open intervals covering I such that every two open intervals in the chain either don’t meet or their intersection is an open interval each, hence we can find a finite series 0 < t1 < s1 < t2 < s2 < t3 < . . . tn < sn < 1 such that if we set I0 = [0, s1), I1 = (t1, s2), ..., Ii(ti, si+1) ...In = (tn, 1] with trivializations −1 ∼ ϕi : p (Ii) − Ii × Fi. Starting with I0 and I1 we can the construct a global trivialization by the procedure described above by induction in n steps. → Let us mention that the proof of the last proposition can be applied to much more general situations. In order to do the construction of a trivialization of a cover p : Y X we only need a family local trivialization with respect to a nice type of open covering that allows us to do the Sn induction. The covering only needs to fulfil two requirements: it must be finite→ X = i=1 Ui Sk−1 and we need that Uk∩ i=1 Ui is connected. This allows us to formulate the following corollary. S Corollary 3.1. Let X be a topological space. Suppose that for every open cover X = λ∈Λ Uλ there Sn exists a finite open cover X = ρ=1 Vρ such that for every ρ there exists λ ∈ Λ with Vρ ⊂ Uλ, and such Sk−1 that Vk ∩ ρ=1 Vρ is connected for any k. Then every cover on X is trivial. n n n 2 This corollary applies for example to I = [0, 1] ⊂ R for any n. Let us illustrate this with I . S Assume that we are given an open cover Uλ of I×I. For any (s, t) ∈ I×I we can find intervals t t t t Is 3 t and Js 3 s such that Is ×Js is contained in one of the Uλ. Let us first fix t ∈ I. Since [0, 1] is s s , s , . . . s Jt [0, 1] compact we can chose among the finitely many 1 2 n such that the si cover , and by reordering the intervals we may assume that Jsi intersects only non-trivially with Jsi−1 and Tn t Jsi+1 . Now set It = i=1 Is. Next we see that we only need finitely many It to cover [0, 1] and we by reordering the intervals we may assume that Iti intersects only non-trivially with Iti−1 and Iti+1 . Hence we get Proposition 3.5. Let p : Y X be a covering space. −1 Let γ :(I, 0) (X, x0) be a path and y0 ∈ p (x0). The γ can be lifted in a unique way to p with → respect to y0 ∈. → If γx0 : x0 x0 is the constant path, then γ˜ x0 (t) = y0 for all t ∈ I.

If γ0, γ1 :(I, 0) (X, x0) are homotopic paths, then γ˜ 0(1) = γ˜ 1(1) and γ˜ 0, γ˜ 1 are homotopic.

Proof. The first part→ follows from the lemmas, the second part is obvious by the unicity of the lift. For the third part consider a homotopy h : γ0 γ1. Then h :(I × I, (0, 0)) (X, x0) can be lifted in a unique way to h˜ :(I × I, (0, 0)) (Y, y0). We will show that h lifts to a homotopy h˜ : γ˜ 0 γ˜ 1. → → 21 • First let us show that h˜ (0, t) = y0 for all t ∈ I. Define j0 : I I × I by j0(t) = (0, t), then ˜ ˜ h ◦ j0 = γx0 and h ◦ j0 is a lift of γx0 , hence h(0, t) = y0 for all t ∈ I by (ii) and the unicity of the lift. →

• Set it(s): I I × I as it(s) = (s, t) and γt = h ◦ it. By (a) we get that h˜ is a lift of h :(I × I, (0, t)) (X, x0) for all t, hence h˜ ◦ it is a lift of γt and we get h˜ (s, t) = γ˜ t(s) for all (s, t) ∈ I ×→I. → • Finally we need to show that h˜ (1, t) does not depend on t. Set y1 = h˜ (1, 0) = γ˜ 0(1). Then p(y1) = ph˜ (1, 0) = h(1, 0) = x1 and h˜ is the unique lift of h :(I × I, (1, 0)) (X, x1) to ˜ p :(Y, y1) (X, x1). Now set j1 : I I × I by j1(t) = (1, t) then h ◦ j1 = γx1 and h ◦ j1 is ˜ a lift of γx1 . By unicity we get that h(1, t) = γ˜ x1 (t) = y1. → → → ˜ ˜ Define j0 : I I × I by j0(t) = (0, t), then h ◦ j = γx0 and h ◦ j is a lift of γx0 , hence h(0, t) = x0 for all t ∈ I. This implies that if we define it(s): I I × I as it(s) = (s, t). Then h ◦ it = γt and h˜ ◦ it is a lift→ of γt hence h˜ (s, t) = γ˜ t(s). Finally → Let p : Y X be a covering space and γ : x0 x1 a path. Then γ can be lifted to p and if we fix the base point of the lift, then the list if uniquely determined. Hence, the proposition − assures us→ that for every point y ∈ p 10 the exists a unique liftγ ˜ y of γ to p, and we may set −1 −1 −1 µ(γ)(y) = γ˜ y(1) ∈ p (x1). We have thus constructed a function µ(γ): p (x0) p (x1) Proposition 3.6. We have µ(γ ⊥ δ) = µ(δ) ◦ µ(γ). → 1 1 1 Proof. Let us define i : I I by i(t) = 2 t and j : I I by j(t) = 2 t + 2 . Clearly we have (γ ⊥ δ) ◦ j = δ and (γ ⊥ δ) ◦ i = γ. y ∈ p−1(x ) → (γ^⊥ δ) ◦→i γ y Let 0 0 . By definition we have y0 is a lift of with respect to 0, so by γ (1) = (γ^⊥ δ) ( 1 ) y = γ (1) = µ(γ)(y ) unicity we get ˜ y0 y0 2 . Set 1 ˜ y0 0 , we get immediately that (γ^⊥ δ) ◦ j δ y δ˜ (1) = (γ^⊥ δ) 1) y0 is a lift of with respect to 1, hence by unicity we get y1 y0 µ(δ)(µ(γ)(y0)) = µ(δ)(y1) = µ(γ ⊥ δ)(y0) which shows the proposition.

3.3 Group actions and coverings

Definition 3.6. Let G be a group and S a set. A group action of G on S is a group homomorphism

µ : G Aut(S)

(S) S µ G S where Aut is the set of bijections of . A group→ action of on is also called a (set-theoretical) representation of G. ∼ Hence a representation of G is the data of a set S and for any g ∈ G a bijection µ(g): S − S −1 −1 such that µ(gh) = µ(g) ◦ µ(h) Note that we have also µ(g ) = µ(g) and µ(e) = IdS. → Definition 3.7. Let µ1 : G Aut(S1) and µ2 : G Aut(S2) be two representations. A morphism µ1 µ2 is given by a map ϕ : S1 S2 such that µ1(g) ◦ ϕ = ϕ ◦ µ2(g) for all g ∈ G: → → ϕ → → S1 / S2

µ1(g) µ2(g)   S1 ϕ / S2

22 Recall that a group G is nothing but a category G with a single object {∗} whose endomorphisms are given by G. Note that a representation µ : G Aut(S) is the same as a functor µ : G Set such that µ(∗) = S, and a morphism of representations is nothing but a morphism of such func- tors. This immediately leads to the slightly more→ general definition: →

Definition 3.8. Let G be a groupoid and C a category. A representation of G in C is a functor µ : G C.A morphism of representations is a morphism of functors. We denote by Rep(G, C) the category of representations of G in C. → A trivial represntation is a represntation µ that sends every automorphism of G to an identity morphism in C.

Definition 3.9. Let p : Y X be a cover. We define the monodromy functor µ : Π1(X) Set by the following µ(x) = p−1(x) for any x ∈→X → −1 −1 µ(γ): p (x0) p (x1) is the unique map defined above.

Let p1 : Y1 X, p2 : Y2 X be two coverings on X and f : Y1 Y2 be a morphism of → −2 −1 coverings. Then f defines maps fx = f| −1 : p (x) p (x). p1 (x) 1 2 → → → Lemma 3.8. We have µ(p2)([γ]) ◦ fx0 = fx1 ◦ µ(p1)([γ])→. In other words

f −1 x0 −1 p1 (x0) / p2 (x0)

µ(p1)([γ]) µ(p2)([γ])   −2 −1 p1 (x1) / p2 (x1) fx1

−1 Y1 Y1 Y2 Proof. Let y ∈ p1 (x0). Letγ ˜ y be the unique lift of γ to Y1 such thatγ ˜ y (0) = y andγ ˜ f(y) the Y2 unique lift of γ to Y2 such thatγ ˜ f(y)(0) = f(y). Y1 Y1 Y1 Then p2 ◦ f ◦ γ˜ y = p1 ◦ γ˜ y = γ and f ◦ γ˜ y (0) = f(y) and by unicity of the lift we get Y ˜ 2 Y1 (γ)f(y) = f ◦ γ˜ y . Therefore

˜ Y2 µ(p2)([γ])(f(x0)(y)) = (γ)f(y)(1) = f(γ˜ y(1)) = fx1 (µ(p1)([γ])(y)).

Therefore f : Y1 Y2 defines a morphism of representations µ(f): µ(p1) µ(p2). Clearly we get µ(g ◦ f)x = (g ◦ f)x = (g ◦ f)| −1 = g| −1 ◦ f| −1 = µ(g) ◦ µ(f) → p1 (x) p2 (x) p1 (x) →

And µ(idY) = idµ(p). Consequently we have shown that we have defined the monodromy functor µ : Cov(X) − Rep(Π1(X), Set) Lemma 3.9. The monodromy functor is faithful,.and if X is locally connected then it is conservative. →

Proof. If f, g : Y1 Y2 are two maps of coverings such that µ(f) = µ(g), then f| −1 = g| −1 p1 (x) p1 (x) S −1 for each x ∈ X and since Y1 = x∈X p1 (x) we get that f = g. Hence µ is faithful. Let f : Y1 Y2 be→ a morphism of coverings such that µ(f) is an isomorphism. Then f| −1 : p1 (x) 23 → −1 −1 −1 p1 (x) p2 (x) is a bijection and since Yi is the disjoint union of the sets pi (x) we get that f is bejective, hence a homeomorphism since X is locally connected. Proposition→ 3.7. Let X be locally arcwise connected, then µ is full.

−1 Proof. Let φ : µ(p1) µ(p2) be a morphism of representations. It defines φx : p1 (x) −1 −1 p2 (x) for all x ∈ X, since Yi is the disjoint union of pi (x) φ defines a map ϕ : Y1 Y2 such that ϕ| −1 = φx. In particular p2 ◦ϕ = p1. We have to show that ϕ is continuous. Since this is p1 (x) → → a local property we can assume without loss of generality that X is arcwise connected→ and Y1,Y2 are trivial coverings Yi = X × Fi. Then ϕ : X × F1 X × F2 is defined as ϕ(x, λ) = (x, φx(λ)). 0 0 We shall show that φx(λ) does not depend on x. Let x ∈ X and chose a path γ : x x . Then φx 0 ◦ µ(p1)([γ]) = µ(p2)([γ]) ◦ φx, but p1 and→p2 are trivial, hence µ(pi)([γ]) = id. Hence ϕ = id × φx is continous. Corollary 3.2. Let X be simply connected and locally arcwise connected. Then every covering on X is trivial.

3.4 Universal coverings and fundamental groupoid

Definition 3.10. A universal covering on X is a covering p˜ : X˜ X such that X˜ is simply connected.

Let X be a topological space and x0 ∈ X. Let us define the universal representation µ˜ (which → depends on the choice of x0) by

(i) µ˜ (x) = π1(X, x0, x) for any x ∈ X

(ii) µ˜ ([γ]) : π1(X, x0, x1) π1(X, x0, x2);[ρ] 7 [ρ][γ] for any path γ : x1 x2

Since we haveµ ˜ ([γ][δ])([ρ]) = [ρ][γ][δ] = µ˜ ([δ]) ◦ µ˜ ([γ])([ρ]) for any γ : x1 x2, δ : x2 x3 and → → any ρ : x0 x1 we see thatµ ˜ is indeed a representation. −1 We shall see that for any covering p : Y X we get for any choice of y0 ∈ p (Y) a morphism of representationsµ ˜ µp, and that this morphism is an isomorphism if Y is simply connected. → Lemma 3.10. Let p : →Y X be a covering. Then π1(p) is faithful.

Proof. Let y0, y1 ∈ Y and→ γ, ρ : y0 y1 two paths such that π1(p)([γ]) = π1(p)([ρ]). Then [p ◦ γ] = [p ◦ δ], hence p ◦ γ and p ◦ ρ are homotopic, But γ is the unique lift of p ◦ γ such that γ(0) = y0 and ρ is the unique lift of p ◦ ρ such that ρ(0) = y0. Hence γ and ρ are homotopic, hence [γ] = [ρ].

In particular this means that p induces an isomorphism of π1(Y, y0) to a subgroup of π1(X, x0). We shall identify this subgroup with π1(Y, y0), hence the class of a loop γ : y0 y0 is iden- tified with its image [p ◦ γ] ∈ π1(X, x0) or, in other words, an element [ρ] ∈ π1(X, x0) satisfies

[ρ] ∈ π1(Y, y0) if and only if its unique liftρ ˜y0 is a loop at y0.

−1 Proposition 3.8. Let p : Y X be a covering and x0 ∈ X. Then any choice of y0 ∈ p (x0) defines a map −1 αx :→π1(X, x0, x) − p (x)[ρ] 7 µ([ρ])(y0).

→ →

24 0 (i) For any γ : x x the following diagram commutes

αx −1 π1(X, x0, x) / p (x)

·[γ] µ([γ])   0 −1 0 π1(X, x0, x ) / p (x ) αx 0

Hence the family αx defines a morphism α : µ˜ µp. −1 (ii) αx(ρ) = αx(δ) if and only if [ρ][δ] ∈ π1(Y, y0) , in particular π1(Y, y0) is the stabilizer of y0 for −1 → the group action of π1(X, x0) on p (x0). (iii) If Y is simply connected then α is an isomorphism. In particular, any to universal coverings are isomorphic.

Proof. (i) µ([γ]) ◦ αx([ρ]) = µ([γ])µ([ρ])(y0) = µ([ρ][γ])(y0) = αx 0 (·[γ])(y0) −1 (ii) Recall that [ρ][δ] ∈ π1(Y, y0) if and only if (ρ^⊥ δinv)y0 is a loop at y0 and therefore

−1 −1 (µ([δ]) ◦ µ([ρ]))(y0) = µ([ρ][δ] )(y0) = y0.

Composing with the isomorphism µ([δ]) from the left gives the equivalent conditionαx([ρ]) = µ([ρ])(y0) = µ([δ])(y0) = αx([δ]). (iii) Since π1(Y, y0) is trivial the map is clearly injective. Since Y is arcwise connected we may −1 chose for any y1 ∈ p (x) a path ρ : y0 y1,then µ([p ◦ ρ])(y0) = y1 since ρ is the unique lift of p ◦ ρ such that ρ(0) = y0. Hence the map is also bijective.

1 Now we may apply this procedure to calculate Π1(S ).

1 2πit The map p : R 7 S ⊂ C t 7 e is continous. It is easy to check that p is a covering that 1 −1 is trivial on S \{p} where p is any point of the circle. Clearly p (1) = Z. Hence the choice of −1 0 ∈ p (1) = Z→defines a bijection→ 1 Φ : π1(S , 1) − Z [γ] 7 µ(p)([γ])(0) Note that Φ([c ]) = 0 since µ(p) is a group action. Set 1 → → 1 2πint γn : I S ; t 7 e Clearly the lift of γ is given byγ ˜ (t) = tn, hence 0 → →

Φ([γn]) = µ(p)([γn])(0) = (γ˜ n)0(1) = n

1 1 Hence π1(S , 1) = {[γn]}n∈Z ' Z. Let us determine the group structure of π1(S , 1) in order to conclude that Φ is a group isomorphism. Clearly γ0 is the constant path hence the unit, and just as obviously we have γn,inv = γ−n by construction. In order to find the group law, one can actually construct explicit homotopies γn ⊥ γm γn+m, but this is a bit tiresome, and we may also use the machinery of coverings: Let n, m ∈ Z. We have

Φ([γn][γm]) = µ(p)([γn][γm])(0) = µ(p)([γm])(µ(p)([γn])(0) = µ([γm])(n)

We may lift γm by the map t 7 tm + n, hence µ([γm])(n) = m + n. Therefore

Φ([γ ][γ ]) = m + n = Φ([γ ]) → n m n+m 25 Since Φ is bijective we have proved that the group law is given by [γn][γm] = [γn+m]. Next let us construct a universal covering. Let us first assume thatp ˜ : X˜ X is a universal covering on a connected and locally arwise connected space X. Assume that X˜ is trivial on an ∼ arcwise connected open subset U. Let us fix a trivialization ϕ : p˜ −1(U) − →U × F. Consider a loop γ : I U at x ∈ U. Any λ ∈ F defines a lift → (γ) : I U × F −∼ X˜ → ˜ ϕ(x,λ) U X˜ X˜ and since is connected this is necessarily a→ loop in .→ Since is simply connected it is ho- motopic to the constant path on X˜. Therefore γ = p˜ ◦ γ˜ is homotopic (on X, not on U) to the constant path at x. Hence U has the property that every loop on U is homotopic to a constant loop on X. This property can be visualized by the fact that the image of the functor

Π1(U) − Π1(X) is equivalent to the terminal category with only one object and one morphism. In particular → this implies that if Y X is any covering of X then the monodromy of Y|U is trivial and there- fore Y|U is trivial on U. → Definition 3.11. Let X be a connected and locally arcwise connected space. An open subset U ⊂ X is called semi-1-connected if it is arcwise connected and every loop γ : x x on U is homotopic to the constant loop cx on X. The space X is called locally semi-1-connected if it can be covered by semi-1-connected open subsets. We already know that if there exists a universal coveringp ˜ : X˜ X then X can be covered by semi-1-connected open subsets, and moreover the universal cover must be trivial on any such subset. We also know that its monodromy has to be isomorphic to→ the universal representation µ˜ . These two facts allow us to give a general construction:

Theorem 3.1. Let X be a locally semi-1-connected space. Then there exists a universal cover X˜. S Proof. For every x ∈ X chose an open semi-1-connected neighborhood Ux. Then X = x∈X Ux is covered by semi-1-connected open subsets. Chose a base point x0 ∈ X and let

px : Ux × π1(X, x0, x) Ux be the trivial covering. w → For every w ∈ Ux chose paths δx : x w in Ux (since Ux is semi-1-connected there is only one w w w possible choice up to homotopy on X), and for every x ∈ Uxy = Ux ∩ Uy set δxy = δy ⊥ δx,inv. Now set

w ϕxy : Uxy × π1(X, x0, y) − Uxy × π1(X, x0, x);(w, [γ]) 7 (w, [γ][δxy])

w Let us emphasize that this definition does not depend on the choice of the paths δx , in particu- x → → lar we may assume δx = cx for all x ∈ X. Clearly ϕxy is a bijective map such that px ◦ ϕxy = py, its inverse is given by ϕyx. Let us show that it is a local homeomorphism. Chose an arcwise connected open neighborhood V of w ∈ Uxy such that w ∈ V ⊂ Uxy. Let v ∈ V and chose a w v path δ : w v in V ⊂ Ui. Since δx ⊥ δ ⊥ (δx)inv is a loop in Ux, it is homotopic to the constant w v w v path cx on X, hence [δx ][δ] = [δx], and similarly [δy ][δ] = [δy], thus showing

w w −1 w −1 v v −1 v [δxy] = [δy ][δ][δ] [δx ] = [δy][δx] = [δxy]

w w Therefore ϕxy(v, [γ]) = (v, [γ][δxy]) for all v ∈ V which means that ϕij|V = idV × (·[δxy]) which is a homeomorphism onto V × π1(X, x0, y). Hence ϕxy is a bijective local homeomorphism, 26 hence a homeomorphism. Clearly ϕxx = id since all loops in Ui are homotopic (on X) to the constant loop. −1 We get ϕxy = ϕyx by construction. w w w w −1 w w −1 w Let w ∈ Uxyz. Then [δyz][δxy] = [δz ][δy ] [δy ][δx ] = [δxz], hence

w w w ϕxyϕyz(w, [γ]) = (w, [γ][δyz][δxy]) = (w, [γ][δxz]) = ϕxz(w, [γ])

˜ ˜ ∼ By the patching lemma we get that there exists p : X X with isomorphisms Φx : X|Ux Ux × π1(X, x0, x), compatible with the isomorphisms ϕ|xy. We need the following auxiliary lemma to calculate the→ lift of a path. → 0 Lemma 3.11. Let γ : x x be a path on X, and [ρ] ∈ π1(X, x0, x). Let γ˜ be the unique lift of γ such that Φx(γ˜ (0)) = (x, [ρ]). Then t Φγ(t)(γ˜ (t)) = (γ(t), [ρ][γ ]) where γt(s) = γ(st) for s ∈ [0, 1].

Proof. We shall show that the subset J = {t ∈ [0, 1] | the assertion is true for any s 6 t} is non- empty, open and closed, hence equal to [0, 1]. Clearly the lemma holds for t = 0 so J is non-empty. ˜ Let t0 ∈ J. Chose an open interval J0 around t0 such that γ(J0) ⊂ Uγ(t0). Since X is trivial on t0 Uγ(t0) the pathγ ˜ has to take values in the same sheet, hence Φγ(t0)(γ˜ (t)) = (γ(t), [ρ][γ ]) for all t ∈ J0 since this is true for t = t0. We have

γ(t) γ(t) Φ (γ˜ (t)) = ϕ Φ (γ˜ (t)) = (γ(t), [ρ][γt0 ][δ ]) = (γ(t), [ρ][γt0 ][δ ]) γ(t) γ(t)γ(t0) γ(t0) γ(t)γ(t0) γ(t0)

γ(t) Since any path in U from γ(t0) to γt is homotopic to δ , we may replace this path by γ(t0) γ(t0) γt γt (s) = γ(t + s(t − t ) t0 defined by t0 0 0 . Finally we note that there is an obvious homotopy γt0 ⊥ γt ∼ γt t ∈ J J t0 thus showing that the formula holds for all 0. Hence is open. Now assume that the assertion does not hold for t0, i.e. there exists s 6 t0 such that the formula is false. If s < t0 then ]s, 1] is an open neighborhood of t0 in [0, 1] \ J, so let us assume s = t0.

t Φγ(t0)(γ˜ (t0)) 6= (γ(t0), [ρ][γ ])

Chose J0 3 t0 such that γ(J0) ⊂ Uγ(t0), chose t < t0, t ∈ J0 and suppose that we have t Φγ(t)(γ˜ (t)) = (γ(t), [ρ][γ ]), then

γ(t) γ(t) Φ (γ˜ (t)) = ϕ (γ(t), [ρ][γt]) = (γ(t), [ρ][γt][δ ]) = (γ(t), [ρ][γt][δ ]) γ(t0) γ(t0)γ(t) γ(t0)γ(t) γ(t0)

γ(t) Similarly to the case before we may replace δ by any path in U from γ(t) to γ(t0), so γ(t0) γ(t0) t0 t t0 t0 we may take γt (s) = γ(t + s(t0 = t)). It is easily checked that γ ⊥ γt ∼ γ , hence

t0 Φγ(t0)(γ˜ (t)) = (γ(t), [ρ][γ ] ˜ Since X is trivial on Uγ(t0) we know that Φγ(t0)(γ˜ (t)) is in the same sheet as Φγ(t0)(γ˜ (t0)) which is a contradiction. Hence ]t, 1] is an open neighborhood of t0 where the formula is false, there- fore I is closed.

−1 −1 Now we can show that X˜ is arcwise connected, let y1 ∈ p (x1) and y2 ∈ p (x2), then

Φx1 (y1) = (x1, [γ]) for some path γ : x0 x1 and Φx2 (y2) = (x2, [δ]) for some path δ : x0 x2, set ρ = γinv ⊥ δ and letρ ˜ be the unique lifting such that Φx1 (ρ˜(0)) = (x1, [γ]). Then −1 Φx1 (ρ˜(1)) = (x2, [γ][γ] [δ]) = (x2, [δ]). 27 Let us show that X˜ is simply connected. Since X˜ is arcwise connected it is enough to show γ x = Φ−1(x , c ) γ = p ◦ γ that every loop ˜ at ˜ 0 x0 0 x0 is homotopic to the constant loop. Set ˜ , then γ ∈ π1(X, x0), andγ ˜ is the unique lifting of γ. In particular by our lemma we get

t Φγ(t)(γ˜ (t)) = (γ(t), [γ ])

Thus Φx0 (γ˜ (1)) = (x0, [γ]) = (x0, [cx0 ]) sinceγ ˜ is a loop, hence γ is homotopic to the constant loop on X and by the homotopy lifting theoremγ ˜ is homotopic to the unique lift of the constant loop which is the constant loop cx˜ 0 .

It is useful to keep in mind the particular construction of a universal covering that we have just done.

Corollary 3.3. Let X be a locally semi-1-connected topological space and x0 ∈ X a base point. For every ˜ S universal covering space p˜ : X X there exists an open cover X = x∈X Ux where Ux is a semi-1- ˜ ∼ connected open neighborhood of x and isomorphisms of coverings Φx : X|Xx − Ux ×π1(X, x0, x) such that the bijections → −1 ∼ Φx : p (x) − π1(X, x0, x) → ∼ define an isomorphism µp˜ µ˜ . → ˜ Corollary 3.4. Let X be locally semi-1-connected, p˜ : X X a universal covering and µX˜ its mon- −1 x˜ odromy representation. Let→x0 ∈ X and x˜ 0 ∈ p˜ (x0). For each x˜ chose a path δ˜ : x˜ x˜ 0 and set x˜ x˜ δ = p˜ ◦ δ˜ : p˜(x˜) x0. →

(i) For each [γ] ∈ π1(X, x0) the map

x˜ x˜ f[γ] : X˜ − X˜ ; x˜ 7 µ([δ ][γ][δ ])(x˜)

p [δx] is a automorphism of ˜ that does not→ depend on the choice→ of the paths (ii) The map ˜ π1(X, x0) − AutCov(X) (X);[γ] 7 f[γ] is an isomorphism of groups. → → S (iii) Suppose that X = x∈X Ux is an open covering where Ux is a semi-1-connected open neighborhood ˜ ∼ of x, and suppose that we have isomorphisms of coverings Φx : X|Xx − Ux × π1(X, x0, x) such that the bijections −1 ∼ Φx : p (x) − π1(X, x0, x) → ∼ define an isomorphism µp˜ µ˜ . Further assume that Φx0 (x˜ 0) = (x0, [cx0 ]) and Φx(x˜) = (x, [δ]). then we have → → Φx(f[γ](x˜)) = (x, [γ][δ])

Proof. By the proposition µX˜ is isomorphic toµ ˜ defined as

(i) µ˜ (x) = π1(X, x0, x) for any x ∈ X

(ii) µ˜ ([γ]) : π1(X, x0, x1) π1(X, x0, x1);[ρ] 7 [ρ][γ] for any path γ : x1 x2 Let us first calculate all automorphisms ofµ ˜ . → → Let [γ] ∈ π1(X, x0), then [γ] induces an automorphism F[γ]of µ by

(F[γ])x : π1(X, x0, x) − π1(X, x0, x);[ρ] 7 [γ][ρ].

→ 28 → Since (F[γ])x0 ([cx0 ]) = [γ] we get that F[γ] = F[δ] if and only if [γ] = [δ].

Now let F be an automorphism of µ and set [γ] = Fx0 ([cx0 ]). Now let [ρ] ∈ π1(X, x0, x1), then we get the commutative diagram

F π1(X, x0) / π1(X, x0)

µ([ρ]) µ([ρ])   π1(X, x0, x) / π1(X, x0, x) Fx

Hence Fx(µ([ρ])[cx0 ]) = µ([ρ])Fx0 ([cx0 ]) and therefore Fx([ρ]) = [γ][ρ] thus showing that F = F[γ]. It is easily checked that F[γ⊥δ] = F[δ] ◦ F[γ], hence we get a group isomorphism

π1(X, x0) − Aut(µ)

Now assume that we are given a universal coveringp ˜ : X˜ − X together with an open covering S → X = x∈X Ux where Ux is a semi-1-connected open neighborhood of x, and suppose that we ˜ ∼ have isomorphisms of coverings Φx : X|Xx − Ux × π1(X,→ x0, x) such that the bijections

Φ : p−1(x) −∼ π (X, x , x) x → 1 0 ∼ define an isomorphism Φ : µ µ˜ . Then we get a group isomorphism p˜ → π (X, x ) − Aut(µ) − Aut(µ );[γ] 7 f = Φ−1 ◦ F ◦ Φ 1 0 → p˜ [γ] [γ] −1 It is easily verified that this satisfies to the forumla of (i) if we setx ˜ = Φ (x , [cx ]) and it also → → → 0 x0 0 0 gives (iii). We still have to prove (i). Theorem 3.2. Let X be locally semi-1-connected. Then µ is an equivalence of categories.

Proof. We only need to show that µ is essentially surjective. ˜ ˜ ∼ Let us fix x0 ∈ X and a universal coveringp ˜ : X X together with isomorphisms Φx : X|Ux − π1(X, x0, x) as in Theorem 3.1. In particular we have Lemma 3.11. x = Φ−1(x , [c ]) ∈ p−1(x ) Let us further fix ˜ 0 x0 0 x0 ˜ 0→and therefore get a canonical group isomor-→ ˜ ˜ phism π1(X, x0) − AutCov (X, X); γ 7 f[γ]. The automorphism f[γ] satisfies the defining property that if Φx(x˜) = (y, [δ]) then Φx(f[γ](x˜) = (y, [γ][δ]).

Let µ : Π1(X) − →Set be a representation,→ and set Sx = µ(x). We will write S = Sx0 and Y˜ = X˜ × S. Letq ˜ : Y˜ X˜ X be the composition ofp ˜ with the natural projection X˜ × S X˜. Thenq ˜ : Y˜ X is→ a covering of X that is trivial on every semi-1-connected open subset of X. Let us put (x˜ 1, s1) ∼ →(x˜ 2, s→2) ifp ˜(x˜ 1) = p˜(x˜ 2) and there exists [γ] ∈ π1(X, x0) such that→x ˜ 2 = −1 f[γ](x˜ 1) and→s2 = µ([γ]) (s2). It is easily checked that this is an equivalence relation on Y˜, let Y = Y/˜ ∼ be the quotient. Since (x˜ 1, s1) ∼ (x˜ 2, s2) impliesp ˜(x˜ 1) = p˜(x˜ 2) we get a well-defined map p : Y X that sends the class (x,˜ s) top ˜(x˜). Let Ux be the semi-1-connected open neighborhood of x ∈ X such that we have the isomor- ˜ phism Φx →: X|Ux − Ux × π1(X, x0, x). We shall now define an equivalence relation on Ux ×

π1(X, x0, x) × S such that Φx × IdS induces a homeomorphism Y|Ux ' (Ux × π1(X, x0, x) × S)/ ∼. We define a relation→ on Ux × π1(X, x0, x) × S by putting (y1, [δ1], s1) ∼ (y2, [δ2], s2) if and only if −1 y1 = y2 and there exists [γ] ∈ π1(X, x0) such that [δ2] = [γ][δ1] and s2 = µ([γ] )(s1). It is easily verified that ∼ is an equivalence relation. Let us denote by Z(Ux) le quotient. −1 Note that (x˜ 1, s1) ∼ (x˜ 2, s2) if and only if (f[γ](x˜ 1), µ([γ] )(s1)) = (x˜ 2, s2) and if we set (x, [δ]) = Φx(x˜ 1) then this is equivalent to

−1 −1 (x, [γ][δ], µ([γ] (s1)) = (Φx(f[γ](x˜ 1)), µ([γ]) (s1)) = (Φx(x˜ 2), s2), 29 hence to (Φx(x˜ 1), s1) ∼ (Φx(x˜ 2), s2). Therefore homeomorphism Φx × IdS induces a homeomorphism:

Y|Ux − Z(Ux) = (Ux × π1(X, x0, x) × S)/ ∼

→ Let us define

Ψx : Ux × π1(X, x0, x) × S − Ux × Sx ; Ψx(y, [δ], s) = (y, µ([δ])(s))

Note that if (y1, [δ1], s1) ∼ (y2, [δ2], s2) then y1 = y2 and there exists [γ] ∈ π1(X, x0) such that −1 → [δ2] = [γ][δ1] and s2 = µ([γ] )(s1). Hence

−1 −1 Ψx(y2, [δ2], s2) = (y1, [γ][δ1], µ([γ] )(s1)) = (y1, µ([γ][δ1])µ([γ] )(s1)) −1 = (y1, µ([γ] [γ][δ1])(s1) = Ψx(y1, [δ1], s1)

Note that Ψx is open, so it induces an open map

Ψx : Z(U) − U × Sx ; (y, [δ], s) 7 µ([δ])(s)

−1 Let (y, s) ∈ U × Sx. Chose a path→δ : y x0, then the class of→(y, [δ] , µ([δ])(s)) is mapped to (y, s) par Ψx, hence Ψx is surjective. Note that Ψx(y1, [δ1], s1) = Ψx(y2, [δ2], s2) if and only if y1 = y2 and µ([δ1])(s1) = µ([δ2])(s2). −1 Then µ([δ1][δ2] )(s1) = s2. Hence (y1, [δ1], s1) ∼ (y2, [δ2], s2) by the loop [δ1 ⊥ δ2,inv] ∈ π1(X, x0) and Ψx is bijective, continous and open, hence a homeomorphism. By composition we get a trivilaization ∼ ∼ Θx : Y|Ux − Z(U) − Ux × Sx Hence p : Y X is a covering. Let us calculate the monodromy of p : Y →X. We have→ isomorphisms

→ −1 ∼ Θx : p →(x) − {x} × Sx ' Sx ∼ Let us show that we defined an isomorphism Θ→: µp − µ. −1 Let γ : x1 x2 a path and (x,˜ s) ∈ p (x1). Set (x1, [δ]) = Φx1 (x˜). Then → µ([γ])(Θx1 ((x,˜ s))) = µ([γ])(Ψx1 (x1, [δ], s)) = µ([γ])(x1, µ([δ])(s)) = (x2, µ([γ])µ([δ])(s)) ˜ Letγ ˜ : I X be the unique lift of γ such thatγ ˜ (0) = x˜ (which is the same as asking Φx1 (γ˜ (0)) = (x1, [δ])) and set further → γ^ : I Y ;γ ^(t) = (γ˜ (t)), s) Then p ◦ γ^ = p˜ ◦ γ˜ = γ and γ^(0) = (x,˜ s). Therefore we may use Lemma 3.11 to calculate →

Θx2 (µp([γ])((x,˜ s))) = Θx2 (^γ(1)) = Ψx2 (Φx2 (γ˜ (1)), s) = Ψx2 (x2, [δ][γ], s) = (x2, µ([δ][γ])(s))

Hence Θx −1 1 p (x1) / Sx1

µp([γ]) µ([γ])  −1  p (x2) / Sx2 Θx2 ∼ commutes and Θ defines an isomorphism of representations µp − µ. 30 → References

[B] Berger, C., Topologie pour la Licence, http://math.unice.fr/∼cberger/topologie.pdf, 2004. [Bo] Boubaki, N., Topologie générale, Chapitres 1 à 4, Masson, 1990. [G] Godement, R., Topologie algébrique et théorie des faisceaux, Hermann, 1958. [GH] Greenberg, M. J. and J. R. Harper, Algebraic Topology, a first course, Addison-Wesley, 1981. [Sch] Schapira, P., Algebra and Topology, http://people.math.jussieu.fr/∼schapira/lectnotes/AlTo.pdf, 2008. [Z] Zisman, M., Topologie Algébrique Élémentaire, Librairie Armand Colin, 1972.

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