1 Introduction Category Theory Topological Preliminaries Function Spaces Algebraic Preliminaries...

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1 Contents 1 Introduction Category Theory Topological Preliminaries Function Spaces Algebraic Preliminaries Homotopy Basic Results H-groups and H-cogroups The Fundamental Group Free Groups and Free Products The Fundamental Groupoid Covering Spaces The Van Kampen Theorem Homology Basic Results Jordan Curve Theorem Degrees of Maps Finite CW Complexes Examples Topological Properties i

2 ii CONTENTS 5.3 Euler Characteristics Homology of CW complexes Examples Orientation in Balls and Spheres Homology of Products Defining Homology Chain Complexes Cubical Singular Homology Excision Products and Cohomology Tensor Products Cohomology Cup Product Exterior Cohomology Product Cap Product Manifolds 91

3 0 CONTENTS

4 Chapter 1 Introduction 1.1 Category Theory Definition 1.1 A category, C consists of a class Ob, of objects and another class Mor of Morphisms. We assume two functions, dom, cod : Mor C with the following properties: a) If X, Y C, then hom(x, Y ) = {f Mor : dom f = X, cod f = Y } is a set. b) If f, g Mor with cod f = dom g, then there is an g f Mor such that dom g f = dom f and cod g f = cod g. c) If f, g, h Mor with cod f = dom g and cod g = dom h, then h (g f) = (h g) f. d) For each object X, there is a morphism 1 X hom(x, X) with 1 X f = f whenever cod f = X and f 1 X = f whenever dom f = X. If f hom(x, Y ) we write f : X Y. There are many examples of categories. (1) Let Set denote the category whose objects are sets and where the morphisms with domain X and codomain Y are simply the functions from X to Y. Use ordinary composition of functions for composition. 1

5 2 CHAPTER 1. INTRODUCTION (2) Let Top denote the category whose objects are topological spaces and where the morphisms with domain X and codomain Y are the continuous functions f : X Y. Composition is the usual composition of functions. (3) Let Grp denote the category whose objects are groups and where the morphisms with domain X and codomain Y are the homomorphisms f : X Y. Comosition is the usual composition of functions. (4) Let Ab denote the category whose objects are abelian groups and whose morphisms are homomorphisms. (5) Let R be a ring. Let RMod denote the category whose objects are R- Modules and whose morphsims are module homomorphisms with ordinary function composition. (6) Let Top 2 denote the category of pairs (X, A) where X is a topological space and A X. A morphism f : (X, A) (Y, B) is defined to be a continuous map from X to Y with f[a] B. (7) Let ptop Denote the category whose objects are ordered pairs (X, x 0 ) where X is a topological space and x 0 X. A morphism f : (X, x 0 ) (Y, y 0 ) is a continuous function from X to Y with f(x 0 ) = y 0. (8) Let (X, ) be an ordered set. Define a category whose objects are the points of X and where hom(x, y) = {(x, y)}, the ordered pair, exactly when x y. If x y and y z, we define (x, y) (y, z) = (x, z). Then 1 x = (x, x). (9) For any category C, there is an opposite category C op that has the same objects, but where hom C op(x, Y ) = hom C (Y, X) and g C op f = f C g. (10) If M is any monoid (associate operation with identity), we may define C = {x} to be any singleton, Mor = M and f g = fg the operation in M. Suppose that f : X Y in a category C. We say that f is a section, (resp, a retract, resp. an isomorphism) if there is a g : Y X such that g f = 1 X,

6 1.1. CATEGORY THEORY 3 (resp. f g = 1 Y, resp. f g = 1 Y and g f = 1 X ). In other words, f is an isomorphism if and only if f is invertible. In this case, we say that X and Y are isomorphic objects. Notice that f is an isomorphism if and only if it is both a section and a retract. Proposition 1.2 Isomorphism is an equivalence relation on Ob. In the category of sets, sections are the one-to-one functions and retracts are the onto functions. In the category of topological spaces, isomorphism is the same as homeomorphism. Now, if C and D are categories, a (covariant) functor, F : C D consists of a map from the objects of C to the objects of D, where we write F (X) for the object in D corresponding to the object X in C, and a map between morphisms, so that if f : X Y in C, we have F (f) : F (X) F (Y ) in D such that F (1 X ) = 1 F (X) for every object X in C and that F (g f) = F (g) F (f) whenever g f is defined in C. A contravariant functor is similar, but we require that when f : X Y in C, that F (f) : F (Y ) F (X) in D and that F (g f) = F (f) F (g). Notice that the direction of the arrows is reversed for contravariant functors. For example, (1) If C is any of Top, Grp, Ab, we can define the forgetful functor from C to Set by letting F (X) denote the underlying set for X and F (f) be the function. In other words, the forgetful functors forget the topological, group, or other structures. Similarly, there are forgetful functors from R Mod to Ab and from Top 2 to Top that forget the ring multiplication and the second component, resp. (2) Let A be an element of any category C. Define a functor from C to Set by defining F (X) = hom(a, X) and, for f : X Y, defining F (f)(g) = g f. Similarly, we can define a contravariant functor G by letting G(X) = hom(x, A) and G(f)(g) = f g.

7 4 CHAPTER 1. INTRODUCTION (3) In the case where C = RMod, the collection hom(x, Y ) is an abelian group and the functors in the last example can be considered to go into Ab. (4) The identity functor is contravariant from C to C op. Proposition 1.3 Suppose F : C D is a functor. If A, B are isomorphic objects in C, then F (A), F (B) are isomorphic objects in D. More specifically, a covariant functor takes retracts to retracts and sections to sections while a contravariant functor takes retracts to sections and sections to retracts. The primary technique of algebraic topology is to find functors from topological categories into algebraic categories. In this way, questions about the existence of continuous functions turn into questions about the existence of homomorphisms, which are usually easier questions to answer. Definition 1.4 Let X and Y be objects in the category C. A product consists of an object, denoted by X Y and morphisms π X : X Y X and π Y : X Y Y such that whenever Z is an object with morphisms f X : Z X and f Y : Z Y, then there is a unique f : Z X Y such that f X = π X f and f Y = π Y f. Products in the categories Set, Top, Grp all consist of the usual products with the usual projection maps π X and π Y. Definition 1.5 Let X and Y be objects in the category C. A coproduct consists of an object, denoted by X Y and morphisms i X : X X Y and i Y : Y X Y such that whenever Z is an object with morphisms f X : X Z and f Y : Y Z, then there is a unique f : X Y Z such that f X = f i X and f Y = f i Y. Now suppose that C and D are categories and that F, G : C D are two functors. A natural transformation η from F to G consists of an assigment of some η X hom(f (X), G(X)) for each object X of C such that whenever f : X Y is a morphism in C, we have η Y F (f) = G(f) η X. If η X is an isomorphism for every X, we say that η is a natural equivalence.

8 1.2. TOPOLOGICAL PRELIMINARIES Topological Preliminaries We let I = [0, 1], and R denote the closed unit interval and the real line with the usual topology. Also define R n to be the Euclidean space with norm (x 1,, x n ) 2 = x x n 2 and distance d(x, y) = x y. Also, define the unit sphere S n = {x R n+1 : x = 1} and the unit ball B n+1 = {x R n+1 : x 1}. Note that S 0 = { 1, +1} R and B 1 = [ 1, 1] R. We will also and without comment consider R n R n+1 via the identification of x with (x, 0). In this way, for example, S n S n+1. Then each B n is compact and S n is compact and is connected if n 1. Two results we will use frequently are the following: Proposition 1.6 Suppose that X is a topological space and A 1,, A n are closed subspaces of X. Suppose that Y is a topological space and that f : X Y is a function such that f Ai is continuous on A i for each 1 i n. Then f is continuous. Proposition 1.7 Suppose that X is a compact space, Y is a Hausdorff space, and f : X Y is one-to-one and continuous. Then f is a homeomorphism. Suppose that X and Y are topological spaces. We define the product X Y and the disjoint union X Y by setting X Y = {(x, y) : x X, y Y } and X Y = (X {1}) (Y {2}). A basis for the product topology of X Y is given by the set B = {U V : U open in X, V open in Y} and the topology on X Y is given by all sets of the form (U {1}) (V {2}) where U is open in X and V is open in Y. There are continuous functions p X : X Y X, p Y : X Y Y, i X : X X Y, and i Y : Y X Y such that Proposition 1.8 Suppose that Z is a topological space and f : Z X, g : Z Y are continuous functions. Then there is a unique continuous function f g : Z X Y such that p X (f g) = f and p Y (f g) = g. In fact, (f g)(z) = (f(z), g(z)) for z Z. Proposition 1.9 Suppose that Z is a topological space and f : X Z and g : Y Z are continuous functions. Then there is a unique continuous function

9 6 CHAPTER 1. INTRODUCTION f g : X Y Z such that (f g) i X = f and (f g) i Y (f g)(x, 1) = f(x) and (f g)(y, 2) = g(y). = g. In fact, In other words, X Y and X Y are the product and co-product of X and Y in the category of topological spaces. We typically identify X with the subspace X {1} and Y with Y {2} in X Y without further comment. An important operation with topological spaces that we will need frequently is that of forming quotient spaces. Let X be a topological space and an equivalence relation on X. Let X/ denote the set of equivalence classes and let q : X X/ be the quotient map q(x) = [x]. We define U X/ to be open exactly when q 1 [U] is open in X. This defines a topology on X/ making q a continuous function. Furthermore, we have the following: Proposition 1.10 Let X be a topological space and an equivalence relation on X. Suppose that f : X Y is continuous such that x 1 x 2 implies that f(x 1 ) = f(x 2 ). Then, there is a unique ˆf : X/ Y which is continuous and satisfies f = ˆf q, where q : X X/ is the quotient map. An example of a quotient space is given by the projective spaces. For this, for x, y S n, define x y if and only if x = y or x = y. Let RP n = S n / for this equivalence relation. Notice that this space can also be obtained as B n by identifying opposite points on the boundary. A very important example is provided by the situation where A X and the equivalence classes of consist precisely of the set A and the singletons outside of A. In other words, we define x y if and only if x = y or x, y A. In this case, we write the quotient space as X/A and say that X/A is obtained by collapsing A to a point. We can simiarly define the operation of collapsing disjoint sets A 1,, A n to separate points. In this case, x y if and only if x = y or there is an 1 i n such that x, y A i. We denote this by X/{A 1,, A n }. As an example, S 1 is homeomorphic to I/{0, 1}. In general, S n is homeomorphic to B n /S n 1. For X a topological space, define the cone of X by CX = (X I)/(X {1}).

10 1.2. TOPOLOGICAL PRELIMINARIES 7 We notice that the map i(x) = [(x, 0)] is an embedding of X into CX. We typically identify the image of i with X. Similarly, we define the suspension SX by collapsing X {0} to one point and X {1} to another point in X I. So SX = X I/{X {0}, X {1}}. Because of this, SX = CX/X. For convenience, we will sometimes construct SX from X [ 1, 1] by identifying X { 1} to one point and X {1} to another. This will allow us to identify X with X {0}. For both CX and SX, we denote the equivalence class of (x, t) by [x, t]. Next, suppose that X and Y are topological spaces and suppose that A X and f : A Y is a continuous map. Define X f Y = (X Y )/ where we require a f(a) for a A and all other points are in their own equivalence classes. The main result, which is surprisingly tricky to prove is Theorem 1.11 Suppose that X and Y are T 4 spaces and A X is closed. Then, for f : A Y continuous, X f Y is T 4. Proof: For clarity, let i X : X X Y and i Y : Y X Y be the embeddings and let q : X Y X f Y be the quotient. Suppose that C and D are disjoint closed subsets of X f Y. Then, (q i Y ) 1 [C] and (q i Y ) 1 [D] are disjoint closed subsets of Y. By the T 4 property, there is a continuous g Y : Y I such that g Y (y) = 0 if y (q i Y ) 1 [C] and g Y (y) = 1 if y (q i Y ) 1 [D]. Now, let F = A (q i X ) 1 [C] (q i X ) 1 [D], so F is a closed subset of X. For x F, define g F (x) = g Y f(x) if x A, g F (x) = 0 if x (q i X ) 1 [C] and g F (x) = 1 if x (q i X ) 1 [D]. We need to show this definition is consistent. So, for example, suppose that x A (q i X ) 1 [C]. Then, q i X (x) = q i Y f(x) by nature of the equivalence relation defining q, so f(x) (q i Y ) 1 [C]. Hence it follows that g Y f(x) = 1 as required for consistency for the definition of g F. A similar treatment works for x A (q i X ) 1 [D]. Next, use Tietze s extension theorem to find g X : X I with g X (x) = g F (x) for x F. This gives maps g X and g Y which are continuous, so by the basic property of X Y, there is a function g : X Y I such that g i X = g X and g i Y = g Y.

11 8 CHAPTER 1. INTRODUCTION But now, if a A, we have g i X (a) = g X (a) = g F (a) = g Y f(a) = g i Y f(a), so we can factor g through q to find ĝ : X f Y I such that g = q ĝ. But now, since g X = 0 on (q i X ) 1 [C] and similarly for Y and for D, we see that ĝ = 0 on C and ĝ = 1 on D. Since points of X f Y are closed (why?), this shows that X Y is a T 4 space. As an example, let X be T 4, A X closed. If we let Y = {y 0 } and f(a) = y 0, then X f Y is the same as X/A. Hence, X/A is T 4. In the particular case where X = B n, A = S n 1 and f : S n 1 Y we say that X f Y is obtained from Y by attaching a n-cell via the boundary map f. This will be a very useful way to build up spaces in the future. More generally, suppose that f : X Y is a continous function. Then, recalling X CX, we can form the mapping cone of f as CX f Y and the mapping cylinder M f = (X I) f Y where we regard f as a map of X {0} to Y. Notice that X {1} is homeomorphic to X in M f. Proposition 1.12 Let i : A X be the inclusion map of A X. Then X/A is homeomorphic to CA i X/CA. Proof: Let q : CA i X CA i X/CA and p : X X/A be the quotient maps, the collapse points, and j : X CA i X the inclusion map. Then q j[a] =, so there is a k : X/A CA i X/CA with q j = k p. Also, we may define a map φ : CA X X/A by sending everything in CA to and X to X/A via p. This factors through CA i X and then through q to give g : CA i X/CA X/A which is an inverse to k. In the category of pointed topological spaces, the situation is a bit different. Instead of X Y, the coproduct of (X, x 0 ) and (Y, y 0 ) is obtained from the subspace X Y = (X {y 0 }) ({x 0 } Y ) of X Y. This can also be obtained by taking X Y and identifying x 0 with y 0. The maps i X : (X, x 0 ) (X Y, (x 0, y 0 )) and i Y : (Y, y 0 ) (X Y, (x 0, y 0 )) then have the required properties.

12 1.3. FUNCTION SPACES 9 Similarly, the product of (X, x 0 ) and (Y, y 0 ) is (X Y, (x 0, y 0 )) with usual projections. When disussing pointed spaces, we usually use p = (1, 0,, 0) and consider spaces (B n, p), (S n, p), etc. We default to (R n, 0). In this way, spaces such as S n S k are obtained. Another construction for pointed spaces is a bit more complicated. The smash product is defined as (X, x 0 ) (Y, y 0 ) = (X Y )/(X Y ) with the collapse point as the distinguished point. If q : X Y X Y is the quotient map, we write q(x, y) = [x, y] for convenience. Proposition 1.13 If X, Y, and Z are compact, then (X Y ) Z X (Y Z). Proof: In fact, both are homeomorphic to (X Y Z)/A where A = (X Y {z 0 }) (X {y 0 } Z) ({x 0 } Y Z). Proposition 1.14 We have (S 1 S n, ) is homeomorphic to (S n+1, p). Proof: For this, we identify S 1 as [ 1, 1]/{ 1, 1}. Define F : S n [ 1, 1] S n+1 by F (x, t) = p + ((1 t)x + tp) for t 0 and F (x, t) = p ((1 + t)x tp) for t 0. Here, p + (x) = (x, + 1 x 2 ) for x B n+1. Similarly for p. A consequence of this is that S n S m S n+m via induction. We also extend the suspension and cone constructions to pointed spaces by letting, for (X, x 0 ), C(X, x 0 ) = (X I)/(X {1}) ({x 0 } I) and S(X, x 0 ) = (X I)/(X {0}) (X {1}) ({x 0 } I). In both cases, we let the collase point be the special point. It should be notes that (SX, ) is then the same as (S 1 X, ). 1.3 Function Spaces Suppose that X and Y are topological spaces. We set Y X to be the set of all continuous f : X Y. We want to give the set Y X a topology reflecting the topologies of X and Y. While this can be done in a great deal of generality, we will content ourselves to the case where X is a compact space and Y is a

13 10 CHAPTER 1. INTRODUCTION metric space. If d is the metric on Y, we can give Y X the uniform metric by setting d u (f, g) = sup{d(f(x), g(x)) : x X}. Since X is compact and f, g are continuous, this supremum is finite and d u defines a metric, hence a topology, on Y X. In the more general situation, the so-called compact-open topology is used. This is equal to the uniform topology in our case. We then have the following. Proposition 1.15 The evaluation map e : Y X X Y defined by e(f, x) = f(x) is continuous. Proof: Let (f, x) Y X X and ε > 0. Choose open U X such that y U implies that d(f(x), f(y)) < ε/2. Then, if y U and d u (f, g) < ε/2 implies d(f(x), g(y)) d(f(x), f(y)) + d(f(y), g(y)) < ε/2 + d u (f, g) < ε. Proposition 1.16 If X and Y are compact and Z is metric, then (Z X ) Y is homeomorphic to Z X Y via a homeomorphsim defined by T (f)(x, y) = f(y)(x). Proof: We first have to show that the image of T is conained in Z X Y. So, suppose that f (Z X ) Y and let (x 0, y 0 ) X Y and ε > 0. Choose δ > 0 so that d(y, y 0 ) < δ implies that d u (f(y), f(y 0 )) < ε/2. Since f(y 0 ) is continuous, there is a η > 0 so that d(x, x 0 ) < η implies that d(f(y 0 )(x), f(y 0 )(x 0 )) < ε/2. Then d(y, y 0 ) < δ and d(x, x 0 ) < η implies that d(t (f)(x, y), T (f)(x 0, y 0 )) < ε. Next, we show that T is onto. So, suppose that g Z X Y. If we fix y Y, the map x g(x, y) is continous. So, letting f(y)(x) = g(x, y) defines a function f : Y Z X. We need to show this function is continuous. So let ε > 0 and y 0 Y. For each x X, g is continuous at (x, y 0 ), so there are open sets U x and V x in X and Y resp. such that (x, y) U V implies that d(g(x, y), g(x, y 0 )) < ε. But now, X is compact and is covered by the sets U x, so there are finitely many x 1,, x n so that U 1,, U n covers X. Let V = n k=1 V n. Then V is open, y 0 V and if x X, y V, we have that d(g(x, y), g(x, y 0 )) < ε. In other words, d(f(y), f(y 0 )) < ε. Now we have that f (Z X ) Y and clearly T (f) = g. Finally, we show that T is a homeomorphism. Actually, we have even more: T is an isometry. In fact, for f 1, f 2 (Z X ) Y, we have d(f 1, f 2 ) =

14 1.4. ALGEBRAIC PRELIMINARIES 11 sup{d(f 1 (y), f 2 (y)) : y Y } = sup{d(f 1 (y)(x), f 2 (y)(x) : y Y, x X} But this is equal to sup{d(t (f 1 )(x, y), T (f 2 )(x, y) : x X, y Y } = d(t (f 1 ), T (f 2 )). Proposition 1.17 We have that Z X Y is homeomorphic to Z X Z Y via the map where T (f) = (f i X, f i Y ). Proof: Clearly, the image of T is contained in Z X Z Y. By the coproduct property of X Y, T is onto. We identify (Y, y 0 ) (X,x0) with the subspace of Y X of functions with f(x 0 ) = y 0. What we find is that, instead of the product (X Y, (x 0, y 0 )) in the above results, we need the pair (X Y, X Y ). Since this is not a pointed space, we define the smash product X Y = (X Y )/(X Y ). Then (X Y, ) is the appropriate exponent in the pointed versions of the above results. 1.4 Algebraic Preliminaries All rings in this course will be abelian. In fact, the most common will be Z, and Z 2, although Q will occasionally appear. Central to this course is the notion of an R module and of a homomorphism between such modules. In particular, an R module is an abelian group M with a multiplication R M M satisfying 1 m = m, r(m + n) = rm + rn, and r(sm) = (rs)m. A module homomorphism is a homomorphism of abelian groups f : M N such that f(rm) = rf(m) for r R. In this case, ker f = {m M : f(m) = 0} is a submodule of M and im f = {f(m) : m M} is a submodule of N. We always have that im f is isomorphic to M/ ker f as R-modules. We say that a sequence of R modules and homomorphisms M f N g P is exact if im f = ker g. Extending this, a sequence M n f n Mn+1 is said to be exact if im f n = ker f n+1 for each n. Hence, 0 M f N 0 is exact if and only if f is an isomorphism. Similarly, 0 N M i P p 0 is exact if and only if i is one-to-one, p is onto, and im i = ker p. In this case, we have that P is isomorphic to N/f[M]. We say that thi sis a short exact sequence.

15 12 CHAPTER 1. INTRODUCTION If M and N are modules, we let M N = {(m, n) : m M, n N} with coordinatewise operations. There are inclusion maps i M : M M N, i M (m) = (m, 0), projection maps p M : M N M, p(m, n) = m and similar maps i N such that p N such that p M i M = 1 M, p N i N = 1 N, p N i M = 0 = p M i N, i M p M + i N p M = 1 M N. In this way, M N is both a product and a co-product in the category of R modules and homomorphisms. Proposition 1.18 Suppose that N and P are submodules of an R module M. Then M is isomorphic to N P via the map (n, p) n + p if and only if M = N + P and M P = 0. In this case, we say that M is the internal direct product of N and P. We can create a category of short exact sequences of R modules where the objects are short exact sequences 0 N M i P p 0 and a morphism from one short exact sequence 0 N 1 i 1 M1 p 1 P1 0 to another 0 N 2 i 2 M2 p 2 P2 0 is a triple of homomorphisms φ 1 : N 1 N 2, φ 2 : M 1 M 2 and φ 3 : P 1 P 2 such that i 2 φ 1 = φ 2 i 1 and p 2 φ 2 = φ 3 p 1 as in the following diagram. i 0 N 1 p 1 1 M1 P1 0 φ 1 φ 2 φ 3 i 0 N 2 p 2 2 M2 P2 0 Composition of morphisms is defined by composition at each stage. Similarly, we can define a category of long exact sequences. An important result is the following: Theorem 1.19 Suppose that 0 N M i P p 0 is a short exact sequence of R modules. Then the following are equivalent: a) 0 N i M p P 0 is equivalent to 0 N i N N P p P P 0 with identity maps for N and P. b) There is a map k : M N such that k i = 1 N, c) There is a map j : P M such that j p = 1 P.

16 1.4. ALGEBRAIC PRELIMINARIES 13 If any of these happen, we say the exact sequence splits. Notice that every abelian group can be given the structure of a Z module by setting nx = x + x + x (n times) for n > 0 and nx = ( n)( x) for n < 0. Also notice that the above definitions of exactness are good for groups that are not necessarily abelian. A very important result is the following: Theorem 1.20 (The 5 Lemma) Suppose we have a commutative diagram M 1 M 2 M 3 M 4 M 5 f 1 f 2 f 3 f 4 φ 1 φ 2 φ 3 φ 4 φ 5 N 1 N 2 N 3 N 4 N 5 g1 g2 g3 g4 with exact rows. Suppose that φ 1, φ 2, φ 4 and φ 5 are isomorphisms. Then φ 4 is an isomorphism also. The proof of this result is a perfect example of what is known as diagram chasing. We will chase many diagrams in this course. Exercises: 1. Show that a morphism f in a category is an isomorphism if and only if it is both a section and a retract. In particular, show that if f has a left inverse and a right inverse, they are equal and f is invertible. 2. Let C be a category and A 0 an object in C. Define F : C Set by setting F (A) = hom(a 0, A) and for f : A B a morphism in C, let F (f)(g) = f g where g hom(a 0, A) = F (A). covariant functor from C to Set. Show that F is a 3. Let C be a category and A 0 an object in C. Define F : C Set by setting F (A) = hom(a, A 0 ) and for f : A B a morphism in C, let F (f)(g) = g f where g hom(b, A 0 ) = F (B). contravariant functor from C to Set. Show that F is a 4. Let f : G H be a morphism in the category of groups and homomorphisms. Show that f is a section if and only if f is one-to-one and there exists a normal subgroup N of H such that H = f[g] N and f[g] N = e.

17 14 CHAPTER 1. INTRODUCTION 5. Suppose that f : X Y and g : Y Z are quotient maps. Show that g f : X Z is a quotient map. 6. Show that CX/X is homeomorphic to SX. 7. Show that CS n is homeomorphic to B n Show that B n+1 /S n is homeomorphic to S n Show that S(S n ) is homeomorphic to S n Show that SX and CX are connected for any space X. 11. Let S n = {(x 1,, x n+1 ) S n : x 1 0}. Show that S n /S n is homeomorphic to S n. 12. What are products and co-products in the category of topological pairs, Top 2? f g 13. Suppose that R is a field and 0 V 1 V 2 V 3 0 is an exact sequnece of finite dimensional vector spaces over R. Show that dim V 1 dim V 2 + dim V 3 = 0. f 1 f Suppose that R is a field and suppose that 0 V 1 V2 Vn 0 is an exact sequence of finite dimensional vector spaces over R. Show that n+1 i=1 ( 1)i dim V i = Suppose that N 1 f 1 M1 g 1 P1 and N 2 f 2 M2 g 2 P2 are exact. Show that f 1 f 2 g 1 g 2 N 1 N 2 M1 M 2 P1 P 2 is exact. 16. Show that X/{A, B} is homeomorphic to (X/A)/B. 17. Show that S n /S n 1 is homeomorphic to S n S n. 18. Show that S n /S 0 is homeomorphic to (S n 1 S 1 )/(S n 1 p). 19. Regard SX as a quotient of X [ 1, 1] and identify X with X 0. Show that SX/X is homeomorphic to SX SX.

18 Chapter 2 Homotopy Definition 2.1 Suppose that f, g : X Y are two continuous maps between topological spaces. We say that f and g are homotopic if there is a continous H : X I Y such that H(x, 0) = f(x) and H(x, 1) = g(x) for all x X. We write f g and say that H is a homotopy from f to g. If A X and f(a) = g(a) for all a A, we say that f and g are homotopic relative to A if there is a homotopy H from f to g such that H(a, t) = a for all a A and t I. In this case, we write f g(rel A). So homotopic maps are homotopic relative to the empty set. Similarly, if f, g : (X, A) (Y, B) are maps of pairs, we say f and g are homotopic if there is a H : (X I, A I) (Y, B) with H(x, 0) = f(x) and H(x, 1) = g(x) for all x X. Theorem 2.2 Homotopoy of maps is an equivalence relation on the space Y X. We denote the set of equivalence classes of such maps by [X : Y ]. In the case of pairs and pointed spaces, we use the notation [X, A : Y, B] instead of [(X, A) : (Y, B)]. Theorem 2.3 Suppose that f 1, f 2 : X Y are homotopic as are g 1, g 2 : Y Z. Then g 1 f 1 g 2 f 2. This allows us to defne a new category Hop whose objects are topological spaces and whose morphisms are equivalence classes of maps under homotopoy. 15

19 16 CHAPTER 2. HOMOTOPY In other words hom(x, Y ) = [X : Y ]. There is functor Top Hop that is the identity on objects and takes every continuous function to its homotopy equivalence class. Most functors of interest in algebraic topology factor through this forgetful functor. We also get a category of pairs in a similar way which we denote by Hop Basic Results Definition 2.4 We say that a subset A R n is convex if x, y A and t I implies that (1 t)x + ty A. Proposition 2.5 If Y R n is a convex set and f, g : X Y are two continuous functions, then f g. Furthermore, if f = g on A X, then f g (rel A). Lemma 2.6 Suppose that q : X Y is a quotient map and Z is locally compact. Then q 1 : X Z Y Z is a quotient map. Proof: Suppose that U Y Z is such that (p 1) 1 [U] is open in X Z. We need to show that U is open in Y Z. So, let (y 0, z 0 ) U and choose x 0 X with p(x 0 ) = y 0. Then (x 0, z 0 ) (p 1) 1 [U]. Hence, there are open sets V X and W Z such that (x 0, z 0 ) V W (p 1) 1 [U]. Since Z is locally compact, we may even obtain that V W (p 1) 1 [U] and W compact. Now, let V = {x X : {x} W (p 1) 1 [U]}. We claim that V is open. In fact, if x V, there is an open set J containing x such that J W (p 1) 1 [U] by compactness of W. Then, J V. But now, p(v ) satisfies p 1 p[v ] = V, so p[v ] is open in Y. Finally, (y 0, z 0 ) p[v ] W U. Hence, U is open, as required. Theorem 2.7 Suppose that f, g : (X, A) (Y, B) are homotopic as maps of pairs. Then the induced maps ˆf, ĝ : X/A Y/B are homotopic as maps of pointed spaces.

20 2.1. BASIC RESULTS 17 Proof: In fact, let H : (X I, A I) (Y, B) be a hommotopy from f to g. Compose H with the quotient map q : (Y, B) (Y/B, ). The resulting map factors through q 1, which is a quotient map by the lemma, so the induced map (X/A) I Y/B is continuous and is clearly a homotopy from ˆf to ĝ. Definition 2.8 We say that two spaces X and Y are homotopy equivalent if there is an isomorphism f : X Y in the category Hop. In other words, there exists g : Y X such that f g 1 Y and g f 1 X. We write X Y. Clearly, homeomorphic spaces are homotopy equivalent. The converse fails, though. For example, the space B n is homotopy equivalent to a single point. This follows since B n is convex, so the identity function and any constant function are homotopic. For another example, let X = R n \ {0} and Y = S n 1. Define f(x) = and g(x) = x. clearly, f g = 1 Y. But, H(x, t) = tx + (1 t) x x provide a homotopy from g f to 1 X. Here, the crucial point is that the image of H really is in X. Notice that B n \ 0 is homotopy equivalent to S n 1 in the same way. It is often not easy to determine whether two spaces are homotopy equivalent. x x Definition 2.9 We say that a topological space X is contractible if there is a x 0 X such that the inclusion map i : {x 0 } X is a homotopy equivalence. For a pair (X, x 0 ), we require that the inclusion ({x 0 }, x 0 ) into (X, x 0 ) be a homotopy equivalence. In each case, the only possible inverse map is the constant map c : X {x 0 }. But some care is required. In the case of spaces, the homotopy H : X I X need not satisfy H(x 0, t) = x 0 for all t, while for the map of pairs, it is required to do so. See comb space example. As an example, the cone space of X is always contractible since we may define H : (X I) I X I by H(x, t, s) = (x, (1 s)t + s). notice that H(x, 1, s) = (x, 1), so H factors through the quotient map to CX I. So, H ([x, t], s) = [x, (1 s)t + s] is then a homotopy from the identity map on CX

21 18 CHAPTER 2. HOMOTOPY to the constant map to the point [x, 1]. Similarly, the cone space of (X, x 0 ) is contractible to the collapse point. In particular, each B n is contractible. Definition 2.10 We say that a map f : X Y is null-homotopic if it is homotopic to a constant map. Proposition 2.11 A map f : X Y is null-homotopic if and only if it has an extension to some F : CX Y. Definition 2.12 We say that A X is a retract if there is a map r : X A such that r(a) = a for all a A. In other words, we require r i = 1 A. We say that A is a deformation retract if there is a retract r : X A which is a homotopy equivalence. In other words, we require that r(a) = a for all a A and that there is a homotopy H : X I X such that H(x, 0) = x, H(x, 1) = r(x). If the homotopy can be made relative to A, we say that A is a strong deformation retract. Definition 2.13 We say a T 4 space X is an absolute retract if whenever f : X Y is an embedding of X into a closed subspace of a T 4 space, then f[x] is a retract of Y. For example, S n 1 is a strong deformation retract of B n \ {0}. It is a later theorem that S n 1 is not a retract of B n. Proposition 2.14 I is an absolute retract. This follows from Tietze s Extension Theorem. We say that I is an absolute retract: it is retract in any T 4 space in which it is embedded. With a bit more work, we can replace I by I n in this result. It is a fundamental result that S n 1 is NOT a retract of B n. For n = 1, this follows from connectivity of B 1. Definition 2.15 We say that a closed subset A X satisfies the homotopy extension property if whenever f : X Y and H 0 : A I Y are maps with H 0 (a, 0) = f(a) for all a A, there exists a map H : X I Y such that H = H 0 on A I and H(x, 0) = f(x) for all x X.

22 2.1. BASIC RESULTS 19 Theorem 2.16 Suppose that (X, A) has the homotopy extension property and that A is contractible. Then the quotient q : X X/A is a homotopy equivalence. Proof: Let F : A I A be a homotopy from 1 A to the constant map at a 0 A. By the homotopy extension property, there is a map H : X I X such that H(x, 0) = x for all x and H(a, t) = F (a, t) for all a A, t I. How consider q H : X I X/A. This factors through the quotient map q 1 to give G : X/A I X/A with G (q 1) = q H. Also notice that H(a, 1) = a 0 for all a A, so there is a map g : X/A X such that g q(x) = H(x, 1) for all x X. Now, clearly, H is a homotopy from 1 X to g q. On the other hand G([x], 0) = G (q 1)(x, 0) = q H(x, 0) = q(x) = [x], and G([x], 1) = G (q 1)(x, 1) = q H(x, 1) = q g q(x) = q g([x]), so G is a homotopy from 1 X/A to q g. Proposition 2.17 The subset A X has the homotopy extension property if and only if (X 0) (A I) is a retract of X I. As an example, (B n, S n 1 ) has the homotopy extension property. Even more so, it turns out that (X {0}) (A I) is a strong deformation retract of X I. Theorem 2.18 Suppose that (X, A) has the homotopy extension property. Then (X {0}) (A I) is a strong deformation retract of X I. Proof: Let r : X I (X {0}) (A I) be a retract. In other words, r(x, 0) = (x, 0) for all x X and r(a, t) = (a, t) for all (a, t) A I. Define H : (X I) I X I by H(x, t, s) = (π 1 r(x, st), min{1, (1 s 2 )t+sπ 2 r(x, st)}). Then H is clearly continuous, H(x, t, 0) = (π 1 r(x, t) = (x, t), H(x, t, 1) = (π 1 r(x, t), π 2 r(x, t)) = r(x, t), so H is a homotopy from the identity map of X I to the retract r. Even more H(x, 0, s) = (π 1 r(x, 0), sπ 2 r(x, 0)) = (x, 0) and for a A, H(a, t, s) = (π 1 r(a, st), (1 s 2 )t + sπ 2 r(a, st)) = (a, (1 s 2 )t + s(st)) = (a, t), so H is a homotopy relative to X {0}) (A I).

23 20 CHAPTER 2. HOMOTOPY Theorem 2.19 Suppose that (X, A) has the homotopy extension property and that f : A Y is a map. Then (X f Y, Y ) has the homotopy extension property. Proof: let r : X I (X {0}) A I be a retraction. Let q : X Y X f Y be the quotient map. Then (q 1) (r 1) : (X Y ) I (X f Y ) I factors through (X f Y ) I to give the desired retract. Definition 2.20 We say that a closed A X is a neighborhood deformation retract (NDR) if there exists u : X I and H : X I X such that a) A = u 1 [0] b) H(a, t) = a for all a A, t I. c) H(x, 0) = x for all x X. d) if u(x) < 1, then H(x, 1) A. As an example, {0} is a NDR of I. Let u(x) = x/2 and H(x, t) = x(1 t). Proposition 2.21 If X is a metric space A is closed and U is open with A U, then there is a u : X I such that A = u 1 [0] and U = u 1 [0, 1). So, in metric spaces, for a closed set A X to be an NDR requires an open set A U and a homotopy H : X I X that is relative to A such that H(x, 1) A for all x U. Theorem 2.22 If A X is a NDR, then (X, A) satisfies the homotopy extension property. Proof: We give a retract of X I to (X 0) (A I). let u : X I and H : X I X be as given in the definition of NDR. Define ( ( ) ) t H x,, 0 0 t u(x) r(x, t) = u(x) (H(x, 1), t u(x)) u(x) t 1 Continuity of r follows from compactness of I and the constancy of H on A.

24 2.2. H-GROUPS AND H-COGROUPS 21 Proposition 2.23 If A is an NDR in X, then A Y is an NDR in X Y. Theorem 2.24 Let A X be closed and f : A Y a map. Then (X {1}) Y is an NDR in the mapping cylinder M f. Theorem 2.25 Let X be a topological space. Then X is an NDR in CX. Theorem 2.26 Suppose that (X, A) has the homotopy extension property. Suppose that f, g : A Y are homotopic maps. Then X f Y and X g Y are homotopy equivalent spaces. Proof: Let F : A I Y be a homotopy from f to g and consider (X I) F Y. The image of X {0} Y is homeomorphic to X f Y and the image of X {1} Y is homeomorphic to X g Y. But now, suppose that H : (X I) I X I is a homotopy of 1 X I to a retract onto A I X {0}. Let Y I Y be the projection. Then we obtain a map [(X I) Y ] I (X I) Y (X I) F Y. This factors through the quotient to give a homotopy [(X I) F Y ] I (X I) F Y from the identity map to a retract onto X f Y. Hence (X I) F Y is homotopy equivalent to X f Y. By symmetry, (X I) F Y is also homotopy equivalent to X g Y and we are finished. 2.2 H-groups and H-cogroups We now turn to the question of when the set [X, x 0 : Y, y 0 ] can be made into a group. One hint is the following characterization of groups: A group is a pointed set (G, e) with maps m : G G G and ν : G G such that m (1 m) = m (m 1), m (1 c e ) = 1 = m (c e 1), and m (1 ν) = m (ν 1) = c e. Here, c e : G G is the constant map. Definition 2.27 An H-group is a pointed toplogical space (H, e) and maps m : H H H and ν : H H such that m (1 m) m (m 1), m (1 c e ) 1 m (c e 1) and m (1 ν) c e m (ν 1).

25 22 CHAPTER 2. HOMOTOPY Proposition 2.28 Assume that (H, e) is an H-group. Define an operation on [(X, x 0 ) : (H, e)] by f g = m (f g). Then [(X, x 0 ), (H, e)] is a group under this operation. Furthermore, if k : (X, x 0 ) (Y, y 0 ), then ˆk(f) = k f defines a homomorphism of groups ˆf : [Y, H] [X, H]. Clearly any topological group is an H-group. The converse is far from being true. In fact, loop spaces are always H-groups as we now show. Definition 2.29 Let (X, x 0 ) be a pointed metric space. Define Ω(X, x 0 ) = (X, x 0 ) (S1,p) (X, x 0 ) (I,{0,1}). Let ω 0 (t) = x 0, so ω 0 is the constant loop, ω 0 ΩX. Now, suppose that p, q Ω(X, x 0 ) and regard them as maps p, q : I X. We define p(2t) 0 t 1 2 p q(t) = 1 q(2t 1) 2 t 1 Then p q ΩX and we obtain a continuous : ΩX ΩX ΩX. We show that this makes (ΩX, ω 0 ) into an H-group. In fact, for p ΩX, define i(p)(t) = p(1 t). Next, we note that (ΩX, ω 0 ) (Y,y0) ((X, x 0 ) (S1,p) ) (Y,y0) (X, x 0 ) (SY, ). This suggests that it might be simpler to work with maps out of SY rather than into ΩX. This in turn, leads us to the definition of an H-cogroup. Definition 2.30 An H-cogroup is a pointed topological space (H, e) and a comultiplication m : H H H and a co-inversion ν : H H such that (1, m) m (m, 1) m, (1, c e ) m 1 H (c e, 1) m, and (1, ν) m c e (ν, 1) m. Theorem 2.31 If (H, e) is an H-cogroup, then the set [H, e : X, x 0 ] can be given the structure of a group with operations f g = m (f g) and f 1 = f ν. Furthermore, if k : (X, x 0 ) (Y, y 0 ) is a map, then k (f) = k f defines a homomorphism of groups, k : [H, e : X, x 0 ] [H, e : Y, y 0 ]. Proposition 2.32 For any compact space (Y, y 0 ), the space (SY, ) is an H- cogroup.

26 2.2. H-GROUPS AND H-COGROUPS 23 In particular, for each n 1, the space (S n, ) S(S n 1, ) is an H- cogroup. Hence [(S n, p), (Y, y 0 )] is a group. We define the n-th homotopy group of (Y, y 0 ) by π n (Y, y 0 ) = [(S n, p) : (Y, y 0 )] What is more, we have π n (X, x 0 ) π n 1 (ΩX, ω 0 ). There is one special case which does not turn out to be a group and that is π 0 (X, x 0 ) = [S 0, p : X, x 0 ]. Here, S 0 = { 1, +1 = p} so a basepoint preserving map, f, can be identified with the point f( 1) X. A homotopy from f to g is then simply a map p : I X with p(0) = f( 1) and p(1) = g( 1). Such a map is called a pth from f( 1) to g( 1). Hence, π 0 (X, x 0 ) can be identified with the set of equivalence classes where two points are equivalent if there is a path between them. The equivalence classes are then called path components and the space X is said to be path connected if there is exactly one path component. Theorem 2.33 For a map f : (S n, p) (X, x 0 ), the following are equivalent: a) [f] = 0 in π n (X, x 0 ), b) f is null-homotopic, c) There is an extension of f to some F : (B n+1, p) (X, x 0 ) We can obtain another description of π n (X, x 0 ) by noting that B n is homeomorphic to I n and that (S n, p) (B n /S n 1, ) (I n / I n, ), so maps (S n, p) (X, x 0 ) are also represented by maps f : (I n, I n ) (X, x 0 ). With this, identification, the operation on π n (X, x 0 ) is given by f(x 1,, x n 1, 2x n ) 0 x n 1 2 f g(x 1,, x n ) = 1 g(x 1,, x n 1, 2x n 1) 2 x n 1 Theorem 2.34 For n 2, the group π n (X, x 0 ) is abelian. There is a more general construction for triples where we define π n (X, A, x 0 ) = [B n, S n 1, p : X, A, x 0 ]. There is a map B n B n B n where S n 1 S n 1 S n 1 for n 2 which makes π n (X, A, x 0 ) into a group. Proposition 2.35 Let f : (B n, S n 1, p) (X, A, x 0 ) be a map. The following are equivlent:

27 24 CHAPTER 2. HOMOTOPY a) f is null-homotopic b) [f] = 0 in π n (X, A, x 0 ) c) f is homotopic relative to S n 1 to a map into A. We notice that π n (X, x 0, x 0 ) = π n (X, x 0 ). From the inclusion maps i : (A, x 0 ) (X, x 0 ) and j : (X, x 0, x 0 ) (X, A, x 0 ) we get correspnding maps between homotopy groups. We also get a map : π n (X, A, x 0 ) π n 1 (A, x 0 ). Theorem 2.36 For a triple (X, A, x 0 ), we have a long exact sequence π n (A, x 0 ) i π n (X, x 0 ) j π n (X, A, x 0 ) π n 1 (A, x 0 ) Furthermore, this is natural in the sense that a map f : (X, A, x 0 ) (Y, B, y 0 ) induces a commutative diagram between the long exact sequences. So, the end result of our work so far is a collection of functors π n : phom Grp. In general, it is quite difficult to evaluate these functors, even at spheres. This is a major problem in the theory of homotopy groups. However, we will turn our attention to the simplest of these, π 1, which can be computed in many cases. Exercises: 1. Define f, g : S 1 S 1 by f(x) = x and g(x) = x. Show f g. 2. Let n be odd and define f, g : S n S n by f(x) = x and g(x) = x. Show that f g. 3. Suppose that f, g : S n S n and that for no x S n is it the case that f(x) = g(x). Show f g. 4. Show that X and X I are homotopy equivalent. 5. Show that (S 1 S 1 ) \ {(p, p)} is homotopy equivalent to S 1 S Suppose that A is a retract of the Hausdorff space X. Show that A is closed in X.

28 2.2. H-GROUPS AND H-COGROUPS Suppose that A X and f : A Y. Show that f has an extension to X if and only if Y is a retract of X f Y. 8. Show that a retract of a contractible space is contractible. 9. Show that a retract of a retract is a retract. 10. Show that Y is an absolute retract if and only if whenever A X is a closed subspace of a T 4 space and f : A Y is continuous, then f has an extension F : X Y. 11. Show that products of absolute retracts are absolute retracts. 12. Show that a retract of an absoute retract is an absolute retract. 13. Show that {0, 1} is an NDR of I. 14. Suppose that X = A B is the disjoint union of two closed sets. Show that (X, A) has the homotopy extension property. 15. Suppose that X = A B is a union of two closed subsets. Suppose that (A, A B) has the homotopy extension property. Show that (X, B) has the homotopy extension property. 16. Suppose that B A X and that (X, A) and (A, B) both have the homotopy extension property. Show that (X, B) has the homotopy extension property. 17. Suppose that (X, A) has the homotopy extension property. Show that (X Z, A Z) has the homotopy extension property. 18. Let f : X {x 0 } be the constant map. Show that M f is homeomorphic to CX, the cone of X. 19. Suppose that A U where A is closed and U is open. Suppose also that U B where B is closed. Suppose that (B, A (B \ U)) has the homotopy extension property. Show that (X, A) has the homotopy extension property. It is often possible to choose (B, A) to be homeomorphic to a mapping cylinder (M f, A) for some map f : B \ U A.

29 26 CHAPTER 2. HOMOTOPY 20. Suppose that A B X, that (X, A) has the homotopy extension property, and that B is and NDR in X and is contractible. Show that X/A is homotopy equivalent to X SA. Hint: consider the space X i CA. Then take the quotient with respect to two different contractible sets. 21. Show that S n /S k is homotopy equivalent to S n S k+1 for k n Let f : S 1 S 1 S 1 S 1 be the identity map on each component of the domain. Let X = ([ 1, 1] {0}) ({0} [0, 1]) R 2. Show that the mapping cylinder M f is homeomorphic to X S Let (X, x 0 ) be contractible. Show that [X, x 0 : Y, y 0 ] and [Y, y 0 : X, x 0 ] both have exactly one point. In particular, π n (X, x 0 ) = Show that π n (X Y, (x 0, y 0 )) is isomorphic to π n (X, x 0 ) π n (Y, y 0 ) via the map π n (p X ) π n (p Y ) where p X and p Y are the projections. 25. Show that π k (B n, S n 1, p) π k 1 (S n 1, p).

30 Chapter 3 The Fundamental Group 3.1 Free Groups and Free Products Definition 3.1 A monoid is a set with an associative operation with an identity. If M and N are monoids, a homomorphism is a function f : M N such that f(1) = 1 and f(xy) = f(x)f(y) for all x, y M. An equivalence relation on a monoid M is said to be multiplicative if a a and b b implies that ab a b. In this case, M/ becomes a monoid by defining [a]b] = [ab]. Let S be a set and let M(S) = {1} n=1s n. Define multiplication on M by juxtaposition, 1(x 1,, x n ) = (x 1,, x n ) = (x 1,, x n )1 and (x 1, x n )(y 1, y m ) = (x 1,, x n, y 1, y m ). Then M(S) is a monoid. Furthermore, if N is any other monoid and ϕ : S N is a function, there is a homomorphism ˆϕ : M(S) N defined by ˆϕ(x 1, x n ) = ϕ(x 1 ) ϕ(x n ). We call M(S) the free monoid generated by S. We want to find a group with a similar property concerning homomorphisms and maps from S. We want this group to be generated by S as a group, which means that it is generated by S S 1 as a monoid. So consider M(S S 1 ) and let be the smallest multiplicative equivalence relation such that ss 1 1 s 1 s for each s S. Then F G(S) = M(S S 1 )/ is a group since each [s] is invertible with inverse [s 1 ] and every element of F G(S) is a product of 27

31 28 CHAPTER 3. THE FUNDAMENTAL GROUP such elements. Theorem 3.2 Let S be a set. Then F G(S) is a group. Furthermore, if φ : S G is any function from S into a group, there is a unique homomorphism ˆϕ : F G(S) G such that ˆϕ i = ϕ where i : S F G(S) is the map i(s) = [s]. Definition 3.3 Let S be a set and let R F G(S). Let N R be the normal subgroup of F G(S) generated by R. We call the group F G(S)/N R the group generated by S with relations R and denote it by S R. In this group, all words in R are identified to the identity. Now, suppose that G and H are groups. Regard G and H as sets and consider the free group F G(G H). Let i G and i H be the inclusion functions into this free group. Let R consist all all words in F G(G H) of the form i G (g 1 )i G (g 2 )i G (g 3 ) where g 1 g 2 g 3 = e G in G as well as i H (h 1 )i H (h 2 )i H (h 3 ) where h 1 h 2 h 3 = e in H. Also, include in R the words i G (e G ) and i H (e H ). In other words, we identify those words in the free group to the identity that multiply to be the identity in their particular group. This will give the group G H =< G H R >, the amalgum of G and H. Note that there are functions i G : G G H and i H : H G H that factor through the free group. Proposition 3.4 The functions i G and i H are homomorphisms. Proof: First note that i G (e G ) = 1 and i H (e H ) = 1 since both are in R. Next, for g G, gg 1 e G = e G, so i G (g)i G (g 1 )i G (e G ) = 1. This shows that i G (g 1 ) = i G (g) 1. Now, let g = (g 1 g 2 ) 1. Then g 1 g 2 g = e G, so i G (g 1 )i G (g 2 )i G (g) = 1. But this shows that i G (g 1 )i G (g 2 ) = i G (g) 1 = i G (g 1 ) = i G (g 1 g 2 ). Thus, i G is a homomorphism. Similarly, i H is a homomorphism. Theorem 3.5 Suppose that K is a group and that ϕ G : G K and ϕ H : H K are homomorphisms. Then there is a unique homomorphism ϕ : G H K such that ϕ i G = ϕ G and ϕ i H = ϕ H. In other words, G H is a co-product in the category of groups.

32 3.2. THE FUNDAMENTAL GROUPOID 29 Proof: There is a homomorphism ˆϕ : F G(G H) K induced by ϕ G and ϕ H. Let N be the kernel of this map. Clearly, e G, e H N. Also, if g 1 g 2 g 3 = e G in G, we have that ϕ G (g 1 )ϕ G (g 2 )ϕ G (g 3 ) = e K since ϕ G is a homomorphism. Thus, ˆϕ(i G (g 1 )i G (g 2 )i G (g 3 )) = e K. This shows that i G (g 1 )i G (g 2 )i G (g 3 ) N. A similar consideration holds for relations in H. Thus, N N R where R is the collection of relations defining G H. This allows the factorization through G H and gives ϕ. Note that a similar construction holds for a collection {G α } α of groups to give a co-product α G α. 3.2 The Fundamental Groupoid Definition 3.6 A groupoid is a small category (i.e. the collection of objects is a set) such that every morphism is an isomorphism. As an example, let G be a group and let C be the category with one object such that hom(, ) = G with composition being the group multiplication. Conversely, if C is a groupoid and X is an object in C, then hom(x, X) is a group. Suppose that Y is another object in C and that f : X Y is a morphism. Then, the function F : hom(y, Y ) hom(x, X) defined by F (g) = f 1 g f is a homomorphism of groups whose inverse is F 1 (g) = f g f 1. Thus hom(x, X) and hom(y, Y ) are isomorphic as groups. Our primary example of a groupoid is the fundamental groupoid of a topological space π 1 (X). The objects of this category are the points of X and the morphisms from x X to y X are homotopy equivalence classes of maps (I, 0, 1) (X, x, y). The composition is then defined by p q = q p (reversal is required because of the way composition and path composition are defined). As above, we have that this composition is associative up to homotopy and the constant path at x is the identity at x. Furthermore, each morphism has an inverse defined by p 1 (t) = [(1 t). We notice that hom(x, x) = π 1 (X, x) as defined above. This leads immediately to the following result. Proposition 3.7 Let X be a topological space and x, y X. If there is a path

33 30 CHAPTER 3. THE FUNDAMENTAL GROUP from x to y, then π 1 (X, x) and π 1 (X, y) are isomorphic as groups. Definition 3.8 We say that a space is path connected if there is a path between any two points. Definition 3.9 We say that a topological space X is simply connected if X is path connected and π 1 (X, x) is trivial for some (hence all) x X. Proposition 3.10 A space is simply connected if and only if it is path connected and whenever p, q are paths with p(0) = q(0) and p(1) = q(1), then p q (rel {0, 1}). Proof: Let x = p(0) = q(0) and y = p(1) = q(1). Then p p c y p q 1 q c x q q, where we used that p q 1 is a loop at x, so is homotopy equivalent to c x. Proposition 3.11 Any contractible space is simply connected. Proof: This is slightly more subtle that it looks. Suppose that H : X I X is a homotopy from the identity map to the constant map at x X. If we knew that H was a homotopy that fixed x, we would have π 1 (X, x) = π 1 (x, x) is trivial. The problem is that no such homotopy may exist. So let p be a loop with p(0) = p(1) = x In I I, let a(t) = (t, 0), b(t) = (0, t), c(t) = (t, 1), and d(t) = (1, 1 t). Then a and b c d have common endpoints, so are homotpoic relative to {0, 1}. If we define G(s, t) = H(p(s), t) and q(t) = G(0, t) = H(x, t) = G(1, t), we then see that p = G a is homotopic to G (b c d) = q c x q 1 relative to {0, 1}. But now, q c x q 1 q q 1 c x relative to {0, 1}, so [p] is trivial in π 1 (X, x). Theorem 3.12 For n 2, S n is simply connected. Proof: Let p : I S n with p(0) = p(1) = P. If p fails to be onto, we may consider p : I S n \ {x} with the last space homeomorphic to R n, which is contactible. So, in this case, p is null-homotopic.