MAS435 Algebraic Topology Part B: Semester 2

Transcription

1 MAS435 Algebraic Topology Part B: Semester 2 Dr E Cheng School of Mathematics and Statistics, University of Sheffield e.cheng@sheffield.ac.uk Notes taken by Ale Corner Contents 1 Chain complees and homology Chain complees Homology Products of abelian groups Free abelian groups Maps between chain complees Low-dimensional cell-complees Low-dimensional eamples Simplices Singular homology and homotopy invariants Reduced homology Homotopy invariance Abelian groups and abelianisation Abelianisation Finitely-generated abelian groups What happened to van Kampen s Theorem? 15 6 Quotients and relative homology Mayer-Vietoris Sequence Aioms for homology 24

2 CONTENTS 2 8 Further remarks Moore spaces Wedge sums Suspension Introduction We have seen that Algebraic Topology is about studying topological spaces using algebra. In the first part of the course the algebra we used was group theory. We saw how to study a topological space X via its fundamental group π 1 X which helped us some features of spaces but not others. The main limitation was that the fundamental group is constructed from loops in a space, and cannot detect higher-dimensional features. The higher-dimensional versions of the fundamental group are constructed from higher-dimensional loops. That is, where a loop in X is a map a higher-dimensional loop is a map S 1 X S n X for higher values of n; these loops form a group called the nth homotopy group π n X. The trouble with this approach is that it is very difficult to compute. That is why we turn to homology instead. It is easier to compute than higher homotopy groups, but as a trade-off, it is less sensitive. In fact in this course we will not do a lot of higher-dimensional calculation. The idea of homology is to produce, from a space X, an Abelian group H n (X) called its nth homology group. In fact, we proceed in two steps: space chain comple homology groups A chain comple is a well-behaved sequence of abelian groups and homomorphisms, as we ll see. This means that homology can be used to study a wide range of things, not just spaces, as long as some sort of chain comple can be obtained from those things. So the first step above immediately gets us into algebra, and the second step is all within the world of algebra. The study of chain complees and their associated homology group is called homological

3 1 Chain complees and homology 3 algebra and can be thought of as a particularly structured part of the world of abelian groups. For the first step, it helps if we know how our how space is built up dimension by dimension from cells, that is, some form of disk/ball. Homology will then 1. find the places where we could have attached disks, and 2. compare it with the places where we actually did attach disks. Formally, this is done by taking a quotient group. This quotient group measures how holey the space is, because it measures the holes that were not filled in by disks. The slogan of homology is cycles mod boundaries. A cycle is the algebraic version of a hole. A boundary is a hole that we filled in with a disk it has now become the boundary of a disk. 1 Chain complees and homology We need to start by looking at the algebraic objects we ll be using: chain complees of abelian groups. All the groups in this part of the course are abelian, which makes a lot of things easier. 1.1 Chain complees Definition 1.1. A chain comple is a sequence of abelian groups and group homomorphisms δ n+2 δ n+1 δ n δ 2 δ 1 δ 0 C n+1 C n C n 1 C 2 C 1 C 0 0 such that for all n δ n δ n+1 = 0. Equivalently Imδ n+1 kerδ n, as subgroups. Eercise 1.2. Show that these conditions really are equivalent.

4 1.2 Homology 4 Terminology The maps δ are called boundary maps or differentials and a chain comple such as the one above may be labelled as C or sometimes just C. The elements of Imδ n+1 are then called boundaries and those of kerδ n are called cycles. Definition 1.3. If Imδ n+1 = kerδ n for all n then the chain comple is called an eact sequence. Another slogan is that homology measures the failure of a chain comple to be eact, as we will now see from the definition. 1.2 Homology Definition 1.4. The n th homology group of a chain comple C is defined to be the quotient group H n (C) = kerδ n / Imδ n+1. Eercise 1.5. What are two different ways of thinking of the elements of this group? Do we have to worry about whether or not this is a normal subgroup? Thinking of elements of H n (C) as equivalence classes of cycles, we call them homology classes. Two equivalent cycles are then called homologous; the condition for being homologous is as follows: f g f g Imδ n+1. Eercise 1.6. Find the homology groups of the following chain complees Z Z Z Z Z Z Z Z Z 0 0 δ n = Z Products of abelian groups We will be taking many products of abelian groups so it is important to be clear about how this works. The fact that the groups are abelian is crucial here non-abelian groups work quite differently. Abelian groups should be easier

5 1.4 Free abelian groups 5 Recall that given two abelian groups A and B we can form their direct product A B, elements of which are pairs (a,b) where a A and b B. This is again an abelian group, inheriting its group operation from A and B: (a,b)+(a,b ) = (a +a,b+b ). As A B is abelian here, the group operation will be written additively. Eercise 1.7. Show that this really is abelian. Similarly, recall the direct sum of A and B, the abelian group A B. This has elements a b where again a A and b B. The group operation is given by (a b)+(a b ) = (a+a ) (b+b ), again using the group operations of A and B. There is then an obvious isomorphism of abelian groups A B = A B. Eercise 1.8. Construct this isomorphism. You should construct the map, show it s a group homomorphism, and show it s an isomorphism. Remark 1.9. Note that A B is the coproduct of A and B in the category Ab of abelian groups and group homomorphisms. This is not the case for arbitrary groups in the category Gp of groups and group homomorphisms. In Gp the coproduct of groups G and H is the free product G H. However in both Ab and in Gp the product of two groups is the same, i.e. it is the direct product. 1.4 Free abelian groups The free abelian group on one generator is isomorphic to Z. For two generators it is isomorphic to Z Z. Carrying on in this way, for k generators, it is isomorphic to Z k. If we write the generators/basis elements as e 1,...,e k then a general element of Z k is Σ k i=1 λ ie i, where λ i Z. All of our chain complees will involve free abelian groups generated by cells of the space. Eample Consider the chain comple 0 δ 2 Z Z δ 1 Z δ 0 0 α a β a where δ 2 and δ 0 are the usual zero maps and δ 1 maps the generators of Z Z as shown. Notice that kerδ 0 = Z and Imδ 1 = Z, so H 0 = Z/Z = 0. Also notice that the kernel of δ 1 is generated by one element, α β, and that Imδ 2 = 0, thus H 1 = Z/0 = Z. The higher homology groups, for n 2, are then all the trivial group 0.

6 1.5 Maps between chain complees Maps between chain complees Just as there are maps between abelian groups, we can define maps between chain complees. Definition A chain map of chain complees f : A B is a collection of group homomorphisms f n : A n B n, for all n, such that the following diagram commutes. A n δ n A n 1 f n f n 1 B n ǫ n B n 1 Proposition A chain map as above induces a group homomorphism H n (f) : H n (A) H n (B), for all n. Proof. Given the chain map f, as above, we need to define a group homomorphism H n (f) : kerδ n / Imδ n+1 kerǫ n / Imǫ n+1. Given a kerδ n, is f n (a) kerǫ n? Yes, by commutativity of the following diagram. δ n+1 δ n A n+1 A n A n 1 f n+1 f n f n 1 B n+1 B n B ǫ n 1 n+1 ǫ n That is to say f n 1 δ n (a) = f n 1 (0) = 0 = ǫ n f n (a). So we have a map kerδ n kerǫ n. Now consider an element a Imδ n+1, so a = δ n+1 () say. We need that f n (a) Imǫ n+1. Howeverf n (a) = f n δ n+1 () = ǫ n+1 f n+1 () bycommutativity of the diagram. Thus f n (a) Imǫ n+1. Thus the map kerδ n kerǫ n restricts to a map Imδ n+1 Imǫ n+1 and so we can construct the map H n (A) H n (B), as required. Note that for each n, H n is a functor from the category ChCp of chain complees and chain maps into the category Ab of abelian groups and group homomorphisms. The above proposition gives us the action on morphisms. Eercise Check that this really is a functor.

7 2 Low-dimensional cell-complees 7 2 Low-dimensional cell-complees Now that we ve seen how to produce homology groups from a chain comple, we need to see how to produce a chain comple from a topological space. There are various different ways of doing this, like different recipes. They do not all produce the same answer, but the clever part is that the resulting homology groups are the same. This is a profound and amazing result that is beyond the scope of this course. One way of producing a chain comple from a space is to start with a cell comple structure on the space. We ve already seen that you can build the same space using different cell comple structures, but again, this doesn t matter because you ll get the same homology groups at the end, no matter what cell comple structure you choose. 2.1 Low-dimensional eamples Definition 2.1. Starting from a cell comple, we produce a chain comple C with: C n = the free abelian group generated by the n-cells of our cell comple, δ n = boundaries, considering orientation in a way that we ll see. Actually the definition of orientation is quite complicated in high dimensions, so we ll just do some low dimensional eamples. For 1-cells the boundary is head tail. For 2-cells the boundary is the sum of all the boundary 1-cells, taking orientation into account (a bit like when we did van Kampen s theorem before, and were reading off the boundary of a cell). So if a 1-cell is pointing backwards it gets a minus sign. Eample 2.2. A possible cell comple for the Torus is the following. b a α a b

8 2.1 Low-dimensional eamples 8 This has one 0-cell, so C 0 has one generator, two 1-cells a and b, so C 1 has two generators a,b, one 2-cell α, so C 2 has one generator α, and no higher dimensional cells than these, so C n is 0 for higher n. Recall that the free abelian group on one generator is isomorphic to Z, and with two generators it is isomorphic to Z Z. The boundary of both a and b is head tail i.e. = 0. The boundary of α is a+b a b = 0. The associated chain comple is then: 0 0 Z 0 Z Z 0 Z 0 0 α 0 a b 0 0 Eercise 2.3. Below are possible cell complees for the circle S 1, the sphere S 2, the Klein bottle and the real projective plane RP 2. Find the associated chain complees and work out the homology. a b a a α y a α a α a b a S 1 S 2 Klein bottle RP 2 Eercise 2.4. Try some different cell comple structures for the same spaces and make sure you get the same homology groups. This is a useful way of checking your answers. To provethat we get the same homologyfor different cell structures on the same space, it is easier to use singular homology, which is more general.

9 2.2 Simplices Simplices Simplices are topologically the same as discs and balls but are geometrically different and more combinatorial. They look more complicated than discs and balls, but the etra structure makes it easier 1 to define boundaries and orientation at higher dimensions. In practice we won t use this much in this course as we ll stick to lower dimensional things. 0-simple 1-simple 2-simple 3-simple The above eamples show that simplices are based around sticking together triangles. The 2-simple is homemorphic D 2 the disc whilst the 3-simple is homemorphic to D 3 the ball. The standard eamples of discs and balls are the unit discs and unit balls. For eample, D 2 = {(,y) R 2 2 +y 2 1}. There are also standard simplices which have a similar description. Definition 2.5. The standard n-simple n is a subset of R n+1 given by {(t 1,...,t n+1 ) R n+1 t i 0,Σ n+1 i=1 t i = 1}. 0 R 1 R 2 2 R 3 1 (0, 1) (0, 1, 0) (1, 0) (1, 0, 0) (0, 0, 1) Given any n+1 points in R m that do not lie in a hyperplane of dim < 2, these span a simple. That is, they are the vertices of an n-simple. To specify an n-simple we just give its vertices and, given an ordering on the vertices, there is a canonical linear homomorphism from any n-simple to any other. We can refer to an n-simple as [v 0,...,v n ], i.e. just its vertices. If we delete one of the v i s then we get n vertices spanning an (n 1)-simple, called a face. Thus an n-simple has n faces. Definition 2.6. A -comple on a space X is like a cell-comple structure but built from simplices instead of spheres and balls. See Section 2.1 in Hatcher. 1 Note that easier here means we don t have to introduce profound new technology. However it can still be fiddly.

10 2.2 Simplices 10 Eample 2.7. The simplices below represent the torus and the Klein bottle. b b a α c β a a α c γ a b Torus b Klein Bottle We can make chain complees from -complees, in an analagous fashion as with cell-complees. The C n are still the free abelian groups on n-cells. But what are the boundaries? δ n+1 δ n n+1(x) n(x) n 1(X) The boundaries here are the sum of the faces, taking into account the orientation. Definition 2.8. The i th face of an n-simple n is obtained by omitting the i th verte, denoted n 1 i. The boundaries use ( 1) i and the face inherits its ordering from the larger simple. Using the other notation, if we let n = [v 0,...,v n ], then the i th face is n 1 i = [v 0,...,v i 1,v i+1,...,v n ]. Sometimes this may be written as [v 0,...,v i 1, ˆv i,v i+1,...,v n ]. [1] [0,1] [1,2] [0,1, 2] [0,2] [0] [2] Definition 2.9. Let X be a topological space with a -comple structure on it. The simplicial boundary map δ n : n (X) n 1 (X) is a group homomorphism given by δ n (σ) = Σ n i=0 ( 1)i σ i where σ n (X) is an n-simple and σ i n 1 (X) is the i th face of σ.

11 3 Singular homology and homotopy invariants 11 3 Singular homology and homotopy invariants The idea of singular homology is that we don t have to start with a specific - comple structure. Recall that a path in a space X is a continuous map I X. A singular n-simple in X is then a continuous map n X where n is the standard n-simple. We can the form the singular chain comple for X where C n (X) is the free abelian group generated by all the singular n-simplices. The boundary map is the same as before and the singular homology of X is then the homology of the singular chain comple. NB. The 0-simple is a point, so a singular 0-simple in X is any point of X. Thus C 0 (X) is the free abelian group generated by all of the points in X. Eample 3.1. We will compute the singular homologyof a point. For all n 0 there is preciselyone map n { }. The boundarymap δ n : C n ( ) C n 1 ( ) is given as before. So an n-simple σ C n (X) is sent to Σ n 1 i=0 (i1)i σ i. This is 0 when n is odd and σ i when n is even. Thus we have a singular chain comple Z Z Z Z 0 and so the singular homology is given by H 0 ( ) = Z and Z/Z = 0 if n > 0, odd H n ( ) = 0/0 = 0 if n > 0, even 3.1 Reduced homology When we consider the homology of the point we still end up with the 0 th homology group being Z which of course just doesn t feel quite right. It would be nice if H 0 ( ) was the trivial group 0. We can achieve this by using an augmented chain comple which replaces the boundary map, for each space X, δ 0 : C 0 (X) 0, with a group homomorphism ǫ : C 0 (X) Z as in the chain comple below. δ 2 C 2(X) δ 1 C 1(X) δ 0 C 0(X) ǫ Z 0 0 Σ i(n iσ i) Σ in i The homology of this chain comple is called the reduced homology of X, written H n (X).

12 3.2 Homotopy invariance 12 Eercise 3.2. Check that this gives 0 for all the homology groups of the point. Proposition 3.3. For any space X, H 0 (X) = H n (X) Z. Equivalently, H 0 (X)/ H n (X) = Z. Proof. We will construct a surjective homomorphism H 0 (X) Z with kernel H n (X). (See the eercises.) Now H 0 (X) = kerδ 0 / Imδ 1 = C 0 (X)/ Imδ 1. We will use the map ǫ : C 0 (X) Z, though we need to check that Imδ 1 kerǫ. Let σ 1 C 1 (X) be the 1-simple generator. Then δ 1 (σ 1 ) = v 1 v 0 and so ǫδ 1 (σ 1 ) = ǫ(v 1 v 0 ) = 1 1 = 0, as required. Now we can use ǫ to induce the following map. C 0(X)/ Imδ 1 [nσ] ǫ Z n By the homework eercises, ker ǫ = kerǫ/ Imδ 1 = H 0 (X). So by the first isomorphism theorem for groups, we have Im ǫ = H 0 (X)/ker ǫ, i.e. Z = H 0 (X)/ H 0 (X). 3.2 Homotopy invariance Something else we would like from homology is that it be a homotopy invariant. That is, homotopy equivalent spaces should have isomorphic homology groups. More precisely, given a map of spaces f : X Y we get group homomorphisms H n f : H n (X) H n (Y). Then if f is a homotopy equivalence then H n f is a group isomorphism, for all n. Note that this is for singular homology. So far we have a way of taking a space X and creating a chain comple C(X). We also have a way of taking a chain comple C(X) and getting homology groups H n (X). The idea now is then to take the idea of homotopy through these processes. Top ChCp Ab f f H n f X α Y C(X) P C(Y) H n(x) H n(y) g g H n g homotopy chain homotopy equality

13 4 Abelian groups and abelianisation 13 There is one thing in this picture that we haven t come across yet. Definition 3.4. Let A and B be chain complees, with respective boundary maps δ n and ǫ n for all n, and let f,g : A B be chain maps. A chain homotopy P : f g is, for all n, a group homomorphism P n : A n B n+1 such that ǫ n+1 P n = f n g n P n 1 δ n for all n. This is sometimes written in the shorthand form δp = g f Pδ. δ n+1 A n+1 A n A n 1 δ n f n+1 g n+1 P n f n g n P n 1 f n 1 g n 1 B n+1 B n B n 1 ǫ n+1 ǫ n 4 Abelian groups and abelianisation From the eamples we have computed so far, it might be clear that when the first homology group H 1 (X) of a space is abelian then it is the same as the fundamental group π 1 (X). In fact, there is a way to take any group and abelianise it, i.e. make it abelian in the nicest possible way. We will see that H 1 (X) is always the abelianisation of π 1 (X). 4.1 Abelianisation Definition 4.1. Let G be a group, not necessarily abelian, and let a,b G. The commutator of a and b is defined to be [a,b] = aba 1 b 1. The commutator subgroup of G is defined to be [G,G] = [a,b] a,b G, i.e. the subgroup generated by all possible commutators. Eercise 4.2. For any group G, the commutator subgroup [G, G] is normal. Definition 4.3. Let G be a group, again not necessarily abelian. The abelianisation of G is defined to be G ab = G/[G,G]. Any group homomorphism G ab A, where A is abelian, corresponds eactly to a group homomorphism G A. That is to say there is a universal property at play. Equivalently, we can say that given any homomorphism f : G A, where A is abelian, there is a unique homomorphism f : G ab A which makes the following diagram commute.

14 4.2 Finitely-generated abelian groups 14 G G ab f! f A The map along the top of the triangle is the quotient map G G/[G,G] = G ab. An easy trap to fall into is to think that if G ab = H ab then G = H. A good countereample to keep in mind is that (Z Z) ab = Z Z = (Z Z) ab but certainly Z Z Z Z. To proceed we will need to recall some facts about finitely-generated abelian groups that makes them particularly easy to work with. 4.2 Finitely-generated abelian groups Definition 4.4. An abelian group A is said to be finitely generated if there eist finitely many elements a 1,...,a k A such that every elements a A can be written as a = n 1 a n k a k where each n i Z. The set {a 1,...,a k } is then said to be a generating set for A. Remark 4.5. Note that we are finitely generating A as a Z-module and this epression is not necessarily unique. It is unique if A is a free group on the generators {a 1,...,a k }. In fact, every group can be epressed in terms of (maybe infinitely many) generators and relations. There is a problem, called the word problem for groups, which states that, given generators and relations for a group, there is no systematic way to tell if two different representations of elements are the same. The following theorem is sometimes known as the fundamental theorem of finitely-generated abelian groups. Theorem 4.6. Every finitely generated abelian group is isomorphic to Z n Z q1 Z qk, where n 0 and each q i is a power of a (not necessarily unique) prime and Z q denotes the integers mod q.

15 5 What happened to van Kampen s Theorem? 15 Eample 4.7. Z 6 = Z 2 Z 3 If n has a prime factorisation as p k1 1 pk2 2...pkm m, then Z n = Z k1 p 1 Z k2 p 2... Z km p m. So abelian groups are much easier to deal with as we are able to classify the finitely generated ones. Remark 4.8. If you re given a (non-abelian) group in terms of generators and relations, you can often work out what its abelianisation is by just looking at the relations and collapsing them using commutativity. Because the idea is, informally, that the abelianisation makes everything commute. Eample 4.9 (Fundamental group of surfaces). Let X be an orientable surface of genus g. Then π 1 (X) = a 1,b 1,...,a g,b g a 1 b 1 a 1 1 b a g b g a 1 g b 1 g. Informally, the relation just vanishes if everything commutes, as everything will cancel out with its inverse. So we releft with all the generatorsand no relations. Thus (π 1 (X)) ab = Z 2g = H 1 (X). Similarly, let Y be a non-orientable surface of genus g. Then Thus π 1 (Y) = a 1,...,a g a 2 1 a2 2...a2 g. (π 1 (Y)) ab = a 1,...,a g 2a a g = a 1,...,a g 1,a a g 2a a g = Z g 1 Z/2 = H 1 (Y). 5 What happened to van Kampen s Theorem? We used van Kampen s theorem a lot when working out fundamental groups. It primarily involved two things: disjoint unions and quotients. We already know how to deal with disjoint unions, we just take direct sums: H n (X Y) = H n (X) H n (Y). However, we re not sure what should be true about quotients. Given a space X with a subspace A we might like to have that: H n (X/A) = H n (X)/ H n (A).

16 5 What happened to van Kampen s Theorem? 16 Fortunately this is not true. If it were, we would have a definite problem. Given any space A we can make the cone on A: X = A I/A { }. Now this is contractible and so H n (X) = 0. So if H n (A) H n (X), then we have H n (A) = 0 for all n. Oops. Instead, if A includes into X nicely, e.g. if everything is a CW-comple then we get the following. 1. There is a relationship between H n (X), H n (A) and H n (X/A) via an eact sequence involving all dimensions of homology at the same time. 2. There is a relationship between H n (X,A) and relative homology, i.e. the homology of the chain comple C(X)/C(A) involving the groups C n (X)/C n (A). The first point means that we can do calculations inductively, or implicitly. The second means that we can sometimes do direct calculations in small eamples. The aim is that given CW-complees i : A X and j : X X/A then we can get an eact sequence: i H n(a) H n(x) H n(x/a) β j i H n 1(A) H n 1(X) H n 1(X/A) β j H 0(X/A) 0 Note that if j were surjective then the β would be zero maps. This long eact sequencemeasuresthefailureofthej tobe surjective. Theslightlycomplicated part here is figuring out what the β should be these maps are sometimes called the Bockstein homomorphisms. We first claim that we can make a chain comple of the following form: δ n+1 δn C n+1(x)/c n+1(a) C n(x)/c n(a) C n 1(X)/C n 1(A)

17 5 What happened to van Kampen s Theorem? 17 This works since δ n (C n (A)) C n 1 (A). This is called the relative chain comple, the elements of C n (X)/C n (A) are called the relative chains and the homology is called the relative homology, H n (X,A). Our second claim is, for all n, H n (X,A) = H n (X/A). Note that we don t have to reduce relative homology, it happens in the process. (Think about what happens for the point.) Eample 5.1. Let X = I and A = {0,1}. Then X/A = S 1 and, for all n 0, H n (X,A) = H n (S 1 ). Now our third claim is that we have a short eact sequence of C n s: 0 C n(a) C n(x) C n(x)/c n(a) 0 This unravels to a long eact sequence of homology which includes the relative homology groups and the Bockstein homomorphisms. H n(a) H n(x) H n(x,a) β H n 1(A) H n 1(X) H n 1(X,A) Then in nice spaces this gives us the long eact sequence for quotients. In fact we also have a short eact sequence of chain complees which we will use after proving the following lemma. Lemma 5.2 (Snake Lemma). Consider the following diagram of abelian groups and group homomorphisms, for which the central diagram is commutative and the middle two rows are eact.

18 5 What happened to van Kampen s Theorem? 18 kerκ kerφ kerψ β f g A B C 0 κ φ ψ 0 A B C f g A / Imκ B / Imφ C / Imψ Then there is an eact sequence β kerκ kerφ kerψ A / Imκ B / Imφ C / Imψ. Proof. We will construct the map and show that it is well-defined. Eactness of the sequence is left as an eercise. First we construct the homomorphism. Take an element c kerψ C. As g is surjective, then c lifts to some element b in B, i.e. c = g(b). By commutativity of the diagram, g φ(b) = ψg(b) = ψ(c) = 0. Thus φ(b) kerg = Imf by eactness and as f is injective, there is a unique element a A such that f (a ) = φ(b). Thus we define our map as β(c) = a + Imκ = [a ]. In defining the map we had to choose a lift of the element c kerψ to an element b B. To check that β is well-defined we must show that given two different lifts to B that these give the same result after applying β. So suppose that c lifts to b and b, i.e. g(b) = c = g(b ). Thus b b kerg = Imf by eactness and so b b = f(ā) for some ā A. By commutativity and eactness f κ(ā) = φf(ā) = φ(b b ) = φ(b) φ(b ) = f (a) f (a ) = f (a a ). As f is injective we then have that κ(ā) = a a and so [a a ] = [κ(ā)] = [0], thus [a] = [a ] and so β is well-defined. The short eact sequence of chain complees is shown in the following diagram, where C n = B n /A n, the i s are inclusions and the j s are quotients.

19 5 What happened to van Kampen s Theorem? δ A δ A A n+1 A n A n 1 i i i δ δ B n+1 B n B n 1 j j j δ δ C n+1 C n C n Now that we have the snake lemma and the short eact sequence of chain complees we can construct the connecting homomorphism on homology. Recall that and β : H n (X,A) H n 1 A H n (X,A) = ker δ n / Im δ n+1 H n 1 A = ker(δ n 1 A )/ Im(δ n A ). For clarity in the following eplanation we will define, for any group homomorphism f : A B, the cokernel of f as cokerf = B/ Imf. We can apply the snake lemma to the following diagram. ker(δ n A) kerδ n ker δ n i j 0 C na C nx C nx/c na 0 δ n A δ n δn 0 C n 1A C n 1X C n 1X/C n 1A 0 i j coker(δ n A) cokerδ n coker δ n

20 6 Quotients and relative homology 20 This gives a homomorphism β : ker δ n coker(δ n A ) = C n 1 A/ Im(δ n A ). Now recall the way in which we define the map β. We start with an element ker δ n and lift it to ˆ C n X. This is sent to C n 1 X via δ n and is uniquely lifted to δ nˆ C n 1 A. Now δ n 1 A δ nˆ = δ n 1 δ nˆ = 0 and so δ nˆ kerδ n 1 A. Thus β is in fact a map ker δ n kerδ n 1 A / Imδ n A = H n 1 A. Now as Im δ n+1 ker δ n then we can form a new group homomorphism β : ker δ n / Im δ n+1 = H n (X,A) H n 1 A. 6 Quotients and relative homology In this section we will look more closely at the relationship between relative homology and the homology of quotient spaces. Theorem 6.1. Given CW-complees A X, then 1. The quotient maps X X/A and A A/A = { } induce an isomorphism H n (X,A) H n (X/A,A/A) = H n (X/A, ) on homology groups. 2. In general, H n (Y, ) = H n (Y) if Y is a CW-comple. 3. There is an isomorphism H n (X,A) H n (X/A). Eample 6.2. Let X = I and let A be the disjoint union of four points. To use the previous results we must epress X as a cell comple in such a way that A is a subcomple. 1 a 2 b 3 c 4 The chain complees for X and A, with the obviousmaps, as well as the relative chain comple are then as follows. C(X) 0 Z 3 Z 4 0 C(A) 0 0 Z 4 0 C(X,A) 0 Z The relative homology groups can then be worked out to be Z 3 if n = 1 H n (X,A) = 0 if n 1

21 6 Quotients and relative homology 21 The quotient space X/A is the wedge sum of three circles S 1 S 1 S 1 which has the following augmented chain comple. δ 1 ǫ 0 Z 3 Z Z 0 The boundary map δ 1 is a zero map, so the reduced homology of X/A is then the same as the relative homology, for all n. Eample 6.3. LetX = D 2 andletabetheboundarys 1. Thechaincomplees for X and A, with the obvious maps, as well as the relative chain comple are as follows. C(X) 0 Z Z Z 0 C(A) 0 0 Z Z 0 C(X, A) 0 Z The relative homology groups can then be worked out to be: Z if n = 2 H n (X,A) = 0 if n 2 The quotient space X/A is the sphere S 2, with the following augmented chain comple. δ 2 δ 1 ǫ 0 Z 0 Z Z 0 The reduced homology groups of X/A = S 2 are then the same as those of the relative homology groups, for all n. Eample 6.4. It was mentioned above that we can use this relative homology for working out quotients only when the spaces are somehow nice enough. The Hawaiian earring (HE) immediately springs to mind as a non-nice space. The Hawaiian earring is not a CW-comple. Compare it with the infinite wedge of circles S 1 S 1... S 1... that all have the same radius and which is a CW-comple. When we build a CW-comple X from cells d α, the topology on X has to satisfy U X is open if and only if U d α is open for all α. (See Hatcher.) Think about the fundamental groupπ 1 (HE). Aloopon the HE cangoround infinitely

22 6 Quotients and relative homology 22 many different circles. Thus π 1 (HE) contains Z N as a proper subgroup as we are then allowed infinitely long words in the generators. A way to build the Hawaiian earring is to start with the interval, so let X = I, and take the subspace of points A = { 1 n n N {0}}. The quotient space is then just X/A. (Draw it!) The chain complees for X and A, as well as the relative chain comple are as follows. C(X) 0 Z N Z N 0 C(A) 0 0 Z N 0 C(X,A) 0 Z N 0 0 Thus the relative homology groups are given by Z N if n = 1 H n (X,A) = 0 if n 1 However we know that Z N is a proper subgroup of π 1 (HE) = H 1 (HE) ab and so H 1 (X,A) H 1 (X/A). As relative homology corresponds, in the right cases, to reduced homology of quotients then it makes sense to have the rest of the long eact sequence to also be reduced. We do the same process as we did above with the short eact sequence of chain complees but we now use the short eact sequence of augmented chain complees. Eample 6.5 (Homology of spheres). The circle, or 1-sphere, S 1 has trivial homology ecept for H 1 which is Z. Likewise the 2-sphere S 2 has trivial homology ecept for H 2 which is Z. Intuitively we d then think that S k has trivial homology ecept for having H k (S k ) = Z. Let X = D k and A = S k 1, so that X/A = S k. The long eact sequence of reduced homology for these spaces is then as follows. Hn(S k 1 ) Hn(D k ) Hn(D k,s k 1 ) β H n 1(S k 1 ) Hn 1(D k ) H0(D k,s k 1 ) 0

23 6.1 Mayer-Vietoris Sequence 23 We know that D k is contractible and so H n (D k ) = 0 for all n. Placing these zeros into the long eact sequence we see that we have many instances of short eact sequences which in turn means that the Bockstein homomorphisms, β, are all isomorphism. I.e. Hn (S k ) = H n (D k,s k 1 ) = H n 1 (S k 1) for all n > 0. Also note that S 0 is a pair of points but otherwise S k is path connected for k > 0. Thus the 0 th homology group of the spheres is given by H 0 (S k Z if k = 0 ) 0 if k 0 The remaining piece to find is the homology groups of S 0 which are plainly given by Z if n = 0, H n (S 0 ) = 0 if n 0. Proceeding by induction we can then prove that H n (S k Z if n = k, ) = 0 if n k. 6.1 Mayer-Vietoris Sequence This is our homological analogue to van Kampen s theorem. Take a space X and find subspaces A and B, with respective inclusions i and j, such that X is covered by the interiors of A and B. (Think that X = A B.) We get a short eact sequence of chain complees of the following form. 0 C n(a B) φ ψ C n(a) C n(b) C n(a+b) 0 i() j() a a b b The group C n (A+B) consists of cells of X which are completely contained in either A or B. The key is that the homology of the chain comple C(A+B) is that of the chain comple C(X) corresponding to X. That is to say, the inclusion C n (A+B) C n (X) induces an isomorphism on homology. It is easy to check the eactness of the sequence. First, ψ is certainly surjective since C n (A+B) is generated by simplices a entirely in A and b entirely in B. We have ψ(a) = a and ψ(b) = b. Second, φ is injective since φ() = (, ), so φ() = (0,0) implies that = 0. It remains to show that kerψ = Imφ.

24 7 Aioms for homology 24 First, ψ(φ()) = (ψ(, )) = = 0, so Imφ kerψ. Now consider (a,b) kerψ, i.e. a+b = 0in C n (A+B). Then a = b, so(a,b) = φ(a) Imφ. Thus the sequence is eact. Using the same techniques as before, this unravels to a long eact sequence of homology. H n(a B) H n(a) H n(b) H n(x) β H n 1(A B) H n 1(A) H n 1(B) H 0(A) H 0(B) H 0(X) 0 Eercise 6.6. Let X = S n and take A and B to be the north and south hemispheres, D n, of X. Use the Mayer-Vietoris sequence to work out the homology of X. 7 Aioms for homology We will now look at the general idea of homology and what we epect it to do. Formally, homology is a functor, for all n, from the category of pairs of spaces (with appropriate maps) to Ab taking a pair (X,A) of spaces to a homology group H n (X,A). This should be homotopy invariant together with, for all n, a map δ n : H n (X,A) H n 1 (A, ) such that for all maps f : (X,A) (Y,B), the following square commutes. H n(x,a) H n f H n(y,b) (δ n ) (X,A) (δ n ) (Y,B) H n 1(A, ) H n 1 f H n 1(B, ) This in fact constitutes what is called a natural transformation. All of the above is also required to satisfy the following aioms.

25 8 Further remarks 25 Aiom 7.1 (Dimension). If X is a point, then H 0 (X) = Z and is the trivial group otherwise. (In fact we could create a homology theory using any G, not just Z.) Aiom 7.2 (Eactness). The following sequence is eact: δ H n(a) H n(x) H n(x,a) H n 1(A) Aiom 7.3 (Ecision). If X is covered by the interiors of A and B then the inclusion(b,a B) (X,A)inducesanisomorphismH n (B,A B) H n (X,A). Aiom 7.4 (Additivity). If (X,A) = i (X i,a i ) then there are inclusions (X i,a i ) (X,A) inducing an isomorphism i H n (X i,a i ) H n (X,A). NB. There would be a fifth aiom (Homotopy) if we weren t restricting to nice spaces. The five aioms are collectively known as the Eilenberg-Steenrod aioms for homology theories. Theorem 7.5. Homology theories are unique. 8 Further remarks 8.1 Moore spaces Given an abelian group G and an integer n 1 we can make a CW-comple X such that G if i = n, H i (X) = 0 if i n. This is called the Moore space M(G,n). The version for homotopy groups is denoted K(G, n), the Eilenberg-Mac Lane spaces. 8.2 Wedge sums Consider a family of based spaces (CW-complees) (X α, α ) and consider the wedge sum of these spaces, α (X α, α ). Theorem 8.1. Let (X α, α ) be a family of based topological spaces, as above. Then for all n, H n ( α (X α, α )) = α Hn ((X α, α )).

26 8.3 Suspension 26 Proof. Put X = α X α and A = α { α }. Then X/A = α (X α, α ). We know that H n (X/A) = H n (X,A), so as required. H n (X,A) = α H n (X α,{ α }) = α Hn (X α ), 8.3 Suspension Given a space X we can form the suspension of X, denoted SX, given by the quotient SX = (X I)/ where the equivalence relation is the identification ( 1,0) ( 2,0) and ( 1,1) ( 2,1). If we take X = S 1 we can picture this attaching a cone above and below S 1. The homology of SX then moves up one dimension. I.e. for all n, Hn (X) = H n+1 (SX). Thus we can make a space Y with H 1 (Y) = Z/k, H2 (SY) = Z/k, H 3 (SSY) = Z/k, etc. GivengroupsG 0,G 1,G 2,..., wecanmakeacw-comple X with H i (X) = G i, for all i. Then we can make a space X = i M(G i,i). Here we have controlled the homology of a space at every dimension. However, this is far from classifying homotopy types. If a given group G = Z m Z/p k Z p, then m is called the kn nth Betti n number and the p ki i are the torsion coefficients.