introduction to algebraic topology MARTIN CADEK 7. Products in cohomology An internal product in cohomology brings a further algebraic structure. The con-travariant functor H* becomes a cofunctor into graded rings. It enables us to obtain more information on topological spaces and homotopy classes of maps. In this section we will define an internal product - called the cup product and a closely related external product - called the cross product. 7.1. Cup product. Let J? be a commutative ring with a unite and let X be a space. For two cochains p> G Ck(X; R) and ip G Cl(X; R) we define their cup product for any singular simplex a : Ak+l —> X. The notation ){o) + {-lf{p U Sip)(a) = Sp(a/[v0,v1,.. . ,vk+1])ip(a/[vk+1,. . .,vk+i+1\) + (-l)kp(a[v0,v1,. .. , Vk\)öi/j(a/[vk,. . .,vk+i+1\) pU ifj E Ck+l(X;R) ((p U tp)(a) = p>(a/[v0, v1,...,vk\) ■ ip(a/[vk, vk+1,vk+l]) Lemma. fc+i E(-!)vwk,- • • •, vk+1])(ip(a/[vk+1,vk+l+1])) i=0 k+l+l ^(-ír^u^wh,... i=0 □ Lemma implies that l 2 (1) If ip and ip are cocycles, then ip U ip is a cocycle. (2) If one of the cochains ip and ^ is a coboundary, then Hk+l(X; R) by the prescription for cocycles ip and ■0. Since U is an J?-bilinear map on Hk(X; R) x Hl(X; R), it can be considered as an J?-linear map on the tensor product Hk(X; R) ®R Hl(X; R). Given a pair of spaces (X, A) we can define the cup product as a linear map U :Hk(X, A; R) ®R Hl(X; R) Hk+\X, A- R), U :Hk(X; R) ®R Hl(X, A; R) Hk+\X, A- R), U :Hk(X, A; R) ®R Hl(X, A; R) Hk+\X, A- R). Moreover, if A and B are open in X or A and B are subcomplexes of CW-complex X, one can define U : Hk(X, A; R) ®R Hl(X, B; R) Hk+\X, AUB;R). Exercise. Prove that the previous definitions of cup product for pairs of spaces are correct. For the last case you need Lemma 3.12. Remark. In the same way as the singular cohomology groups and the cup product have been defined using the singular chain complexes, we can introduce simplicial cohomology groups for A-complexes and a cup product in these groups. 7.2. Properties of the cup product are following: (1) The cup product is associative. (2) If X ^ 0, there is an element 1 G H°(X; R) such that for all a G Hk(X, A; R) lUa = aUl=a. (3) For all a G Hk(X, A; R) and j3 G Hl(X, A; R) aU/3 = (-l)^Ua, i. e. the cup product is graded commutative. (4) Naturality of the cup product. For every map / : (X, A) —> (Y, B) and any a G Hk(Y, B;R), (3 E Hl(Y, B; R) we have />U/3) = /»ur(/3). Remark. Properties (1) - (3) mean that H*(X,A;R) = @°10 H'l(X, A; R) with the cup product is not only a graded group but also a graded ring and that H*(X; R) is even a graded ring with a unit if X ^ 0. Property (4) says that / : (X, A) —> (Y, B) induces a ring homomorphism /* : H*(Y, B; R) —> H*(X, A; R). Proof. To prove properties (1), (2) and (4) is easy and left to the reader as an exercise. To prove property (3) is more difficult. We refer to [Hatcher], Theorem 3.14, pages 215 - 217 for geometrically motivated proof. Another approach is outlined later in 7.8. □ 7.3. Cross product. Consider spaces X and Y and projections p\ : X x Y —> X and p2 : X x Y —> Y. We will define the cross product or external product. The absolute and relative forms are the linear maps fj, : Hk(X, R) H\Y- R) Hk+\X xY;R), \i : Hk(X, A; R) Hl(Y, B; R) -> Hk+l(X x Y, A x Y U X x B; R) given by /i(a/3) =p*1(a)Up*2(/3). For the relative form of the cross product we suppose that A and B are open in X and Y, or that A and B are subcomplexes of X and F, respectively. (See the definition of the cup product.) The name cross product comes from the notation since fi(a <8> (3) is often written as a x (3. Exercise. Let A : X —> X x X be the diagonal A(x) = (x,x). Show that for a,/3e H*(X;R) aU/3 = A* (/i(a®/3)). 7.4. Tensor product of graded rings. Let A* = 0^=o An and B* = 0^=o 5n be graded rings. Then the tensor product of graded rings A* <8> B* is the graded ring C* = ®n=oCn where Cn = 0 A* £J' with the multiplication given by (ax 6x) • (a2 62) = (-l)|6lMa2|(ai • a2) (&x • 62). Here |&i| is the degree of b\ E B*, i.e. &i G -B'6lL If A* and B* are graded commutative, so is A* 5*. Lemma. The cross product \i : Hk(X, A; R) #'(y, B; R) -> x F, A x F U X x 5; i?) a homomorphism of graded rings. Proof. Using the definitions of the cup and cross products and their properties we have fi((a x b) ■ (c x d)) = (-l)|6Mc|/x((aUc) (6Ud)) = (-l)|6|'|c|pt(a U c) U p*2{b U d) = (-l)|bMc|pJ(a)UpJ(c)Up;(6)Up;(d) = p*(a) Up2(6) Up*(c) Up2(d) = fi(a (g) 6) U /i(c d). □ 4 7.5. Kiinneth formulas tell us how to compute the graded J?-modules H*(X x Y; R) or H*(X x Y; R) out of the graded modules H*(X] R) and H*(Y; R) or H*(X; R) and H*(Y; R), respectively. Under certain conditions it even determines the ring structure of H*(X x Y;R). Theorem (Kiinneth formula). Let (X, A) and (Y,B) be pairs of CW-complexes. Suppose that Hk(Y, B; R) are free finitely generated R-modules for all k. Then \i : H*(X, A; R) H*(Y, B; R) -> H*{X x Y, A x Y U X x B; R) is an isomorphism of graded rings. Example. H*(Sk x Sl) = Z[a,/3]/X where X is the ideal generated by elements a2, /32, a[3 = (—l)kl[3a and dega = k, deg/3 = /. Proof. Consider the diagram H*(X, A) ®R H*(Y)-- H*{X) ®R H*(Y) H*(A x Y) where the upper and the lower triangles come from the long exact sequences for pairs (X, A) and {X xY,AxY), respectively. The right rhomb commutes as a consequence of the naturality of the cross product. We prove that the left rhomb also commutes. Let ip and ip be cocycles in C*(A) and C*(Y), respectively. Let $ be a cocycle in C*(X) extending ip. Then p\§ U p*2tp E C*(X x Y) extends p\ip U p*2ip E C*(AxY). Using the definition of the connecting homomorphism in cohomology (see Remark 5.6 B) we get [ids* ® id)(M ® m) = As® ®i>]= pim u pm, 6*{fi([])) = 5*[p*lVup*2^} = [5(piup*M = pim uP*M. First, we prove the statement of Theorem for a finetedimensional CW-complex X and A = B = 0 using the induction by the dimension of X and Five Lemma. If dimX = 0, X is a finite discrete set and the statement of Theorem is true. Suppose that Theorem holds for spaces of dimension n — 1 or less. Let dimX = n. It suffices to show that fi : H*(Xn,Xn-1)(g)H*(Y) -> H*(Xn x Y.X^1 x Y) is an isomorphism and than to use Five Lemma in the diagram above with A = Xn~x to prove the statement for X = Xn. Xn/Xn~1 is homeomorphic to |Ji D™/ \_\i dD™. 5 To prove that /x: h* (V sj*) ® #*(y) -H-* (V s? x y) is an isomorphism, we use again the diagram above for X = |_|^ D™ and ^4 = |_|^ dD™ and the induction with respect to n. So we have proved the theorem for X a finite dimensional CW-complex and A = 5 = 0. Using once more our diagram and Five Lemma, we can easily prove Theorem for any pairs (X, A), (Y, 0) with X of finite dimension. For X of infinite dimension, we have to prove Hl(X) = Hl(Xn) for i < n which is equivalent to Hl(X/Xn) = 0. We omit the details and refer the reader to [Hatcher], pages 220 - 221. □ 7.6. Application of the cup product. In this paragraph we show how to use the cup product to prove that S2k is not an H-space. A space X is called an H-space if there is a map m : X x X —> X called a multiplication and an element e G X called a unit such that m(e, x) = m(x, e) = x for all x G X. Suppose that there is a multiplication m : S2k x S2k —> S2k with a unit e. According to Example after Theorem 7.5 H*{S2k x S2k-1) = Z[a,/3]/X where X is the ideal generated by relations a2 = 0, /32 = 0 and a/3 = /3a. The last relation is due to the fact that the dimension of the sphere is even. Moreover, a = 7 (g) 1 and (3 = 1 <8> 7 where 7 G H2k{S2k;'L) is a generator. Let us compute m* : H*(S2k;Z) -> iJ*(,S2fc x S2fc;Z). We have m*(7) = aa + 6/3, a, 6 G Z. Since the composition g2fc id xe g2fc x g2fc rn g2k is the identity, we get that a = 1. Similarly, 6=1. Now compute m*(72): 0 = m*(0) = m*(72) = (m*(7))2 = (a + /3)2 = 2a/3 ^ 0, a contradiction. Does this proof go through for odd dimensional spheres? 7.7. Kiinneth formula in homological algebra. Consider two chain complexes (C*, dc), 9d) of .R-modules. Suppose there is an integer N such that Cn = Dn = 0 for all n < N. Then their tensor product is the chain complex (C* <8> -D*, d) with (C»(8)D»)n= 0 q ® A and the boundary operator on C{ <8> -Dj «9(c (g> d) = 9cc (8) d + (-1)^ (8) dDd. It is easy to make sure that dd = 0. 6 Next we can define the graded J?-module * as (C**D*)n= 0 Torf(a,A')- i+j=n A ring R is called hereditary if any submodule of a free J?-module is free. Examples of hereditary rings are Z and all fields. Theorem (Algebraic Kiinneth formula). Let R be a hereditary ring and let C* and be chain complexes of R-modules. If C* is free, then the homology groups of C* <8> -D* are determined by the splitting short exact sequence 0 -> (H>(C) <8> H,(D))n 4 Hn(C* <8> D*) -> (H>(C) * #*(£>))„-i -> 0 where /([c] <8> [d]) = [c <8> d]. 27iis sequence is natural but the splitting is not. Notice that for the chain complex _ f 0 for n 7^ 0, n ~ |G for n = 0 the Kiineth formula gives the universal coefficient theorem for homology groups, see Theorem 6.11 B. The proof of the Kiinneth formula is similar to the proof of the universal coefficient theorem and we omit it. 7.8. Eilenberg-Zilbert Theorem. To be able to apply the previous Kiinneth formula in topology we have to show that the singular chain complex C*(X x Y) of a product X x Y is chain homotopy equivalent to the tensor product of the singular chain complexes C*(X) <8> C*(Y). Theorem (Eilenberg-Zilbert). Up to chain homotopy there are unique natural chain homomorphisms $ :C*{X) c*(y) -> C*(X x Y), :C*(X x Y) -> C*{X) <8> C*{Y) such that for 0-simplices a and r <&(ct r) = (a, r), ^(a, r) = a r. Moreover, such chain homomorphisms are chain homotopy equivances: there are natural chain homotopies such that ^$ ~ i&c«x®c«{Y), ^ ~ ida(xxy) • For the proof of this theorem see [Dold], IV. 12.1. The chain homomorphism ^ is called the Eilenberg-Zilbert homomorphism and denoted EZ. It enables a different and more abstract approach to the definitions of the cross and cup products. The cross product is /x([a] <8> [/3]) = [(a /3) o EZ] 7 for cocycles a G C*(X; R) and (3 G C*(Y; R) and the cup product is ([ip] <8> []) = [(v? 8) ) o EZ o A*] for cocycles Lp,ip G C*(X; J?) and the diagonal A : I 4 I x I. In our definition in 7.1 we have used for EZ o A* the homomorphism a ->• o-/[v0, vi,.. •, vk] a/[vfc,.. •, vn]. The properties of EZ can be used for a different proof of the graded commutativity of the cup product. 7.9. Kiinneth formulas in topology. The following statement is an immediate consequence of the previous paragraph. Theorem A (Kiinneth formula for homology). Let R be a hereditary ring. The homology groups of the product of two spaces X and Y are determined by the following splitting short exact sequence 0 -> (H>(X; R) <8> H*(Y; R))n \ Hn(X xY;R)^ (H>(X; R) * H*(Y; i?))n_i -> 0 where /([c] <8> [d]) = [c <8> d\. This sequence is natural but the splitting is not. For cohomology groups one can prove Theorem B (Kiinneth formula for cohomology groups). Let R be a hereditary ring. The cohomology groups of the product of two spaces X and Y are determined by the following splitting short exact sequence 0 (H*(X; R) H*(Y; R))n A Hn(X xY;R)^ (H*(X; R) * H*(Y; R))n+1 -+ 0. This sequence is natural but the splitting is not. For the proof and other forms of Kiinneth formulas see [Dold], Chapter VI, Theorem 12.16 or [Spanier], Chapter 5, Theorems 5.11. and 5.12. CZ.1.07/2.2.00/28.0041 Centrum interaktivních a multimediálních studijních opor pro inovaci výuky a efektivní učení ■f^GS ne® ^2^^fc I f00. "L„ 1-;-1 MINISTERSTVO ŠKOLSTVÍ, OPVaHMvánr »0^0 M fond V ČR EVROPSKÁ UNIE MLÁDEŽE A TĚLOVÝCHOVY prokonkurcmcnchopiiort ''-lNA»* INVESTICE DO ROZVOJE VZDĚLÁVÁNÍ