M8130 Algebraic topology, tutorial 06, 2017 30. 3. 2017 Exercise 1. Prove the following equalities (assuming some conditions): H>(\JXi) = ($H,(Xi) i=l i=l oo oo H,(\JXi) = ($H,(Xi) i=l i=l Solution. Denote z the distinguished point of X V Y. For the pair (X V Y, X) we have the following long exact sequence ----> Hi+1(X V Y,X) A Hi(X) -> Ht(X V Y) -> Ht(X V Y,X) A H^X) Thus we have short exact sequence, which splits, because we have continuous (cts) map idVconstz: XVY ^ X (which maps Y to z). Thus we have H^XVY) ^ H*(X)®H*(XV Y, X). Now it remains to prove H*(X V Y, X) = HAY). If X V Y is a CW-complex and X its subcomplex, it is known that H^X VF,I) = Hi(X V Y/X) = Hi(Y). More generally, let U be some (sufficiently small) neighborhood of z in X. From excision theorem we have: Hi(XVY,X) = Hi(XVY \ (X \ \ (X \ £/)) = Hi(UVY,U). Because U should be1 contractible, H^U V F, £/) = H^Y, z) = Hi(Y). The second equality we get from the first by induction. Let us prove the third equality. Denote Yn =I1VI2V---Vl„ and Y = V^Li Yn and denote z the distinguished point of Y and Yn for every n. We have the following diagram (where each arrow is an inclusion): Since Ak is compact, every continuous (cts) map Ak —> Y has image in some Yn, thus it is easy to prove C*(Y,z) = colim C*(Fn, z), thus Cif(Y1,z)^Cif(Y2,z)^--- C*(Y,z) H*(Y) = colim H>(Yn) = colim 0X, = □ i=l i=l 1It is true at least for X locally contractible. It is not true generally. M8130 Algebraic topology, tutorial 06, 2017 30. 3. 2017 Let X be a topological space with finitely generated homological groups and let H^X) = 0 for each sufficiently large i. Every finitely generated abelian group can be written as Z © Z © • • • © Z © Tor; where Tor denote torsion part of the group. The number k is k—times called the rank of the group. Euler characteristic x of X is defined by: x(X) = ^(-l)irank^(X) i=0 {Z % = n n Thus x(Sn) = 1 - (-l)n. 0, otherwise. Exercise 2. Let (C*,<9) be a chain complex with homology iJ*(C*). Prove that x{C*), where X (C„) = ^(-l)*rankQ. i=0 Solution. We have two short exact sequences: 0 -> Zi ^ d 4 -> 0 0^Bt^Zt^ Zt/Bt = Ht^0, where Ci, cycles Zt and boundaries Bt are free abelian groups, thus rank(7« = rankZj rank5j_i and rankiJj = rankZj — rankBi. Thus we have = ^(-iyrankZ, + ^(-l)*rank5,_1 j=0 i=0 OO OO = ^(-l^rankZi - ^(-1)* rank^ = X(X). □ j=0 i=0 Let X be a topological space with finitely generated homological groups and let H^X) = 0 for every sufficiently large i. Let f: X —>• X be a continuous map. Map f induces ho-momorphism on the chain complex /*: C*(X) —> C*(X) and on £/«e homologiy groups H*f: H*(X) —> H*(X), where H*f (Tor H*(X)) C ToriJ*(X). TTras i£ induces homomor-phism HJ: H*(X)/TorH*(X) ^ H*(X)/TorH*(X). Since H*(X)/Tor H*(X) = Z © Z © ■ ■ ■ © Z; mop H*f can be written as a matrix, thus we rankH»(X) can compute its trace. So we can define the Lefschetz number of a map f: L(f) = YJ(-iytrHJ. i=0 M8130 Algebraic topology, tutorial 06, 2017 30. 3. 2017 Similarly to the case of the Euler characteristic, it can be proved that1 oo oo X)(-l)ťtr^/ = X;(-l)ťtr/ť. Theorem. If L(f) ^ 0, then f has a fixed point. Exercise 3. Use the theorem above to show, that every cts map f on Dn and RPn where n is even has a fixed point. ' % ~ Because H0f: H0(Dn) = Z -> 0, otherwise. Z = H0(Dn) can be only the identity, we have L(f) = I, thus / has a fixed point. {Z, i = 0, Z/2, i M which satisfies X(t,x) = v(X(t,x)) for every x E M and X(0,x) = x. There exists t0 such that X(t0,x) ^ x. Denote f(x) = X(t0,x), thus / has no fixed point, thus L(f) = 0. Because / is homotopic to id and tiH{íá = rankiJj(M), we get from homotopy invariance 0 = L(f) = L(id) = x(M). □ Exercise 5. Use Z/2 coefficients to show, that every cts map f: Sn —> Sn satisfying f(—x) = —f(x) has an odd degree. Solution. The map / induces a map g: RPn —> RPn, since f({x, —x}) C {f(x), —f(x)}. We have the short exact sequence3 a i-o"i + o"2 i-2o" = 0 0 C*(MPn, Z/2) C*(Sn, Z/2) C*(MPn, Z/2) 0, where a: A'1 —> RPn is an arbitrary element of C*(RPn), a±,a2 are its preimages of a projection: Sn I A*--*■ RPn 2h: d(X)^Ci(X) 32cr = 0 because of the Z/2 coefficient. M8130 Algebraic topology, tutorial 06, 2017 30. 3. 2017 From the short exact sequence we get the long exact sequence Hi(RPn; Z/2)-*- Hi(Sn;Z/2)-*■ Hi(RPn; Z/2)-*■ Hi_1(RPn; Z/2)-*- 0 ■Hi(Sn; Z/2) Hi(RPn;Z/2) ■Hi_1(RPn;Z/2) 0 Hi(RPn;Z/2) — Because H0(RPn; Z/2) = Z/2 and g0 on -^o(^-Pn; ^/2) is an isomorphism, we can show by induction, that Hi(RPr\'L/2) = Z/2 and g,i is an isomorphism for every i < n — 1. An induction step is shown on the following diagram (three isomorphisms imply the fourth): 0- ■ Z/2 Z/2 0- l l = t ■Z/2- For i = n we have the following situation (the vertical isomorphisms were proved by induction): ■Z/2 Z/2 —Z/2 Z/2 —Z/2 ■ 0 Z/2- ■Z/2- ■Z/2- ■Z/2- 0 Thus /* (the arrow marked by ?) has to be an isomorphism for Hn, thus it maps [1]2 to [1]2, hence / has degree 1 mod 2. □