M8130 Algebraic topology, tutorial 06, 2024
31. 10. 2024
Exercise 1. 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)j. We have the short exact sequence1
a h
■ 0"! + 0"2 H
2a = 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*(BLPn), a±,a2 are its preimages of a projection:
Sn ' 1
A* -—RPn 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-1/2) —
Because H0(RPn; Z/2) = Z/2 and g0 on iJ0(^-Pn; Z/2) is an isomorphism, we can show by induction, that Hi(RPn,Z/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-I
t
•Z/2-
For i = n we have the following situation (the vertical isomorphisms were proved by induction):
0-
■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. □
Exercise 2. Let <p E Ck(X;R),tp e Cl(Y;R). Prove 5(<p U tp) = 5p U + (-l)Vu^. Use r = [e0,..., ek+l+1] G Cfc+z+i(X).
12cr = 0 because of the Z/2 coefficient.
M8130 Algebraic topology, tutorial 06, 2024
31. 10. 2024
Solution. Easily work out
k+l+l
% U ^)(r) = (<p U i)){5r) = (<p U ^) (       (-l)V[eo, • • •, eV • •, efc+z+1]) =
i=0
k
= ^(-l)V(T/[eo, ...,eh..., ek+1])ijj(T/[ek+1,ek+l+1}) +
i=0
k+l+l
+       (-^V^/N, • • •, ek])ijj(T/[ek, ek+l+1]).
i=k+l
Now, the right hand side of the formula, the first part gives
(dip U tp){r) = <fy>(r/[e0,ek+l}) ■ i/j(r/[ek+1,ek+l+1}) = = ip(5r/[e0,ek+1\) ■ ip(r/[ek+1,ek+l+l}) =
k+l
= ^(-l)V(^/[eo, ek+1}) ■ ijj(T/[ek+1,ek+l+1]).
i=0
The second part is
(-l)k(V U <ty)(r) = (-l)V(r/[e0, • • •, ek])5^(r/[ek,ek+l+1}) = i+i
= ^(-l)J+AV(<fr~/[eo, • • •, ek]) ■ i/j(r/[ek,ek+j,ek+l+1]).
3=0
Now, the last summand of the first part plus the first summand of the second part yields
(-l)fc+V(r/[e0,..., efc])^(r/[efc+i,..., ek+l+1}) + + (-l)kip(T/[e0,ek])ijj(T/[ek+1,ek+l+1}) = 0,
and we are done, LHS = RHS.
□
Exercise 3. Compute the structure of graded algebra H*(Sn x Sn;Z) for n even and n odd. Use the following:
If Hn(Y; R) is free finitely generated group for all n and (X, A), Y are CW-complexes, then
x: H*(X,A;R)(g)H*(Y;R) -> H*(X xY,AxY;R) is an isomorphism of graded rings.
Solution. We will omit writing the Z coefficients.
Now, H*(Sn) ® H*(Sn) -> H*(Sn x Sn) and we know that for spheres H° = Z with generator 1 and Hn = Z, denote generator a. Also, a Li a £ H2n = 0, a U a = 0, so we
M8130 Algebraic topology, tutorial 06, 2024
31. 10. 2024
get Z[a]/(a2} and deg(a) = n. We can write the same for the second, so denote the other generator b and have deg(6) = n and we have Z[6]/(62).
Now we compute tensor product Z[a]/(a2} <g> Z[b]/(b2}, we have four generators: la ® lb, a <g> If,, la <8> 6, a <S> 6, we will denote them l,c,d,c ■ d. Compute
(a ® lb) ■ (la (8) 6) = (-l)°"°(a • 1„) ® (16 • 6) = a ® 6,
because 0 is an idempotent element, i.e. 0-0 = 0, and (—l)n = 1 for n even, again, as in the first exercise, we use Evenness of Zero. (We refer the reader to "Principia Mathematica" Whitehead, Russell,(1910,1912,1913).) Continue with computation
(la <8> b) • (a <g> lb) = (-l)n'n(la • a) ® (6 • 16) = (-l)na ® b,
so the algebra we get is H*(Sn x S*n) = Z[c, d]/(c2, <i2, dc — (—l)ncd). For n even we have dc = cd. □
Exercise 4. Prove that there is no multiplication on even dimensional spheres. Multiplication on the sphere Sn is a map m: Sn x Sn —)• Sn such that there is an element 1 G Sn satisfying m(x, 1) = x, m(l, rr) = x.
Hint: compute m*: H*(Sn) —)• H*(Sn x S'n); describe two rings.
Solution. We have #*(£") = Z[7]/(72) and iJ*(,Sn x 5n) = Z[a,/3]/(a2,/32) = #*(£") ® H*(Sn), because we already know, that a/3 = /3a. Our situation can be described with two diagrams:
gn-%SnxSn       H* (S") H* (Sn X Sn)
id
H*(S
m
Take m*(^y) = aa + b(3 with a, b G Z and prove first, that a = b = 1. Use mo %x = id, so «*(m*7) = 7. This gives i*(aa + 6/3) = 7 and since i*(aa + b/3) = 07, we have a = 1, same for 6. Final computation yields
0 = m*(0) = m*(72) = m*(7 U 7) = m*7 U 771*7 = = (a + 0) U (a + /3) = a2 + a/3 + /3a + /32 = 0 + 2a/3 + 0^0,
and that, my friends, is a contradiction. □
Exercise 5. Show commutativity of the diagram below. Use five lemma to prove that taking any two fi's iso's, the third // is iso as well.
M8130 Algebraic topology, tutorial 06, 2024
31. 10. 2024
H*(X, A)®H*(Y)
H*(X)®H*(Y)
H*(A)®H*(Y)
H*(X xY,AxY)
H*(X x Y)
H*(A x Y)
Solution. We will name the parts of the diagram as follows: back-square, upper-triangle, lower-triangle, left-square, right-square. The triangles come from the long exact sequence of of pairs (X, A) and (X x Y, A x Y), the right-square commutativity comes from an inclusion, back-square commutes as well (topology knowledge). The only problematic part is the left-square and it is exercise on computation of connecting homomorphism S*. □