M8130 Algebraic topology, tutorial 07, 2017
6. I 2017
Exercise 1. (39 in total) Use theorem "f: Sn -> Sn s.t. f(—x) = -f(x) deg(/) is odd" to prove that for f: Sn —>- M. there exists x e Sn: f(x) = f(—x).
Solution. We will use proof by contradiction. Suppose that f(x) — f(—x) is always non-
fix\ _ fl-x)
zero and define g(x) = ,, „, .-—-—. Then obviously g(—x) = —g(x). Since g: Sn —>
\\f(x)~ f(-x)\\
S^1 C Sn is odd , the map g has odd degree, but g(Sn) C Sn which implies deg(g) = 0 (g is homotopic to a constant map). Now we use well known fact that zero is even number, some say that it is the evennest number of them all (why?) and conclude desired contraction. □
Exercise 2. (40 in total) Let ip E Ck(X; R), tp G Cl(Y; R). Prove 5((p U ijj) = 5ip U f + (-l)kipU5ip. User= [e0,...,ek+l+1] € Ck+l+1(X).
Solution. Easily work out
k+l+l
5{<p U </0(t) = (cp U iP)(5t) = {<p U ^) ( ^ (-^Vteo, • • •, e„ ..., efc+z+i;
j=0 k
= ^(-l)V(T"/[eo,..., ej,..., efc+i])^(r/[efc+i,..., efc+m]) +
j=0
+ Yl (-1)tv(T/\eo>--->eM(T/\ek,--->eh--->ek+i+i])-
i=k+l
Now, the right hand side of the formula, the first part gives
(5(p U ip){r) = 5ip(r/[e0,ek+1]) ■ ip(r/[ek+1,ek+l+1}) = = (p(5r/[e0,ejfe+i]) • ip(r/[ek+1,efc+z+i]) =
k+l
= ^2(-iy^(ST/[e0, ...,e,t,.. . ,ek+l]) ■ f(r/\ek+1,.. .,ek+i+l}).
i=0
The second part is
(-l)k(V U <ty)(r) = (-l)V(T/[e0, • • •, ek])5^(r/[ek,ek+l+1]) = i+i
= ^2(-iy+k^(Sr/[e0,ek]) ■ i/j(r/[ek,ek+j,ek+l+1]).
j=0
Now, the last summand of the first part plus the first summand of the second part yields
{-l)k+lip{r/[e0,.. .,ek})Hr/[ek+u      ek+l+1}) + + (-l)V(T/[eo, • •. ,efc])^(r/[efc+i,... ,ek+l+1\) = 0,
and we are done, LHS = RHS.
□
M8130 Algebraic topology, tutorial 07, 2017
6. I 2017
Exercise 3. (41 in total) Prove aUb = A*(a x 6), where a e Hk(X), b e Hl(X),A: X Ixl,ih>(i,x) and x is cross product defined a x(3 = p*xaUpY(3 (px,Py o,reprojections from X x Y).
Solution. By the following diagram
X-——^X x X
X X
we have that pxA is id. So, compute
A* (a x b) = A*(p*xa U p*Yb) = A*(p*xa) U A*(p*Yb) = (pxA)*a U (pYA)*b, and we are done. The thing is that cup product is natural. □
Exercise 4. (42 in total) 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)®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 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> lb, la <g> b, a <g> b, we will denote them 1, c, d, c ■ d. Compute
(a ® lb) ■ (la ® b) = (-l)°"°(a • la) ® (lb • 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 ® b) ■ (a ® lb) = (-irn(l„ • a) ® (6 • lb) = (-l)na ® 6,
so the algebra we get is H*(Sn x S*n) = Z[c, d]/(c2, d2, dc — (—l)ncd). For n even we have dc = cd. □
M8130 Algebraic topology, tutorial 07, 2017
6. I 2017
Exercise 5. (43 in total) 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,x) = x.
Hint: compute m*: H*(Sn) —> H*(Sn x Sn), describe two rings.
Solution. We have H*(Sn) = Z[7]/(72) and H*(Sn x Sn) = Z[a, f3]/(a2, /32) = H*(Sn) ® H*(Sn), because we already know, that a/3 = /3a. Our situation can be described with two diagrams:
H*(Sn)^-H*(Sn x Sn)
Sn x gn
id
H*(Sr
Take m*(^y) = aa + b(3 with a, b G Z and prove first, that a = b so «*(m*7) = 7. This gives i\(aa + 6/3) = 7 and since i*(aa + b/3) same for 6. Final computation yields
1. Use moii = id, 07, we have a = 1,
0
m*(0) = m*(72) = m*(7 U 7) = m*7 U m*7 =
(a + /3) U (a + 0) = a2 + a/3 + /3a + /32 = 0 + 2a/3 + 0^0,
and that, my friends, is a contradiction.
□
Exercise 6. (unfinished) (44 in total) Use five lemma to prove that taking any two fi's iso's, the third // is iso as well. (Proving five lemma might be a homework.) Also, show commutativity of the diagram.
H*(X, A)®H*(Y)
H*(A)®H*(Y)
H*(X xY,AxY)
8*
H*(A x Y)
H*(X)®H*(Y)
H*(X 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*. □