M8130 Algebraic topology, tutorial 13, 2017
18. 5. 2017
Exercise 69. Let (X, A) be a pair. Show that the following diagram commutes:
nn(X, A,x0)-^nn(A,x0)
Hn(X,A)
■ Hn_i(A)
where d is the boundary homomorphism, h is the Hurewicz homomorphism and <9* is the connecting homomorphism.
Solution. Take [/] G irn(X,A,x0), that is /: (Dn, Dn~x, s0) -> (X,A,x0). Then d[f] = [f/S12-1] and hd[f] = hlf/S12-1] = (f /S12-1)^), where b is a generator in H^S12-1). (We recall the definition of the Hurewicz homomorphism: if g: S11^1 —?■ A, g*: Hn_i(Sn~l) —?■ Hn-i(A) and h[g] = g*(b) G Hn^(A).)
Let a G Hn(Dn, S'^1) be a generator such that <9*a = b. We proceed using commuta-tivity of the following diagram:
Hn-i(A)
Hn(X,A) —
Now d*h[f] = <9*(/*a) = (f/Sn~1)^(d^a) = hd[f] which concludes the proof.
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Exercise 70. Homology version of Whitehead theorem: If X and Y are simply connected CW-complexes and f : X —>• Y induces an isomorphism on all homology groups, then f is a homotopy equivalence. Show that /*: nn(X) —> nn(Y) is an iso for all n and then use the homotopy version of Whitehead theorem.
Solution. First, for consider / : X H F an inclusion. For pair (Y, X) we have the following long exact sequences connected by Hurewicz homomorphisms:
Kn+l(Y,X) h
h
■7Tn(X)
■Kn(Y,X)
n+l
(Y, X)        Hn{Y)        Hn{X)        Hn(Y, X)
Since Hn(X) —> Hn(Y) are isomorphisms, all Hn(Y,X) = 0. Since Y and X are simply
h
connected, we have ni(Y,X) = 0. Now, 7i2(Y,X) —> H2(Y,X) is an iso by Hurewicz theorem. Since H2(Y,X) = 0, n2(X,Y) = 0 as well. With the same argument we can proceed inductively to get nn(Y, X) = 0 for all n. Then /* : 7rn(X) —> nn(Y) are isomorphisms.
Now, for general map /: X —> Y we use the argument with mapping cylinder Mf. Do you remember this one from the first lecture? This construction you thought would not be interesting? Well, hold your hats, its usage is advantageous here!
M8130 Algebraic topology, tutorial 13, 2017
18. 5. 2017
Y
We know that /*: Hn(X) —?■ Hn(Y) is an iso and (as iY is homotopy equivalence) iY*: Hn(Y) —> Hn(Mf) is an iso. Thus, we conclude that ix* '■ Hn(X) —> Hn(Mj) is an iso. That is,    is homotopy equivalence and therefore / is also homotopy equivalence. (Let us remark that for this exercise you should revisit Whitehead theorem, at.pdf study text is also recommended) □
Exercise 71. (application of Whitehead theorem) Show that (n — \)-connected compact manifold of dim n is homotopy equivalent to Sn (n > 2).
Solution. Take manifold M as two parts: disk and its complement. Map the disk to Sn = Dn/Sn~1, that is the identity on the interior of the disk and the complement goes to one point. We have /: M —?■ Sn and for i < n by Hurewicz theorem (M is (n — 1)-connected) we get an iso /* : Hi(M) = Hi(Sn) = 0. The same iso we obtain for i > n (both Hi are zero).
We conclude our application with this spectacular diagram where all arrows are iso, reasoning goes from the bottom arrows by excision, then with definiton of fundamental class we get the top arrow (our main focus) an iso. (we denote interior of B as B°)
Hn(M)-^Hn(Sn)
Hn(M, M - (Bn)°)--        Hn(Sn, Sn - (Bn)°)
Hn(disk, ddisk)--—*■ Hn(disk, ddisk)
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Remark. (Story time) We have got that a compact manifold 3-dim (compact) which is 2-connected is homotopy equivalent to S3. There is another result: Every 3-dim manifold simply connected compact manifold is homeomorphic to S3. This latter result (it might seem we are close to proving it) is actually famous Poincare conjecture, one of Millenium Prize Problems and it was already solved by Grigori Perelman in 2002. Interesting story and interesting mathematician for sure. It is well known (as we like to say) that Perelman declined Fields medal (among other prizes).
Exercise 72. Find a map f with Hopf invariant H(f) = 2.
M8130 Algebraic topology, tutorial 13, 2017
18. 5. 2017
Solution. We study a space X with a basepoint e. Denote construction J2(X) = X x X/ ~, where (x, e) ~ (e, x). Apply this idea to Sn. We get a projection p: Sn x Sn —>• J2(Sn). On the left we have one 0-cell, two n-cells and one 2n-cell, while on the right we have one of each. We get that J2(Sn) has to be a space of the form Cf, so Hn(J2) = Z given by a and H2n{ J2) = Z given by 6 and Fn(Sn x Sn) = Z©Z (generators ai, a2) and F2n(Sn x Sn) = Z (with b0). Now, p*: ^(J2) -> #*(Sn x Sn) and p*(a) = ax + a2,P*(&) = &o-
H(f)b
H(f)p*(b) H(f)bo
H(f)a1a2 H(f)a1a2 2
because b0 = a\a2 and by evenness of the dimension H\H2 — H2H\.
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