M8130 Algebraic topology, tutorial 11, 2017
I 5. 2017
Exercise 57. Recall definitions of n-connectness and n-equivalence. Prove the following lemma: Inclusion A ^ X is n-equivalence if and only if (X, A) is n-connected.
Solution. "<^=" Take long exact sequence:
f f
-> 7ln(A,x0) 4- 7ln(X,x0) -> 7ln(X,A,x0) -> 7Tn_i(A,X0) 4 7Tn_i(X)        7Tn_i(X, A) ->
and use the assumption that iti(X, A, x0) = 0 for i < n. Then we get that f\ is epimorphism and /2 is isomorphism.
Reasoning is the same as in the other direction, the only thing we need to realize
is n0(A, x0) ^ n0(X, x0). □
Exercise 58. Show nk(S°°) = 0 for all k, where S00 is colimSn.
Solution. We have S1 c S2 c S3 c • • • C Sn c • • • S°°. Take element in 7rk(S°°), that is /: Sk —> S00. We know that f(Sk) is compact in S00. Consider other CW-complex structure than e° U ek for Sk that is Sk = [jki=0 e\ U e\ (two hemispeheres). Then the following holds: S°° = {JZoe\ u 4- So f(Sk) C (S°°)W = SN for some N where (S°°)W is iV-skeleton of S00. Now,
/: Sk ^ SN ^ SN+1 ^ S°°,
so the composition Sk —)• S'Af+1 is a map that is not onto and therefore / is homotopic to constant map. (map into a disc is homotopic to constant map, disc is contractible) Then we have [/] = 0. Thus we have proved nk(S°°) =0. □
Exercise 59. Compute homotopy groups ofRP°°.
Solution. Suprisingly use previous exercise: We can view IRP°° as lines going through origin in S00, or...just take S00/z/2, where the action is x H> —x. So we work with the following fibration (we don't write the distinguished points as they are not needed) Z/2 —> S00 —> ]RP°° and the long exact sequence
7rn(Z/2)     nn(S°°) -> nn(RP°°) 4 nn^(Z/2) -+ n^S00),
d
where for all n > 2 we have all zeroes, for n = 1 consider 0 —> 7Ti(IRP00) —> 7r0(Z/2) —> noiS00). Since n0(S°°) = 0 and 7r0(Z/2) = Z/2, we get that the homomorphism d (it is homomorphism, really, we did it in previous tutorial, but it's still a homomorphism independetly on whether we did it or not) is an isomorphism of groups. By connectness we also know the n0 group. So the final results are:
nn(RP°°) = 0 for n > 2, n^RP00) = Z/2, n0(RP°°) = 0.
□
Exercise 60. Show that the spaces S2 x RP°° and RP2 have the same homotopy groups but they are not homotopy equivalent.
M8130 Algebraic topology, tutorial 11, 2017
I 5. 2017
Solution. Here, also use pre2vious exercise. It is known that nn(X x Y) = 7rn(X) x nn(Y). With this we can compute
7r0(,S2 x MP00) = 0, tt^S2 x MP00) = miS2) x tt^MP00) = {0} x Z/2 = Z/2,
for n > 2nn(S2 x IRP°°) = 7rn(S2) x {0} = 7rn(S2. Now consider MP2 as S2/z/2, work with the fibration (scheme as follows)
Z/2-- S2-- MP2
7Tn(Z/2) --tt^S2) --7Tn(MP2) -^7Tn_i(Z/2)
n > 2 0 7rn(,S2)        = 7rn(MP2) 0
71 = 1 0 tti(MP2) =Z/2 0
and 7r0(]RP2) = 0. Thus we showed that these two spaces have the same homotopy groups. Marvelous. How to show, that they are not homotopy equivalent? Use cohomology group! That's right. It is well known (or we should already know) that
iT(MP2; Z/2) =Z/2[a]/(a3),a E H1 and
H*(S2 x RP°°; Z/2) = H*(S2;Z/2) ® H*(RP°°;Z/2) = Z/2[/3]/(/32) ® Z/2 [7], /3 e H2.
The former space obviously has no non-zero elements of order 4, while the latter has a non-zero element of order 4. This is impossible for homotopy equivalent spaces. We are done.
□
Exercise 61. Extension lemma: Let (X, A) be a pair of CW-complexes, Y a space with 7rn_i(F, y0) = 0 whenever there is a cell of dimension n in X — A. Then every map f:A^-Y can be extended to a map F: X -^Y.
Solution. Set X_i = A, X0 = X^UA, Xk = X^UA, and / = /_i: X_i Y, f0: X0 -+ Y which extends       fo(x0) to any point in Y.
We have fk-i ■ Xk-i —> Y and want to define    : Xfc —>• Y. Also, nk-i(Y, •) = 0.
Consider following diagram:
(D^S^-^iX^X^)
fk-l
Y
The map fk-i o 99 is in nk-i(Y, •) so it is homotopic to constant map, so we define fk on Dk as a constant map. Now,
M8130 Algebraic topology, tutorial 11, 2017
I 5. 2017
Dk x {ljU^-1 x I-Y
Dk x I
and go back to the first diagram:
{D\Sk-l)-^{Xk,Xk^)
We extend fk-i to fk: Xk —> Y and proceed to infinity (and beyond), as we always do. □
Exercise 62. Compare nn(X, x0) and nn(X,xi), when distinguished points (are / are not) path connected. Use proof with pillows.
Solution. First the case where X = S1 U S2 and x0 G S1 and xx G S2. Then n1(X,x0) = 7r1(S'1) = Z but n1(X,x1) = n1(S2) = 0. If distinguished points are not path connected, homotopy groups can be different, so consider now uj a curve connecting x0 and xi, u: I —> X, uj(0) = x0 and uj(1) = xi. We have nn(X,x0) —> nn(X,xi) f:In^-X, dln H> x0, g: In —> X, dln H> x1: as seen in Figure 1, x0 — x\ segments. Proofs with pillows!
Figure 1
Denote action / H> u ■ /, then fi ~ f2 =>• u ■ f\ ~ u ■ f2. We are not satisfying algebraist at the moment, only geometers. Let us try to do something about that.
Figure 2 shows other pillow and that uji ~ uj2 =>• di ■ f ~ co2 ■ f, we can imagine the segment as in Figure 2, two curves.
M8130 Algebraic topology, tutorial 11, 2017 4. 5. 2017
Map given by uj is a bijection uj : x0 —> xi, uj2(uJif) ~ (^1^2)/, w"H^/) ~ — />
so the map 7rn(X, rr0) —> nn(X,x1) is bijection. Figure 3 tries to explain homomorphism.
We get that x0,xi are in the same path component and if uj : x0 —> xx is a curve, then nn(X,xq) —> nn(X,xi) is an isomorphism. In particular, if X is simply connected, then every curve gives the same isomorphism. □