M8130 Algebraic topology, tutorial 10, 2017
27. I 2017
Exercise 53. Long exact sequence of the fibration (Hopf) S1 —> S3 —> CP1 = S2.
Solution. This is an important example of a fibration, it deserves our attention. First we write long exact sequence
^(S1) ^ n3(S3) 4 n3(S2) A ^(S1) -> n2(S3) -> n2(S2) ->
-> ir^S1) -> tt^S3) -> ir^S2) -> ■ ■ ■
and also for Z -> R -> S1 = R/Z, that is 7rn(Z,0) -> 7rn(M) -> ir^S1) -> 7rn_i(Z) for n > 0. Since Z is discrete, (Sn, s0) —)• (Z, 0), s0 goes to a base point 0, therefore the map is constant . Hence we get 7rn(Z, 0) = 0 for n > 1. Also, 7rn(IR) is zero as well, because R is homotopy equivalent to point. We get that this whole sequence are zeroes for n > 2. What we are left with is
0 -> 7Ti(S'1) -> 7T0(Z) -> 7T0(M) = 0,
so 7ri(S'1) = Z. Continue now with the updated long exact sequence:
and 7r1(S'3) = 0, because nk(Sn) = 0 for k < n considering the map Sk —> Sn that can be deformed into cellular map and so it is not surjective. So we get
n3(S3)^n3(S2),    n2(S2) - n^S1) = Z.
It implies that tt^S2) = Z.
Let us remark that if n3(S3) = Z, then n3(S2) = Z. Later we will prove that
nn(Sn) ^ Hn(Sn) ^ Z.
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Remark. (Story time) Eduard Cech was the first who defined higher homotopy groups (1932, Hoherdimensionale Homotopiegruppen) but the community od these groups didn't support the study as they were considered not interesting. Were they mistaken? The rest of this remark is left as an exercise for the reader.
Remark. (For geometers) For G Lie group and H its subgroup we have a fibre bundle H^G^ G/H.
As an example consider otronormal group 0{n), we have inclusions 0(1) C 0(2) C • • • 0(n). Then
0{n — £;)—>• 0{n) -+ 0{n)/0{n - k) = Vn,k
is a fibre bundle, we call Vn^ Stiefel manifold, k-tuples of orthonormal vectors in Rn.
Also, we can take 0{k) —> 0(n)/0(n — k) —> 0(n)/(0(k) x 0(n — k)), Grassmannian manifold (k-dimensional subspaces in Rn):
0{k) ->      -> Gn,k,
is also a fibre bundle.
M8130 Algebraic topology, tutorial 10, 2017
27. I 2017
Exercise 54. Long exact sequence of the fibration F —>• E —>• B ends with
tti(5)     n0(F,x0)     n0(E,x0) n0(B). Show exactness in no(E).
Solution. We showed nn(B) = nn(E,F) for n > 0 which gave us the exactness of the fibration sequence from the exactness of the sequence of the pair (E,F) till n0(E,x0). Denote S° = { — 1,1} and consider the composition
({-1,1},-1) -> (F,x0)=p-1(b0) -> (E,x0) -> (B,p(x0) = b0),
where —1 goes to x0 and 60 and 1 goes to F but that is p_1(&o) so we get the constant map.
For the other part of the exactness we will use homotopy lifting property. Consider / : ({ — 1,1}, —1) —> (E,x0) such that pf is homotopic to the constant map into b0. It means that pf(l) is connected with b0 by a curve. Have a diagram
(1)-~E
s p
(1) x J->B
and remark that /(l) is connected by a curve with x G E such that p(x) = b0. So x e F and / is homotopic to the map g : ({ — 1,1}, —1) —> (F,x0) which maps —1 into x0 and 1 into x.
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Exercise 55. Covering: G —> X —> X/G, with action of G on X properly discontinuous, where X is path connected. Take
m(G, 1) -> m(X) -> miX/G) -> n0(G) -> tt0(X),
that is
0       7Ti(X)       7Ti(X/G) A 7r0(G) 0.
<S7iow;       d is a group homomorphism.
Solution. First note that n0(G) is G taken as a set. We recall that for F —> E —> B we had
nn(E)-3-i- nn(E, F, x0)        7rn_i(F)
^ p. nn(B,b0)
We showed the = finding a map going from nn(B, b0) —> nn(E, F,x0). Also,
/: (r\dr\r-l)^(E,F,xQ),
M8130 Algebraic topology, tutorial 10, 2017
27. 4. 2017
d[F] = [f/dln]. The prescription for d: nn(B,b0) -> 7rn_i(F,x0): take /: (In,dln) -> (B, b0) and make a lift . (F,x0). The diagram is:
(B,b0) and make a lift F: (In, dln, Jn_1)      (E,F,x0), d[f] = F/I"-1: (In~1, <9/n_1
J™-1 x {1} U <9J     x J
const
f
Tn—l
X J-
■ E xq
p
B b0
and pFil*1-1 x {0}) = b0, so F^-1 x {0}) C p^fa) = F.
d
Now, iti{X/G) —)• ^(G), /: (I,dl) —)• (X/G, [x0]). Take two closed curves w,r at [rr0] and lift them to curves w, f in J starting in x0. We denote the end point of uj by gix0 and the end point of f by g^o- Then the operation • (that is not a composition o) is:
uj ■ gtr
UJ	r
UJ	t
UJ	t
UJ	t
And it's a homomorphism.
Exercise 56. Van Kampen theorem - Applications.
Klein bottle K. Model as a square with identified sides as seen in Fig 1,2,3.
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boundary ^-
Fig 1
Fig 2
Fig 3
We denote open sets U\,XJ<i (disc) and point x0 as in figure 1 and 2. (some of the notation in the solution is established in the theorem)
Solution. Set U\ is homotopy equivalent to the boundary (Fig 3), and this boundary is in fact a wedge of two circles, so U\ ~ S1 V S1.
We can compute: 7r1(f/2,^o) = {1} by contractibility, ni(Ui,x0) = n1(S1 V S1,x0) = free group on two generators a, (3 as was already shown in lecture.
Then ni(Ui, x0)*n1(U2, x0) = free group on two generators a, (3, also ^(l^nl^, x0) = 7h.
M8130 Algebraic topology, tutorial 10, 2017_27. 4. 2017
Now, the intersection U\ fl U2 ^ (S,1,rr0) and we take generator uj, «2,1* = 1 in
So, kernel of <p (the map from the theorem) is generated by element tvi(K) is the group with two generators a, (3 and one relation af3a^1f3 = 1 □