M8130 Algebraic topology, tutorial 10, 2019 29. I 2019 Note: This tutorial originates in 20171. 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) -> Tries'3) -> tt^S2) -> ■ ■ ■ and also for Z -> R -> S1 ^ R/Z, that is 7rn(Z,0) -> 7rn(M) -> tt^S1) -> 7rn_i(Z) for n > 0. Since Z is discrete, (S'n, 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 -> tt^S1) -> 7T0(Z) -> 7T0(M) = 0, so 7r1(S'1) = Z. Continue now with the updated long exact sequence: 0 -> 7T3(,S3) -> 7T3(,S2) -> 0 -> 7T2(,S3) -> 7T2(,S2) -> Tr^S1) = Z -> t^3) and 7ri(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), 7r2(S2)-7r1(S1)=Z. It implies that ^(S2) = Z. Let us remark that if n3(S3) = Z, then n3(S2) = Z. Later we will prove that nn(Sn) ^ Hn(Sn) ^ Z. □ Remark. (Story time) Eduard Cech was the first who defined higher homotopy groups (1932, Hoherdimensionale Homotopiegruppen) but the community didn't support the study of these groups 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 orthonormal 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 1see https://is.muni.cz/el/sci/jaro2019/M8130/um/68045051/ M8130 Algebraic topology, tutorial 10, 2019 29. I 2019 is a fibre bundle, we call Vn^ Stiefel manifold, k-tuples of orthonormal vectors in M.n. 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 W"1): 0(k) Vn,k Gn,k, is also a fibre bundle. 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 n0(E). Solution. We showed nn(B) = nn(E,F) for n > 1 which gives 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(6o),zo) -> (E,x0) -> (B,p(x0) = bo), where —1 goes to x0 and b0 and 1 goes to F = 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 and denote the homotopy by H. It means that pf(l) is connected with b0 by a curve. Have a diagram S°x{0}U{-l}x7«E g ^ ' " S° x I-^-- B and remark that imG(—, 1) C F due to the fact H(—,l) = b0. Thus a desired preimage of [/] in is G(-,l). □ Exercise 55. Covering: G —> X —> X/G, with action of G on X properly discontinuous, where X is path connected. Take 7n(G, 1) -> Tn(X) -> tt^X/G) -> n0(G) -> tt0(X), 0 7Ti(X) 7Ti(X/G) 4 7ro(G) 0. X as i(g) = g ■ x0 and identify g with g ■ x0 £ X - this is possible due to the properly discontinuous action): 7T! (X) -^ 7T! (X, G,x0) 7T0 (G) m(X/G, N) M8130 Algebraic topology, tutorial 10, 2019_29. 4. 2019 Let's recall how the isomorphism p* is defined for this particular case: {0}^-^X3x0 (1) ^ - " p —--X/G 9 [x0] uj ■ ř(í) A group structure on 7ri(X, G, x0) is transferred by the isomorphism p*. Take general elements [uj], [r] £ n1(X/G, [x0]) and lift them to elements [uj], [r] £ n1(X,G,x0) using the diagram (1). Since uj(1) = r(l) = [x0] it must be ZJ(l),ř(l) G G.For simplicity, we denote ZZJ(l) = gi • xQ and r(l) = g2 ■ x0. Consider the following representative of some element in 7ri(X, G,x0): 'ZZJ(2Í) iG[0,i], ^•Ť(2t-1) ÍG[i,l], The above map is continuous by the pasting lemma and a definition of G-action. We define [uj] ■ [t] := [uJ ■ r]. It is easy to see that uj ■ r satisfies the diagram (1) for the bottom arrow uj * r. As p* is an isomorphism we obtain p*([ZJ] • [ř]) = p*([čU • r\) = [uj * r] = [uj] * [r] = p* ([ZZJ]) * p* ( [t] ). It remains to show that d is a homomorphism what is quite straightforward: dip] • [Ť]) = dip • Ť]) = (uj • ř)(l) = 9l(g2 ■ xo) = o7(l) • ř(l) = d([u]) ■ Š([ř]). We denote open sets Ui,U2 (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. M8130 Algebraic topology, tutorial 10, 2019 29. I 2019 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. Now, the intersection U\ fl U2 ^ (S1,x0) and we take generator uj, i2,i*(^~1) = 1 in ^i(U2,x0), h,2*(u) = ct(3a~l(3. So, kernel of