M8130 Algebraic topology, Tutorials 2017
Remark. All sets are assumed to be topological spaces and all maps are assumed to be continuous unless stated otherwise. The symbol '=' will denote that two topological spaces are homeomorphic. The closed unit interval will be denoted by I or J.
Exercise 1. Prove that being homotopic is an equivalence relation (on the set of continuous maps between topological spaces).
Solution. Let /, g, k : X —> Y be such that / ~ g, g ~ k, i.e. there exist maps h,h' : X x I —?■ Y such that h(x, 0) = f(x), h(x, 1) = g(x), h'(x, 0) = g(x), h'(x, 1) = k(x).
• Reflexivity: the map Hi : X x I —>• Y defined by Hi(x,t) := f(x) for all t E I is a homotopy between / and itself.
• Symmetry: the map H2 : X x I —?■ Y defined by H2(x,ť) := h(x, 1 — t) for all t E I is a homotopy between g and /.
Transitivity: the map H-$ : IxI —y Y defined by Hs(x,t) : = is a homotopy between / and k.
h(x, 2t) for 0 < t < \ h'(x,2t) for|<i<l
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Exercise 2. Let ~ be an equivalence relation on a topological space X. Prove that the map f : X/ ~—> Y is continuous ifffop-.X^Y is continuous, where p : X —> X/ ~ is the canonical quotient projection.
Solution. The direction "=^" follows from the facts that p is continuous (in fact, the quotient topology is the final topology with respect to p) and the composition of continuous functions is again continuous. For "<=", let U CY be open. Then
p-\rl{u)) = {foP)-\u)
is open by continuity of / o p, so f^iU) must also be open by the definition of quotient topology and we are done. □
Exercise 3. Show that Dn/Sn~1 = Sn using the map f : Dn —^ Sn given by
f(xi,.. .,xn) = (2a/1 - ||a:||a:,2 ||a;||2 - 1).
Solution. It's easy to see that / is continuous. Moreover, its restriction to the interior of Dn gives a bijection to S'n\{(0,..., 0,1)} (the inverse function is given by (y, z) H> -^===y) and
we have fiS*1-1) = {(0,... ,0,1)}, so we can define /' : Dn/Sn-1 -+ Sn by f([x])2= /(x). Then /' is a bijection, and by the previous exercise it is continuous. Finally, both Dn/Sn~1 and Sn are compact (Hausdorff) spaces (since both Dn and Sn are closed bounded subsets of W1 and Ka+1, respectively, and S'^1 C Dn is closed), so /' must be a homeomorphism (a general fact for continuous bijections between compact spaces). □
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Exercise 4. Let f : X —>• Y and Mf = X x /Ujxi Y. Moreover, let lx ■ X —> Mf be given by x ^ (x,0), iy : Y —> Mf be given by y ^ [y] and r : Mf —>• Y be given by r(y) = y, r(x,t) = f(x). Show that
i) Y is a deformation retract of Mf,
U) roLX = f,
Hi) tyo/~tj(.
Solution.
i) Geometrically, the deformation retraction is realized by pushing X along I towards Y.
ii) We have r o lx{x) = r(x, 0) = f{x) for all x G X.
in) The required homotopy h : X x J —>• Mf is given by h(x, s) = [(x, s)].
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Exercise 5. Show that the pair (Mf,X) has the homotopic extension property (HEP), i.e. Lx is a cofibration.
Solution. Let g : / x J —>- {0} x JU/x {0} be any retraction such that g(0, s) = (0, s) and 0(1, s) = (1,0). Then the map r : Mf x J -> X x {0} x J U Mf x {0} defined by r(rr, t, s) = (x, g(t, s)) and r(y, s) = (y, 0) is the required retraction. □
Exercise 6. The smash product between two based spaces is defined by
(C, Co) A (D, d0) := (C x D)/ (C x {d0} U {c0} x D). Show that Xj A A Y/B = (X x F)/(X xBUAxF).
Solution. Let pi : I x F -> X/A x F/5 be given by pi(x,y) = ([x], [y]) and p2 : X/A x Y/B —> X/AAY/B be given by p2 ((N, [y])) = ([ N ], [ [y] ]) ■ Then the composition p2 ° Pi is continuous and factors through (X xY)/(X x BLiAxY), which implies that the canonical bijection between (1x7)/(X xBUAx Y) and X/AAY/B is continuous (using exercise 2). Using the definition of quotient topology several times, it can be shown that this bijection is also open, hence a homeomorphism. □
Exercise 7. Let A = U {0}} C R. Show that (I, A) does not have the HEP, i.e. the inclusion A ^ I is not a cofibration.
Solution. If A x J U / x {0} was a retract of / x J, the retraction would have to preserve connected subsets. But fx J is locally connected while AxJU/xjOjis not, a contradiction.
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Exercise 8. Show that (Sm,*) A (Sn,*) ^ (Sm+n,*).
Solution. Using exercises 3 and 6 from the previous tutorial, we have
(Sm, *) A (Sn, *) 2* (Dm/S™1^1) A (Dn/Sn-1) ^
^ (Dm x Dn)/(Sm-1 x DmUDn x S™-1) = ^ (Jm x In)/(dlm xf U <9Jn x Jm) =
_ jm+nJ^Q^Jm+n^ ^ jym+njQj-ym+n t^im+n
Exercise 9. Show that CPn is a CW-complex. Solution. Clearly CP0 is a point. Next, we have
Cpn = Cn+1 y ^Qy^ ^Xv,\eC\ {0}} ^ S2n+1 \ {0}/{v ~ \V, |A| = 1}
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= {(w, v/l - \w2\) e Cn+1,w e D2n}/{w ~ Aw for \w\ = 1} = = (D2n U .S2"-1)/!^ ~ Aw for w e S2'"-1} = D2n Uf C.
Now taking the canonical projection S2'^1 —> CPn_1 = S2'^1/ ~ as the attaching map / yields a CW-complex with one cell in every even dimension till 2n and none in the odd ones. □
Exercise 10. From the lecture we know that A := {^,n G N} U {0} as a subspace ofM. is not a CW-complex. Show that X := I x {0} U A x I is not a CW-complex either.
Solution. Suppose that X is a CW-complex. Then it cannot contain cells of dimension > 2, because it becomes disconnected after removing any point. In fact, the space obtained after removing any point (a, 0) with a G A has more than two connected components (three for a > 0, to be exact), so these points cannot lie inside a 1-cell. Therefore these points must form 0-cells, but we already know that A does not have discrete topology, a contradiction. □
Exercise 11. Show that the Hawaiian earring given by
1 1
X = Ux, y) G IR2, (x--)2 + y2 = — for some n\
n n2
is not a CW-complex.
Solution. Suppose that X is a CW-complex. Using similar arguments as in the previous exercise, we can see that (0, 0) must be a 0-cell and that X must have either infinitely many 0-cells, or infinitely many 1-cells. But since X is compact, exercise 5 implies that X can have only finitely many cells, a contradiction. □
Exercise 12. Prove that every compact set A in a CW-complex X can have a nonempty intersection with only finitely many cells.
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Solution. X is comprised of cells that are indexed by elemnts of some set J. Let B be a set containing exactly one point from each intersection A fl e13, f3 G J. We need to show that B is closed and discrete, which will imply that B is compact (since B C A) and discrete, hence finite. We know that a set C C Xn is closed iff both C fl Xn_1 and C fl e™ for each a G J are closed, because Dn U/ Xn_1 is a pushout. Using induction, this implies that C C X is closed iff C fl ea is closed for each a G J. Since 5 fl ea contains at most one point for any a G J and X is Tx (even Hausdorff), this shows that B is closed. Using the same argument, B with any one point removed is closed. Therefore B is also discrete and we are done. □
Exercise 13. Show that for a short exact sequence O^A^B^C^Oof abelian groups (or more generally modules over a commutative ring) the following are equivalent:
(1) There exists p : B —>• A such that pf = id^.
(2) There exists q : C —>• B such that gq = idc.
(3) There exist p : B —>• A and q : C —>• B such that fp + qg = id^.
(Another equivalent condition is B = A® C, with (p,g) and f + q being the respective inverse isomorphisms.)
Solution.
(1) =>. (2) and (3):
Since g is surjective, for any c G C there is some b G B such that g(b) = c. Moreover, for any other b' G B such that also g(b') = c, we have b — fp(b) = b' — fp(b'), since b — b' G ker g = im/, so that b — b' = f(a) and
fp(b-b') = fpf(a) = f(a) = b-b'. This shows that we can correctly define q(c) := b — fp{b) for any such b. Then we have
9Q(c) = gip) - gfpib) = g{b) = c (since gf = 0), which shows that gq = idc, and also qg(b) = b — fp(b), hence fp + qg = id#. (3) =>■ (1) and (2):
Applying / from the right to the equation fp + qg = idB yields fpf = f (since gf = 0), which together with the fact that / is injective implies pf = idA. Similarly, applying g from the left yields gqg = g, which together with the fact that g is surjective implies gq = idc. □
Exercise 14. Let 0 —> A* A B* A C* —> 0 be a short exact sequence of chain modules. We have defined the connecting homomorphism <9* : Hn{C) —> Hn-i(A) by the formula <9*[c] = [a], where dc = 0, f(a) = db and g{b) = c. Show that this definition does not depend on a nor b.
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Solution. We have da = 0 iff f(da) = 0 (using injectivity of /) iff 0 = Of (a) = ddb, and the last condition is true.
Now let b,b' e B be such that g(b) = g(b!) = c with a, a' G A such that f(a) = b, f(a') = b'. Then b — b' e ker g = im/, so b — b' = fia) for some a G A. Therefore f(da) = db — db' = f(a — a') and the injectivity of / implies da = a — a', hence [0] = [da'] = [a] — [a'] and we are done. □
Exercise 15. Show dd = 0. Use formula £%n+1°£i = ejn^1oein, where i < j. The definition foraeCn(X), a: An ^X, is
n
^ = ^(-i)Vo4
«=0
Solution. Easily workout
n+l n+l
d(da) = d(J2(-lY a o 4+1) = o 4+1) =
i=0 i=0
n+l
i=0 j<« j>i
n+l it
= E E(-1)'(-1)V ° £n+l ° ^ + E E(-1)i+'"V ° £n+l ° z=l j<j i=l j>i
now, with proper reindex and shift, this yields sln+1 o e3n = 4+i ° e%n = en+i ° en_1> both sums are of the same elements but with opposite signs. Hence, dd = 0. □
Exercise 16. Simplicial homology of dA2.
Solution. Chain complex of this simplicial homology is C0 = Z[v0,Vi,V2] = Z © Z © Z, Ci = Z[ao, ai, o2] = Z © Z © Z. So
0 -+ Ci A C0 -> 0,
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where we want to determine d and we know da0 = v2 — vi, dai = v2 Using simple linear algebra, we study generators ker d and im<9:
v0, da2 = v1- v0.
1	0	0	0	-1			( 1	0	0	0	-1	1
0	1	0	-1	0	1	~i	0	1	0	-1	0	1
0	0	1	-1	1			V i	-1	1	0	0	0
therefore ker<9 has a generator a0 — a± + a2 and im<9 has two generators —v\ + v2 and
-v0 + w2. We get H0 = —-■-■--      = —-■-■--       = Z[v0] = Z and
£[-Vi + v2, -v0 + v2\      £[-Vi + v2, -v0 + v2\
Hi = ker d = Z[ao — Oi + a2] = Z. □ Exercise 17. Simplicial complex, model of torus, compute differentials and homology.
Solution. Again, we get simplicial chain complex formed by free abelian groups generated by equivalence classes of simplicies. Note a±,a2 are actually one generator, same for bi,b2. All the vertices are also equivalent. We choose the orientation and fix it.
Thus we get C0 = Z[v] =Z,d = Z[a, b, c] = Z©Z©Z, C2 = Z[e, /] = Z©Z, C3 = 0, and the following holds: da = 0, db = 0, dc = 0, as well as de = a+b—c, df = c—a—b, d(e+f) = 0, so we get ker d = Z[e + /], im<9 = Z[a + b — c].
Let T be the torus. Then
H2{T) =ker<92 = Z[e +/] = Z,
iJi(T) = Z[a, b, c]/Z[a + b - c] = Z^b' ° ^ 6 ~ ^ = Z[a, b]=Z®Z,
Z[a + b- c]
H0(T) = ker d0 = Z.
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First criterion of homotopy equivalence (Hatcher)
Let (X, A) be a pair that satisfies HEP, i.e. A ^ X is a cofibration. Let A be contractible in itself. Then q: X —> X/A is a homotopy equivalence.
Exercise 18. S2 V S1 ~ S2/S° (using First criterion)
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X
Solution. In the picture (hopefully) above, A is a segment as well as B, so contractible in itself. Clearly S2 V S1 = X/B and S2/S° = X/A and X ~ X/A and X ~ X/B by criterion, therefore X/A ~ X/_B and we are done. □
Exercise 19. Let i: A ^ X is a cofibration, show X/A ~IU CA = Ci. (using First criterion)
Solution. We know CA 4lU CA is a cofibration using homework 1, exercise 2, with Y = CA. Then by criterion X U CA ~IU CA/CA Also X/A is homeomorphic to X U CA/CA (see picture above), which concludes the result.
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Exercise 20. Prove the first criterion of homotopy equivalence.
Solution. We take h: Ax I —> A, on Ax {0} it is identity on A and constant on A x {1}.
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X x I
->■ X
qxidj <7 "4^ _ "V*
X/A x J —X/A
and find g: X/A^X. Define f(x,t) = f(x,t), f([x],t) = [f(x,t)]. If we define X/A A, [x]     f(x, 1), then it is well defined. Now we want to show, that the compositions are homotopy equivalent to the identities.
goq ~ idx: g(q(x)) = g([x]) = f(x, 1), just the way we defined it, so / is the homotopy, as /(-, 0) = idx and /(-, 1) = g o q,
q o g ~ idx/A- q(g([x])) = q(f(x, 1)) = [f(x, 1)] = f([x], 1) and idx/A = /(M,0), so in this case the map / is homotopy. □
Exercise 21. Application of the criterion: two types of suspensions, unreduced and reduced. Unreduced suspension: SX = X x I/ ~; where (rri,0) ~ (x2,0), (xi, 1) ~ (x2,2). Reduced suspension: SX = SX/{x0} x I = (X,x0) A (S1, s0) (this might be a homework)
The criterion says, that if {x0} ^ X is a cofibration, then SX ~ SX.
I ~ {(x0,t),t eljcsx —> sx/{x0,t),te 1} = sx
Exercise 22. There is a lemma, that says: Given the following diagram, where rows are long exact sequences and m is iso,
K,
k
Kr.
n—l
n—l
-^n-1
we get a long exact sequence
u t   a^JF   1®K T  honr\°9 u
This might be a homework.
Exercise 23. There is a lemma, that says: Given the following diagram, where rows are long exact sequences and m is an iso
f
K
Lr.
Ln-i
n—l
n—l
n—l
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we get a long exact sequence
u   (.Mi t   mTr   III T   ioisoo~j _v
We can denote d = ho rrT1 o j.
Show exactness in Ln © Kn and also in Ln.
Solution. We have (g — i) o (i, /) = if — gi = 0 obviously. For x G Ln,y G Kn we have (g — i)(x,y) = 0, so g(x) = i(y). Now, let x be such that j(x) = 0, then there is z G Kn such that i(z) = x. Then, suppose g(x) = a G Ln, then by m being iso we know j(a) = 0, so exists y G Kn such that i(y) = a. Since f(z) and y have the same image, their difference has a preimage, i.e. exists b G Mn+1 such that b H> y — f(z). By iso then there exists c H» z, or denote h(c) = z. Now, all of this is much easier with a picture (that I don't draw). Compute now:
f(z + c) = f(z) + y — f[z) = y and i[z + h(c)) = i(z) = x, and we are done.
Exactness in Ln is easier. It holds d o (g — i) = 0, so take x G ker<9 (also, x G Ln). Now, x H> a, by iso there is 6 in the upper row that maps to zero. Then there exists y such that y H> b. Now we can work with x — g(y). There exists also z such that, obviously, mi- g(y) H> a — a = 0. Get x = g(y) + i(z) = g(y) — i(—z), that is we needed to express x as this difference, hence we are done. □
Exercise 24. There is a long exact sequence of the triple (X, A, B), i.e. (B C A C X):
• • • -> tfn(A, 5)       #n(X, 5) ^> #n(X, A) ^> tfn_i(A, £)->••-,
with Hn(X,A) —Hn_i{A) —> Hn_i{A, B). We get this sequence from a special short exact sequence of chain complexes. Show that it is exact and that the triangle commutes, that is     = ja ° <9*.
Solution. The chain complex of a pair is a quotient. We have
C,{A)   », C.(X)   j C.(X)
C* (5)       C* (5) C*(A)
Take c G C*(A), then j«[c] = j[ic] = [jic] = [c], but seen as a different class. So, j o i = 0 and inclusion im j C ker j holds. The other inclusion ker j C im i is obvious.
Analogous: 0 -> C*(A) -> C*(X) -> -> 0.
L*{A)
Now, show £>* equals the composition: Note, if you are familiar with the definition, it's clear.
Take a chain complex in C*(X) with boundary 0, chain in C*(A), take preimage, boundary in C*(B), we have [c] G Cn(X,B) and its image [c] G Cn(X,A), same representative, but different equivalence. Take dc G Cn_i(X, B), it has preimage [dc]^ G Cn_i(A, 5). (Drawing diagram and chase helps.) The equality of -D*, that it is composition, holds, basically thanks to Ja being inclusion. □
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Exercise 25. Apply previous exercise to the triple (Dk, Sk 1, *), where * is a point. Solution.
■ ■ ■ Hn{Sk-\ *) -> Hn(Dk, *) -> Hn(Dk, S'-1) 4 Hn^{Sk-\*) -> tfn_i(£>fc, *)->•••,
and, note Hn(*) = Z for n = 0 and Hn(*) = 0 otherwise.
We also work with reduced homology groups: Hn(X) = Hn(X,x0), H*{X) = H*(X) © #*(*). Since L>fc is contractible, Hn(Dk,*) = Hn(Dk) = 0 and Hn^1(Dk,*) = 0, so we have
• • • Hn(Sk~\ *) -> 0 -> S"8-1) 4 tfn_i(S*-\ *)->()->•••,
where </? is «so.
Reduced homology for pairs is the same as unreduced:
Hn(Dk, S"'1) = Hn(Dk, S"'1) = H^S"-1).
With this we know that Hn(Dk, S^1) = Hn(Dk, S^1) = Z for n = k and it is 0 for n ^ k. Note also, that (Dk, S*-1) ^ (Ak, dAk), this might be useful later on. □
Exercise 26. Show that the chain in Ck(Ak,dAk) given by id: Ak —> Ak is the representative of the generator of
Hk(Ak,dAk) =Z.
(Use induction and the long exact sequence for triple.)
Solution. First denote Afc_1 boundary without interior of one face. Then work with the triple Ak, dAk, Afc_1, so the sequence needed is as follows:
0 -> Hk(Ak, dAk) = Z -> Hk_1(dAk, A*-1) -> 0,
where we have the zeroes because Ak, Afc_1 are contractible to points. Use excision theorem, (X — C, A — C) = H*(X,A), where C is the boundary with bottom cut out (imagine upper part of the letter A, i.e. triangle). We know that Hk-i(Ak~1,dAk~1) = Z and this is isomorphic (by excision theorem) to Hk-i(dAk, Afc_1). So, everything is Z. We want to show that images of idfc_i and id^ are the same, this is actually the inductive step.
Suppose that the generator in ii/'fc_1(Afc~1, <9Afc_1) is given by idfc_x: Afc_1 —> Afc_1. Then idfc: Ak —> Ak is cycle again, represents element [idfc] G Hk(Ak,dAk). The Beginning of the Induction (coming to theaters this summer):
^([-1,1], {-1,1}) ^ ^0({-l, 1}, {1}), take id: [—1,1] —> [—1,1] as a chain complex, 1 — (—1) = [—1], [—1] generator. □
Exercise 27. Using the Mayor-Vietoris exact sequence compute the homology groups of the torus, (note: Vietoris died in 2002, aged 110, remarkable)
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Solution. It goes as union, intersection, pair and we will want to determine the union. First draw two disks with holes (these glued together give a torus). Call one interior of the disk A and the other B. We work with X = A U B, it is not a problem, that we need to work with A, B open, as from the point of view of homology it doesn't matter. The sequence is
----> Hn(A H5)^ Hn(A) © Hn(B) -> Hn(X) -> Hn^(A n B) -> • • • ,
review: X = A U B is torus, A n B = S1 U S1 disjoint union, Hn(A) = Hn(B) = Z for n = 0,1 and 0 otherwise, Fn(A flB) = #„(A) © = Z © Z for n = 0,1 and 0
otherwise. We can therefore continue with this sequence:
H2{A) © H2(B) -> H2(X) -> HX(A n B) A H±(A) © H±(B) -> H±(X) ->
->• H0(A nB)4 #0(^) © H0(B)     H0(X) 0, and we can rewrite it as
0 -> iJ2(X) ^Z©z4z©Z^ H^X)
where we use the fact, that torus is connected, so H0(X) = Z. Now we want to compute H2(X) and H^X).
We know H2(X) is ker /, Z © Z Z © Z, (a, 6) >->• (a + 6, a + b), then (a, -a) is in the kernel, so H2(X) = Z[a, —a] = Z.
For the Hi(X) group use the fact, that kerg is Z (it has same idea, basically). Now consider the sequence
0     Z ^ H^X)     ker ^ = Z 0,
which splits, so H1(X) = Z © Z. We are done. Sphere with two handels might be a homework.
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Exercise 28. Prove that Sn has a nonzero vector field if and only if n is odd.
Solution. First note, that we have v: Sn —> Rn+1 such that v(x) _L x. Consider the case of S1 and (xo,xi) i—> (xi,—x0). Take (x0, xi, x2, x3,..., x2n+i) G S2n+1 C ]R2n. Then we get (xi, —xo,x3, —x2,... ,x2n+i, —x2n) as image and there is nothing more obvious than that the product is zero, i.e. it's perpendicular.
Note, for S1 C C it is z H> ez, where e is the complex unit, usually denoted as i.
Now we want to prove that if Sn has a nonzero vector field, then n is odd. We use the fact that deg(id) = 1 and deg(-id) = (-l)n+1. Take v: Sn -> Sn. If we show id ~ —id, then 1 = (—l)n+1 =>• n is odd. The homotopy is h(x,t) we are looking for is h(x,t) = cos(t)x + sin(t)v(x), where t G [0,7r]. Also note || h(x,t) ||= cos2t + sin2t = 1. We are done. □
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Exercise 29. Prove deg(/) = Yi=i^e&(f/
Solution. We have /: Sn —>• Sn and y G Sn has a neighborhood (nbhd) V. Denote / 1{y) = {xi,... ,xk} and U nbhd's of Xi, UiC\U3■ = 0. Let degree of f/xi be rii. We have/: (Ui,Ui — Xi) -> (V,V - y) and (f/U^: Hn(UhUi - Xi) = Z -> #n(V, V - y) = Z, so z ^ n^. Consider the following diagram, its evolution (i.e. drawing order) was revealed in the tutorial class:
0 Hn(Uh Ut - xt)(^^ Hn(V, V-y)
Hn(Sn,Sn
Hn(Sn,Sn-f-1(y))
'Hn(Sn
■Hn(Sn,Sn-y)
3y*
~^Hn(Sn)
We have jy* iso and h is iso by excision theorem.
So, we have £"-/(?/))     ®Hn{UhUi-Xi) ^ H^OJ^U-x,)). Iso between
Hn(Sn, Sn — f(y)) and Hn(]J^{Ui, U — x,j)) by excision. So k is also iso.
We can take jy* = h o Y(f/U)* ° k^1 o j* and for generator 1 G Hn(Sn) we have (tiy)* ° /*)(!) = deg(/) • 1- We are done.
Y,de9(f/xi
deg( f/xi
□
Exercise 30. Compute homology groups ofWPn using CW-structure.
Solution. Let X be the CW-complex, we have C%w = 0Z[e™], Hn(C^w,d) = Hn(X). Also, d[e™] = Y/3 ^a[e/j_1]> where oPa is the degree of the map
Sn-1 Xn-1 _^ Xn-ll(Xn-1 - enfV) = S^1.
We know that IRPn = e° U e1 U ■ • • U e", cell in every dimension. Attaching map is
/: S'1     (RP)W = RV = S'l/(x ~ -x).
Then have
O^Z^Z^ —^z^z^z^o,
where Z appears n + 1 times. Now, Si 4 KF -> KPVKF-1 = S\ hence we can apply the attaching map. Class [y] has two preimages y, —y, there are nbhd's, deg(f/y) = ±1.
M8130 Algebraic topology, Tutorials
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Note that f / — y = f/y o (—id). Now we can compute the degree: deg(/) = deg(f/y) + deg(// — y) = ±(1 + (—1)*+1). It is ±2 if i is odd and 0 otherwise. So, we can add to the sequence above:
o ^ z ^ z ^ —^zAzAz^o,
and compute for X = RFn, n even: H0(X) = Z, H±(X) = Z2, H2(X) = 0, H^X) = 0 for i even, H^X) = Z2 for i odd. The homology is zero for i > 0. For n even the only difference is Hn(X) = Z.
(Note: this is because for even they are not orientable, for odd it is oriented space.) □
Exercise 31. Compute homology groups of oriented two dimensional surfaces. What about nonorientable?
Solution. Denote Mg surface of genus g (it is the same as sphere with g handles, i.e. Mi is torus and M2 is double torus (homework 4). The CW-model is e° U e\ U • • • U e\g U e2 and we have
0 -> z -> 0z -> Z -> 0.
1
Second differential is zero, e° — e° = 0. The first one is zero as well (glue the model, the arrows go with + and then —). Then we get H0 = Z, Hi =       Z, H2 = Z.
For nonorientable surfaces, Ng is modelled by one 2-dimensional disc which has boundary composed with g segments every of which repeats twice with the same orientation. So we have one cell in dimensions 2 and 0 and g cells in dimension one. We get (quite similarly)
0 -> z -> 0z -> Z -> 0.
This equality holds: <i[e2] = 2[e{] + 2[e2] + ..., so H2 = 0, H0 = Z and the only interesting case is
i-77 r 1
g1= rz^'-"'^]   =z2 + ffiz.
Z[2eJ + • • • + 2ei] Y
□
Exercise 32. iifaue f: Sn ^ Sn map of degree k. (such map always exists). Let X -Dn+1 Uf Sn and compute homology of X and the projection p: X —> X/Sn in homology.
Solution. Easy, X = e° U en U en+1 andO^Z^Z^O^----> 0 ^ Z ^ 0, also
Sn _^ X(n) _^ X(n)/^(n) _ gn) = £n
We get Hn+1(X) = 0,Hn= Zk, H0(X) = Z. Note, X/Sn ^ Sn+1. So, for the    we have
p,: Hn+i{X) = 0 A Hn+1(Sn+1) = Z
M8130 Algebraic topology, Tutorials
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and
p*:Hn(X)=Zk^Hn(S" and at H0 it is identity Z —> Z.
Exercise 33. Prove the following equalities (assuming some conditions):
H*(XVY) = H*(X) © H*{Y)
n n
H*{\J Xi) = ($H*{Xi)
□
i=l i=l oo oo
i=l i=l
Solution. Denote z the distinguished point of X V Y. For the pair (X V Y, X) we have the following long exact sequence
----> Hi+1(X V Y,X) A Hi(X) -> Ht(X V Y) -> ^(X V A ^_x(X)
Thus we have short exact sequence, which splits, because we have continuous (cts) map idVconstz : X\/Y ^ X (which maps Y to z). Thus we have H*(X\/Y) = H*(X)®H*(X\/ Y, X). Now it remains to prove H*{X V Y, X) = H^Y). If X V Y is a CW-complex and X its subcomplex, it is known that H^X VF,I) = Hi(X V Y/X) = Hi(Y). More generally, let U be some (sufficiently small) neighborhood of z in X. From excision theorem we have:
Hi(XVY,X) = Hi(XVY \ (X \ \ (X \ £/)) = Hi(UVY,U).
Because £/ should be1 contractible, H^U V F, £/) = Hi(Y, z) = Hi(Y). The second equality we get from the first by induction.
Let us prove the third equality. Denote Yn = X\ V X2 V • • • V Xn and Y = \J™=1 Yn and denote z the distinguished point of Y and Yn for every n. We have the following diagram (where each arrow is an inclusion):
C*(Yuz)^C*(Y2,z)-C*(Y,z)
Since Ak is compact, every continuous (cts) map Ak —> Y has image in some Yn, thus it is easy to prove C*(Y, z) = colim C*(Fn, z), thus
n oo
(Y) = colim H* (Yn) = colim 0^ = 0^- □
i=i i=i
1It is true at least for X locally contractible. It is not true generally.
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Let X be a topological space with finitely generated homological groups and let H^X) = 0 for each sufficiently large i. Every finitely generated abelian group can be written as Z © Z © • • • © Z © Tor; where Tor denote torsion part of the group.   The number k is
k—times
called the rank of the group.
Euler characteristic x of X is defined by:
x(X) = ^(-l)irank^(X)
i=0
{Z     % = n n Thus x(Sn) = 1 - (-l)n. 0, otherwise.
Exercise 34. Let (C*,<9) be a chain complex with homology iJ*(C*). Prove that x{C*), where
X
(C„) = ^(-l)*rankQ.
i=0
Solution. We have two short exact sequences:
0 -> Zi ^ d 4        -> 0 0^Bt^Zt^ Zt/Bt = Ht^0,
where Ci, cycles Zt and boundaries Bt are free abelian groups, thus rank(7« = rankZj rank5j_i and rankiJj = rankZj — rankBi. Thus we have
= ^(-iyrankZ, + ^(-l)*rank5,_1
j=0 i=0
OO OO
= ^(-l^rankZi - ^(-1)* rank^ = □
j=0 i=0
Let X be a topological space with finitely generated homological groups and let H^X) = 0 for every sufficiently large i. Let f: X —>• X be a continuous map. Map f induces ho-momorphism on the chain complex /*: C*(X) —> C*(X) and on the homologiy groups H*f: H*(X) —> H*(X), where iJ*/(Tor H*(X)) C ToriJ*(X). TTras i£ induces homomor-phism
HJ: H*(X)/TorH*(X)^-H*(X)/TorH*(X). Since H*(X)/Tor H*(X) = Z © Z © ■ ■ ■ © Z; mop iJ*/ can 6e written as a matrix, thus we
rankH»(X)
can compute its trace. So we can define the Lefschetz number of a map f:
L(f) = YJ(-iytrHJ.
i=0
15
M8130 Algebraic topology, Tutorials 2017 Similarly to the case of the Euler characteristic, it can be proved that1
oo oo
X)(-l)ťtr^/ = X;(-l)ťtr/ť. Theorem. If L(f) ^ 0, then f has a fixed point.
Exercise 35. Use the theorem above to show, that every cts map f on Dn and RPn where n is even has a fixed point.
'   % ~ Because H0f: H0(Dn) = Z ->
0, otherwise.
Z = H0(Dn) can be only the identity, we have L(f) = 1, thus / has a fixed point.
{Z,       i = 0, Z/2,   i <n,i odd; and Z/2 is torsion, we have L(f) = 1 as in the 0, otherwise,
previous case. □
Exercise 36. Let M be a smooth compact manifold. Prove, that there is a nonzero vector field on M if and only if x(M) = 0.
Solution. We will prove only implication =K Let v be a nonzero vector field on M. Define a map X: [0,1] x M —> M which satisfies X(t,x) = v(X(t,x)) for every x E M and X(0,x) = x. There exists t0 such that X(t0,x) ^ x. Denote f(x) = X(t0,x), thus / has no fixed point, thus L(f) = 0. Because / is homotopic to id and tiH{íá = rankiJj(M), we get from homotopy invariance 0 = L(f) = L(id) = x(M). □
Exercise 37. Use Z/2 coefficients to show, that every cts map f: Sn —> Sn satisfying f(—x) = —f(x) has an odd degree.
Solution. The map / induces a map g: RPn —> RPn, since f({x, —x}) C {f(x), —f(x)}. We have the short exact sequence3
a i-o"i + o"2 i-2o" = 0
0     C*(MPn, Z/2)     C*(Sn, Z/2)     C*(RPn, Z/2) 0,
where a: A'1 —> RPn is an arbitrary element of C*(RPn), a±,a2 are its preimages of a projection:
Sn
I
A*--*■ RPn
2h: d(X)^Ci(X)
32cr = 0 because of the Z/2 coefficient.
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From the short exact sequence we get the long exact sequence
Hi(RPn; Z/2)-*- Hi(Sn;Z/2)-*■ Hi(RPn; Z/2)-*■ Hi_1(RPn; Z/2)-*- 0
■Hi(Sn; Z/2)
Hi(RPn;Z/2)
■Hi_1(RPn;Z/2)
0
iJi(MPn;Z/2) —
Because H0(RPn; Z/2) = Z/2 and g0 on -^o(^-Pn; Z/2) is an isomorphism, we can show by induction, that Hi(RPr\'L/2) = Z/2 and g,i is an isomorphism for every i < n — 1. An induction step is shown on the following diagram (three isomorphisms imply the fourth):
0-
■ Z/2 Z/2
0-
l
l = t
■Z/2-
■Z/2
For i = n we have the following situation (the vertical isomorphisms were proved by induction):
Z/2-
■Z/2-
-Z/2-
■Z/2-
0
Z/2-
■Z/2-
■Z/2-
■Z/2-
0
Thus /* (the arrow marked by ?) has to be an isomorphism for Hn, thus it maps [1]2 to [1]2, hence / has degree 1 mod 2. □
Exercise 38. Use theorem "f: Sn Sn s.t. f(-x) = that for g: Sn —> Rn there exists x G Sn: g{x) = g(—x).
-fix)     deg(/) is odd" to prove
Solution. We will use proof by contradiction. Suppose that g{x) — g(—x) is always non-
g(xj — g(—x)
zero and define f(x) = jt—---,--tt- Then obviously f(—x) = —f(x). Since /: Sn —)•
\\g(x) - g{-x)\\
S'^1 C Sn is odd , the map / has odd degree, but f(Sn) C Sn which implies deg(/) = 0 (/ is homotopic to a constant map). A contradiction. □
Exercise 39. Let ip G Ck(X;R),tp G Cl(Y;R). Prove 5((pUtp) = ^U^ + (-l)Vuif Use r = [e0,..., ek+i+1] G Ck+i+i(X).
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Solution. Easily work out
k+l+l
% U ^)(r) = (<p U i(>)(5t) = (pU^)(YJ (-l)V[eo, • • •, ^ • • •, efc+z+1]) =
i=0
^(-l)V(T/[eo, • • • ,e,,... ,efe+i])'0(r/[efe+i,... ,efc+m]) + ^ (-l)V(T/[e0, • • •, ek])i/j(T/[ek,..., eu ..., ek+l+1]).
i=k+l
Now, the right hand side of the formula, the first part gives
(dip U xJ})(t) = 6ip(t/[e0,ek+i\) ■ ^(r/[efe+i,..., ek+i+l}) = = f{Sr/[e0,.... efe+i]) ■ ^{r/[ek+u • • •, ek+i+i]) =
k+l
= ^(-l)V(^/[eo, • • •, e.„ ..., ek+l}) ■ ip(r/[ek+1,..., ek+l+1]).
i=0
The second part is
U <ty)(r) = (-l)V(r/[e0, .. . , e,])#(r/[efc,..., efc+i+i]) =
= ^{-^y+k^T/\e0,... ,ek]) ■ i/j(r/[ek,... ,efe+j,... ,efc+i+i]).
j=0
Now, the last summand of the first part plus the first summand of the second part yields
(-l)fc+V(r/[eo,..., efc])V'(r/[efc+1,..., ek+l+1}) + + (-l)k^(r/[e0,.. .,ek])4iT/[ek+1,ek+l+l}) = 0,
and we are done, LHS = RHS.
□
Exercise 40. Prove aUb = A* (a x 6), where a e Hk(X), b e Hl{X), A: X -> X x A", x (->■ (x,x) and x is cross product defined a x (3 = p*xa U py/3 (px.Py are projections from X x Y).
Solution. By the following diagram
X-——*■ X x X
Px='P\/
/ PY=Pl
X X
we have that pxA is id. So, compute
A*(a x b) = A*(p*xa U p*Yb) = A*(pxa) U A*(p*Yb) = (pxAfa U (pYA)*b, and we are done. The thing is that cup product is natural. □
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Exercise 41. 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> If,, la <8> 6, a ® 6, we will denote them l,c,d,c ■ d. Compute
(a ® lb) • (la ® b) = (-l)°"°(a • la) ® (16 • 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) = (-l)n'n(la • a) (8) (6 • lb) = (-l)na ® 6,
so the algebra we get is H*(Sn x Sn) = Z[c, d]/(c2, <i2, dc — (—l)ncd). For n even we have dc = cd. □
Exercise 42. 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, rr) = rr.
Hint: compute m*: H*(Sn) —)• H*(Sn x S'n); describe two rings.
Solution. We have #*(£") = Z[7]/(72) and #*(£" x 5n) = Z[a,/3]/(a2,/32) = #*(£") ® H*(Sn), because we already know, that a/3 = /3a. Our situation can be described with two diagrams:
Sn—i^xS"    H*(Sn)^-H*(Sn x Sn)
Sr'
H*(Sn)
Take m*(7) = aa + b(3 with a, 6 G Z and prove first, that a = b = 1. Use mo %x = id, so «*(m*7) = 7. This gives i*(aa + b/3) = 7 and since i\(aa + 6/3) = 07, we have a = 1, same for 6. Final computation yields
0 = m*(0) = m*(72) = m*(7 U 7) = 771*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. □
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Exercise 43. (unfinished) Use five lemma to prove that taking any two fi's iso's, the third [i is iso as well. (Proving five lemma might be a homework.) Also, show commutativity of the diagram.
H*(X, A)®H*(Y)
■H*(X)®H*(Y)
<$*(g)id
H*(A)®H*(Y)
H*(X xY,AxY)
8*
H*(X x Y)
H*(A 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*. □
Exercise 44 (Five lemma). Let
be a commutative diagram of modules with exact rows. Show that the middle homomorphism f is an isomorphism.
Solution. We will show the surjectivity of /, injectivity is dual. Let c E C be arbitrary and suppose it maps to d E D. This d corresponds to some d E D, and since d maps to 0 in E by exactness, d also has to map to 0 E E by commutativity. By exactness, there exists c E C which maps to d. Since c and f(c) both map to d by commutativity, there is (by exactness) some b E B which maps to f(c) — c. This b corresponds to some b E B, which maps to some d E C. Then by commutativity, f(c — d) = f(d) — (f(d) — c) = c, as desired. (Note that instead of the four vertical maps being isomorphisms, we only needed the surjectivity of B —> B, D —> D and the injectivity of E —> E). □
Exercise 45. Show that for a finite CW-complex X and H*(Y) being finitely generated free group in all dimensions, the cross product
H*(X) ® H*(Y) A H*(X x Y)
is an isomorphism. (In fact, the same is true for X being an infinite CW-complex.)
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Solution. First let X = pt be a point. Then H*(pt) = Z with 1 e H°(pt) and pt x Y is homeomorphic to F, hence
H*(pt) ®H*(Y) = Z® H*(Y) = H*(Y) = H*(pt x Y).
Now let X = pi Up2 U • • • Upk be a finite disjoint union of points (i.e, a discrete set). Then H*{X) = 0tfc=1Z, hence
(k     \ fc fc
0Z ] ®H*(Y) = 0(Z®iJ*(F)) = 0iJ*(F) j=i   / j=i j=i
= F(FurU---UF) = x Y)
n times
(We should also show that the isomorphism is indeed given by //, but if ei,..., ek £ H°(X) are such that
1 if i = j 0 iU^j
it's not hard to see that
ei(Pj)
^(e- 0 a) = (0,0,..., ^^a^ , 0,..., 0)
j-th place
using projections and the definition of the cup product.) Now we can proceed inductively:
i) We now know that the theorem is true for X of dimension 0.
ii) Since Dn is homotopy equivalent to a point, the theorem is also true for X = |Ja=i D™.
iii) Suppose that the theorem holds for finite CW-complexes of dimension n — 1. Then it is also true for pairs (|J -D™, |J S'^1).
iv) Since H*(X,A) = H*(X/A) for a subcomplex A C X, the theorem also holds for
v) Now let X = X^ be an n-dimensional CW-complex and consider the diagram
H*(X, A)®H*(Y)
■H*(X)®H*(Y)
H*(A)®H*(Y)
H*(X x Y,A x Y)
8*
H*(X x Y)
H*(A x Y)
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M8130 Algebraic topology, Tutorials
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where the lower triangle represents a long exact cohomology sequence by definition, and the same is true for the upper triangle (since free modules are flat). We know that the theorem holds for X^1), and also for /X^-V = \J S^T1. Therefore we can use the Five lemma after unfolding the diagram in the appropriate dimensions, from which it follows that the theorem also holds for X^n\ This completes the induction.
□
Exercise 46. Compute the cohomology rings of CP2 x S6 and CP2 V S6.
Solution. We have H*(CP2) = Z[w]/(w3} for w G H2 and H*(S6) = Z[a]/(a2} for a G H&, hence
H*(CP2 x S6) = Z[w]/(w3} © Z[a]/(a2} ^ Z[w, a]/(w3, a2}.
Next, it is true in general that H* (X AY) = H* (X) © H* (Y) is an isomoprhism of graded rings (this can be proven proven straight from the definitions, but it takes some time). Since CP2 V Se is connected, we have H*(CP2 V Se) ^ if (CP2) © H*(Se) © Z. Now w U a G Hs = 0 (more generally, we could use that fact that (w,0) U (0, a) = (0,0)). Therefore
H*(CP2y S6) ^Z[w,a]/(w3,a2,wa).
□
Exercise 47. Show that the CP2 V S6 is not homotopy equivalent to CP3.
Solution. It suffices to show that the cohomology rings of these spaces are not isomorphic (note that the additive group structure is not enough to distinguish them). We already know that
iT(CP2 V S6) ^ Z[w, a]/(w3, a2, wa)
and we have H*(CP3) = Z[b]/(b4) for b G H2. Any isomorphism would have to map w and b to ± each other (these are the respective generators in dimension 2), but w3 = 0 while b3    o, so this is not possible. □
Define the n-th homotopy group of the space X with the base point x0 as the group of homotopy classes of the cts4 maps (In,dln) —> (X,x0) with the operation given by prescription:
U     M '    \g(2«1-l,«2,...,i„)   i < tt < 1.
Denote it nn(X,x0).
Exercise 48. Show the operation on nn(X,x0) is associative.
4 continuous
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Solution. We want to show (/ + g) + k ~ / + (g + k). We will find prescription for the homotopy by the following diagram:
In the following notation5, understand /(ti) as f(t1,t2, ■ ■ ■,tn) for all t2,... ,tn E I.
7(its*i) (Mi) el    (*iG[0,± + s±],*g[0,i])
MMi) H 0(4*1-(l + s))  (Mi) en   (tlG [i + si,i + si],sG [0,1])
.^ii-B     (Mi) GUI   (tiG[i + Si,l]>aG[0,l])
□
Exercise 49. Show that the element given by prescription
(-f)(t1,...,tn) = f(l-t1,t2,...,tn) is really the inverse element of f.
Solution. We want to show / + (—/) ~ const. The constant will be (function given by) the point f(2ti) = /(0). Again let us draw a diagram (the square (its boundary) in the middle sign the same value; the wavy line sign one value too).
h(s, t\
mi
(Mi) el
(-f)m (mi)gii
(-/)(2ti-l) = /(2ti)   (Mi) GUI,
			
	I	7	
j	V	n	
where II = {(s,£i) | s e [0,       e fe*, ±±*]).
□
Remark. One can see, that proving by pictures is much more pleasant. There is a long exact sequence:
•••->• 7rn+i(X, A,x0) nn(A,x0) 7rn(X,x0) nn(X,A,x0) 7rn_i(A,x0) ->• .. Exercise 50. Show the exactness of this sequence in irn(X, A,x0) and nn(A,x0).
3 and some other following notations
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Solution. At first we will show the exactness in nn(X, A, x0).
Let us show the inclusion "imj* C ker<9". Take an arbitrary / G nn(X,x0), thus /: (In,dln) -> (X,x0). From definition = jo/: (Jn, dln, Jn_1) -> (X,A,x0), where jn-i _ jn _        anc[ «9([/]) = [/|/n-i] = const, since / is constant on whole <9/n D 7n_1.
n    tit,—1^
'imj* 3 ker<9": Take an arbitrary g G ker<9 C 7rn(X, A, x0), thus g: (In,dIn,J
(X, A, rr0)- Since g G ker<9, there is the homotopy h: (Jn_1, <9Jn_1) x I —y (A,x0) such that h(x,0) = g\jn-i(x) and h(x, 1) = const. Because h(x,t) G A and h(x',t) = x0 for all x G Jn_1, x G <9/n_1 and £ G [0,1], we can take / G 7rn(X,rr0) defined by
f(x,t)
g(x,2t) forte[0,±] h(x,2t-l) forte[±,l].
It is not hard to prove that j*(/) is homotopic to g, see picture below.
homotopic
Now, let us show the exactness in nn(A,x0).
"im<9 D keri*" Let / G keri* C 7rn(y4,x0) be an arbitrary. Because i*f ~ const, we have homotopy h: (In,dln) x/-> (X,x0) such that h(x,0) = f(x) and h(x, 1) = rr0- It holds h G 7rn+i(X, A, xq), since /i(rr, 0) G A, h(x, 1) = rc0 and h(x',i) = x0 for all x E In and rr' G <9/n.
"im<9 C keri*" Let h E irn+i(X, A, xq) be an arbitrary. Denote h\in = f. Then h gives the homotopy i*f ~ const in (X, x0), since /i(rr, 0) = f{x), h(x, 1) = rr0 and h(x',t) = x0 for all x' E dln and x E In. □
^4 map p: E —>• 5 is called a fibration if it has the homotopy lifting property for all
Dn x {0}-E
Dn x J-
5
7/ p is a fibration then it has the homotopy lifting property also for all pairs (X, A) of
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CW-complexes:
X x {0} U A x I-ä- E
X x I-
Recall that p: E —>• B is a fiber bundle with fibre F if there are open subsets Ua such that B = [Ja Ua and the following diagram commutes for all Ua:
Ua x F p-\Ua)
Exercise 51. Show that every fibre bundle is a fibration.
Solution. At first consider a trivial fibre bundle E = BxE. Take an arbitrary commutative diagram of the form:
Dn x {0}
Dn x I
f
B x F
pr
B
Then h(—, 0) = / and we can define H: Dn x I —> B x F by H(x,t) = (h(x, t), g(x)). One can see that the diagram commutes with H too.
Now, let p: E —> B be an arbitrary fibre bundle with fiber F and B = \Ja Ua. We can take In instead of Dn and consider a diagram:
In x {0}-
■E
In x I
B
Because In is compact, we can divide In x I to finitely many subcubes Cj x Ik where h = [jk,jk+i] such that h(Ci x Ik) C Ua for some a. Since Ua x F —^ B makes a trivial bundle, we can use the same approach as above for each subcube. Since we know -^|cjx{o} = /|cjx{o}, we can find the lift H for all cubes in the first "column" (see the picture below) in the same way as for the trivial case:
Ci x {0} UaxF Ci x I0 —-*■ Ua
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Since we know H\
CiX{ji}
now, we can continue with the second "column":
CiX{h}-
■UnxF
H
pr
CiXh-
Thus, we can proceed through all columns in this way until we will get H on the whole In x I. The illustration of this situation6:
In4
Co
					
					
					
					
ft					
					
Jo-0 TQ Ji     h h
□
Exercise 52. Show the structure of the fibre bundle Sr'
lPn.
->• [x]. Now, we want to find a neigh-
Solution. The fibre is S° = { — 1,1}, since x, —x \-bourhood U of [x] such that p~l{U) = U x S°. Set U = {[x + v] \ v e [x]^} then we have homeomorphism <p: U x S° —> C Sn given by <p([x + v], 1) = and
</?([rr + v], — 1) = ij^jji■ We can cover the whole IRPn by the open subsets Ui = {(x0 : x\ :
• • • : ^n) | Xi ^ 0}. □
Exercise 53. Show the structure of the fibre bundle S2n+1 -4 CPn with the fibre S1.
Solution. Let us look on the special case S1 ^ S3 —> CP1 = S2 called "Hopf fibration". Realise that we can consider S3 C C2, so we can (locally) define the projection S3 —> CP1
by (z1,z2) ^ f2.
In the general case, realise that we can consider S2n+1 C Cn+1. Take U0 = {[z0 : z\ :
• • • : zn] | 7^ 0} C CPn. We can consider U0 = {[1 : z\ : • • • : zn]}. Then the map f/0xS4 S2n+1 is given by
[(1 : Zl : ••• : 2n),eit] i—>
(eit,eitz1,.. .,euzn) \(eit,eitz1,.. .,euzn)\
We can do the same for other Ui from the covering of IRPr
□
3it is drawn as planar, but it should be n-dimensional
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Exercise 54. 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
tt^S1) h tt3(S3) 4 tt3(S2) A tt^S1) -> tt2(S3) -> tt2(S2) -> -> ^(S1) -> Tries'3) -> 7Ti(,S2) -> • • •
and also for Z -> R -> S1 = R/Z, that is 7rn(Z,0) -> irn(R) -> tt^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 -> tt^S1) -> 7T0(Z) -> 7T0(M) = 0,
so Tries'1) = Z. Continue now with the updated long exact sequence:
0 -> 7T3(S3) -> 7T3(S2) -> 0 -> 7T2(S3) -> 7T2(S2) -> tt^S1) = Z -> 7Ti(S3)
and 7Ti(S3) = 0, because 7rfc(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
7r3(S3)-7r3(S2), 7T2(S2)-7n(S1)=Z.
It implies that tt2(S2) = Z.
Let us remark that if 7r3(S3) = Z, then 7r3(S2) = Z. Later we will prove that
7rn(Sn) 9* Hn(Sn) ^ Z.
□
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.
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Exercise 55. 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) -> (5,p(rr0) = 60),
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
y 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.
□
Exercise 56. 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) -> n0(X),
that is
0       7Ti(X)       7Ti(X/G) A 7r0(G) 0.
<S7iow;       d is a group homomorphism.
Solution. First note that 7r0(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)
- d
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),
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d[F] = [f/dln]. The prescription for d: nn(B,b0) -> irn^(F,x0): take /: (In,dln) -> (B,6q) and make a lift F: (In,dln, Jn_1) (f,F,rE0), = F/I^1: (i"-1^/""1 -> (F, £0). The diagram is:
/^xmua/^x/^--f x0
In~x x J--*■ B b0
and pF^-1 x {0}) = b0, so F(/n-1 x {0}) C p-^&o) = F.
Now, ni(X/G) -4- ^(G), /: (I,dl) —y (X/G, [x0]). Take two closed curves w,r at [x0] and lift them to curves u,f m X starting in x0. We denote the end point of u by gix0 and the end point of f by #2^0- Then the operation • (that is not a composition o) is:
uj • t = uj • giT u ■ t(0) = rc0 unT(l/2) = g±xo
urrii) = g2(gix0)
And it's a homomorphism.
□
Exercise 57. Van Kampen theorem - Applications.
Klein bottle K. Model as a square with identified sides as seen in Fig 1,2,3.
>-1
boundary
-
Fig 3
We denote open sets U\^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\ ~ S*1 V S1.
We can compute: n1(U2,x0) = {1} by contractibility, tti(Ui,x0) = 7r1(S'1 V S1,x0) = free group on two generators a, (3 as was already shown in lecture.
Then 7r1(f/1, x0)*7r1(f/2, ^0) = free group on two generators a, (3, also n1(U1f]U2, x0) = 7h.
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Now, the intersection U\ fl U2 ^ (S1,x0) and we take generator u, i2ju(co x) = 1 in ^i(U2,x0), «i)2*(w) =
So, kernel of </? (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 □
Exercise 58. 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 f2 is isomorphism.
Reasoning is the same as in the other direction, the only thing we need to realize
is n0(A, x0) ^ 7r0(X, x0). □
Exercise 59. Show nk(S(X') = 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 = Ui=o ei u e2 (^w0 hemispeheres). Then the following holds: S°° = U,=0 ei u 4- So ^ (S°°)(A° = SN for some iV 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 60. Compute homotopy groups ofRP°°.
Solution. Suprisingly use previous exercise: We can view RP°° 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) -> <Kn(S°°) -> nn(RP°°) A -> ir^S00),
d
where for all n > 2 we have all zeroes, for n = 1 consider 0 —> 7ri(MP°°) —> 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.
□
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Exercise 61. Show that the spaces S2 x RP°° and RP2 have the same homotopy groups but they are not homotopy equivalent.
Solution. Here, also use pre2vious exercise. It is known that nn(X x Y) = 7rn(X) x nn(Y). With this we can compute
n0(S2 x RP°°) = 0, tti^2 x RP°°) = m(S2) x miRP00) = {0} x Z/2 = Z/2,
for n > 2nn(S2 x RP°°) = nn(S2) x {0} ^ 7rn(,S2. Now consider RP2 as S2/z/2, work with the fibration (scheme as follows)
Z/2-- S2-- RP2
7Tn(Z/2) --7rn(S2) --7Tn(MP2) -^7Tn_i(Z/2)
n > 2 0 7rn(,S2)        ^ 7rn(MP2) 0
n = l 0 tt^MP2) =Z/2 0
and n0(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 62. 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-±: Xk-± —> Y and want to define fk: Xk ^ Y. Also, nk-i(Y, •) = 0.
Consider following diagram:
{D\Sk-l)-^{Xk,Xk^)
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,
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Dk x {ljU^-1 x I-Y
Dk x I
and go back to the first diagram:
(Dk,Sk-1)^(Xk,Xk^1)
We extend fk-i to fk: Xk —> Y and proceed to infinity (and beyond), as we always do. □
Exercise 63. 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 ■ f, 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 =>- uji • f ~ uj2 • f, we can imagine the segment as in Figure 2, two curves.
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Map given by uj is a bijection uj : x0 —> xi, 002(^1 f) ~ (^1^2)/, ~ (ww 1)/ — f,
so the map 7rn(X, x0) —>• 7rn(X, rri) is bijection. Figure 3 tries to explain homomorphism.
We get that x0,xi are in the same path component and if uj : x0 —> x\ 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. □
Exercise 64. Using homotopy groups show that M.Pk is not a retract ofM.Pn, n > k > 1.
Solution. Retract: The composition RPk A RPn 4 RPk is the identity, so we have the following diagram and want to use it to obtain a contradiction.
7T,(i)
From the long exact sequence of the fibration, we know that 0 = n1(Sk) —> n1(RPk n0(S°) -^0 and ni(Sk) = 7ii(RPk) for k > 2. Then considering diagram
nk(Sk
0
z
we see that we are factoring identity through zero group and that is Mission Impossible: Factor Zero (in theaters never). Continue with k = 1. We use the knowledge of RP1 = S1. Then our diagram is a triangle with Z, Z/2,Z with identity Z —> Z, factoring identity through finite group is Mission Impossible: Group protocol (in theaters maybe one day, one can only hope) ...and we are done. □
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Exercise 65. Consider the map q: S1 x S1 x S1 —>• S3 defined as a map S1 x S1 x S1 —> D3/S2 where D3 is a small disk in the triple torus which is the identity in the interior of D3 and constant on its complement. Further, consider the Hopf map p: S3 —>• S2 = CP1 (described in Hopf fibration S1 —> S3 —> S2). Compute g* and (pq)* in homotopy groups. Show that pq is not homotopic to a constant map.
Solution. Take following diagram and treat it like your own:
S1 x S1 x S1
S1
S3
s2
(fiber)
c C2
CP1
Zl
where p(z1,z2) = —. We know that only nontrivial homotopy group of the triple torus
is n1(S1 x S1 x S1), but 71"! S"3 is is trivial, so g* is zero in homotopy groups and the composition as well. By the following diagram we have H, thanks to homotopy lifting property, of course, (denote (S1)3 the triple torus S1 x S1 x S1)
1\3
(S1)3 x I
H
V (s0)
s3
(S2,s0)
where h(0,—) is constant map, h(l,—) = pq, p(H(0)) = s0,lmH(0) C p 1(s0) = S1 and H(l) = q ~ H(0). However, H(0)* and g* differ in the third homology groups:
HsiS1) = 0
H3((S
1\3
tlH3(S3) = Z
We are trying to factor through zero, a contradiction.
□
Exercise 66. (detail for the lecture 9. 5. 2017) If (X,A) is relative CW-complex such that there are no cells in dimension < n in X \A, then (X, A) is n-connected.
Solution. Recall the definition of n-connectness of a pair. For [/] G n^X, A, x0), i < n, use cell approximation of /: There is a cell map q: (D\ S'1^1, s0) —> (X,A,x0), such that
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q ~ / relatively S'1'1 and q(D{) C X« = A since = X® = ■■■ = X™ = A. Note
the following very useful criterion:
[/] = 0 in n,t(X, A,x0)^f~q relatively S*'1, g(Dl) = A.
Thus [/] = 0 in our case, and we are done. □
Exercise 67. Let [X, Y] denote a set of homotopy classes of maps from XtoY. If (X, x0) is a CW-complex and Y is path connected, then [X,Y] = [(X,x0), (Y,y0)].
Solution. Surely, [(X,x0), (Y,y0)] C [X,Y] and denote the class in the left set as (g) and [g] the class in the [X, Y]. The map (g) H> [g] is well defined and injective. To prove that it is also surjective take [/] G [X, Y]. Using HEP for / : X x 0 —> Y and a curve uj : x0 x I —> Y which connects f(x0) with y0 we get g : X x 1 —> Y such that / ~ g and g(x0) = y0, then [/] = [g] and (g) G [(X,x0), (Y,y0)]. We are done. □
Exercise 68. (application) We know that deg(f) is an invariant of [Sn,Sn] = nn(Sn). Study [S2n~1,Sn] = 7r2n_i(S'n) and describe its co called Hopf invariant H(f).
Solution. Have /: dD2n = S2n-1 -> Sn and Sn Uf D2n. For / ~ g we have Sn Uf D2n ~ SnUgD2n, moreover SnUfD2n = Cf (the cylinder of/). For n > 2 we have Cf = e°UenUe2n. Using cohomology: H*(Cf) = 7L for * G {0,n,2n} and 0 elsewhere. Take a G Hn(Cf) generator, we have cup product. Then a U a G H2n(Cf) and for f3 G H2n(Cf) we have a U a = H(f)f3, where H(f) is the Hopf invariant. □
Exercise 69. (continuation of previous exercise) For n odd, what can we say in this case about Hopf inviariant? And for n even? Thanks.
Solution. Knowing a U /3 = (—1)^^/3 U a we see that a U a = 0. So for n odd Hopf invariant is zero.
For n even consider the Hopf fibration S1 -> S3 -> S2 = CP1. For CP2 = D4 l)f CP1 (recall how CPn is built up from CPn_1) we have Cf = CP2 and H*(CP2) = Z[a]/(a3}, with ae H2. The generator of H4 is a2. We get that H(f) = 1. □
Exercise 70. Let (X, A) be a pair and show that the following diagram commutes:
nn(X,A,x0)^-^nn(A,x0)
Hn{X,A)^-*Hn^{A)
where d is boundary homomorphism, h is 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 generator in H^S12-1), as we recall in the following: g: S'^1 —> A,g*: Hn_1(Sn^1) —> Hn_1(A) and h[g] = g*(b) G
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Hn_i(A). Let a G Hn(Dn,Sn x) be a generator, that is d*a following diagram:
b. We proceed with the
Hn(Dn, S^1] Hn(X,A)-
■tfn-lOS"1-1)
Hn-i{A)
So, now d*h[f] = <9(/*a) = (f/Sn   )*(<9*a). With <9*a = 6 we conclude that the diagram
commutes.
□
Exercise 71. Homology version of Whitehead theorem, (recall both versions) We show that /*: 7rn(X) —>• 7rn(F) is an iso for all n and then use the previous version of the theorem.
Solution. First, for /: X ^ Y and inclusion. For pair (X, A) we have the following long exact sequence (with Hurewicz):
■7rn(y)
■7ln(X)
■7in(Y,X)
nn+i(Y,X)
Hn+1(Y, X) — Hn(Y)        Hn(X)        Hn(Y, X)
The thing is that h is not always iso, we have —> ni(Y, X) —>, for Y, X simply connected.
h
Now, for 7r2 —> H2 is iso by Hurewicz theorem, then H2 = 0 so n2 = 0 and with the same argument we can proceed inductively.
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!
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(Mf) 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 72. (application of Whitehead theorem) Show that (n — \)-connected compact manifold of dim n is homotopy equivalent to Sn (n > 2).
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Solution. Take manifold M as two parts: disk and its complement. Map the disk to Sn = Dn/Sn~1, that is identity on the disk and complement goes to one point. We have /: M —?■ Sn and for i < n by Hurewicz theorem (M is (n — l)-connected) following iso Hi(M) = Hi(Sn) = 0. The same iso 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)
□
Remark. (Story time) We have that manifold 3-dim (compact) which os pmuy simply connected is homotopy equivalent to S3. There is another result: M 3-dim manifold simply connected is homeomorphic to S3. This other 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 73. Find a map f with Hopf invariant H(f) = 2.
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 iJn(J2) = Z given by a and H2n{ J2) = Z given by b and Hn{Sn x Sn) = Z©Z (generators au a2) and H2n{Sn x Sn) = Z (with b0). Now, p*: H\J2) -> H\Sn x Sn) and p*{a) = at + a2,p*(b) = b0.
a2 = H(f)b p*(a2) = H(f)p*(b) (a± + a2)2 = H(f)bo (a2 + a\a2 + a2d\ + a\) = i7(/)a1a2 2aia2 = H(f)aia2 H(f) = 2
because b0 = a\a2 and by evenness of the dimension a\a2 = a2a\. □
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