M8130 Algebraic topology, tutorial 03, 2024
10.10.2024
Exercise 1. Show dd = 0. Use formula eln+1 04 = 4+i o eln, where i < j. The definition foraeCn(X), a: An ^X, is
n
&T = ^(-l)Vo4. «=0
Solution. Easily workout
n+l n+l
d{da) = d{J2(-iy a o = d(a o 4+1) =
i=0 i=0
n+l
= E(-1)t (0 <+i0 £n + E(-1)i_v 0 <+i ° =
i=0 i<« i>«
n+l n
= E D-ift-i)'* 0 <+i ° 4 + E E(-1)J+"V 0 <+i0
i=l j<« i=l j>i
now, with proper reindex and shift, this yields o eJn = s'n^_1 o eln = e3n+1 o e^-1, both sums are of the same elements but with opposite signs. Hence, dd = 0. □
Exercise 2. Simplicial homology of dA2.
Solution. Chain complex of this simplicial homology is C0 = Z[v0,v1,v2] d = Z[ao, ai, a2] = Z © Z © Z. So
z © z © z,
0 -> Ci ^> Co -> 0,
where we want to determine <9 and we know da0 = v2 — v±, da± = v2 Using simple linear algebra, we study generators ker d and im«9:
v0, da2 = v± - v0.
i	0	0	0	-1			( 1	0	0	0 -1	i
0	i	0	-1	0	1		0	1	0	-1 0	i
0	0	1	-1	1		f 1	V1	-1	1	0 0	0
therefore ker d has a generator a0 — ai + a2 and im<9 has two generators —v\ + i>2 and -w0 + w2- We get #0 = - 1 = ^r - ^ / = Z[w0J = Z and
Z[-wi + v2; Hi = ker d = Z[o0 — ai + a2] = Z.
+ v2]      Z[-v1 + v2, -v0 + v2]
□
Exercise 3. Simplicial complex, model of torus, compute differentials and homology.
M8130 Algebraic topology, tutorial 03, 2024
10.10.2024
Solution. Again, we get simplicial chain complex C* formed by free abelian groups generated by equivalence classes of simplicies. Note ai,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 +/] iJi(T) = Z[a,b,c]/Z[a + b H0(T) = ker d0 = Z.
□
Exercise 4. 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}.
X x I--—> X
qxidj Q ■4*" _ -v"
XIA x I —L^. XIA
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] H> 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 = /(N,0), so in this case the map / is homotopy. □
= Z,
Z\a,b,a + b — c]     ^r ^ ^
-c]=   [' '   ,-^ = Z[a,b]=Z(BZ
Z\a + 0 — c
M8130 Algebraic topology, tutorial 03, 2024
10.10.2024
Exercise 5. S2 V S1 ~ S2/S° (using First criterion)
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/5 by criterion, therefore X/A ~ X/_B and we are done. □
Exercise 6. Let i: A ^ X is a cofibration, show X/A ~ X U CA = Ci. (using First criterion)
Solution. We know CA 4IU CA is a cofibration using homework 1, exercise 2, with Y = CA. Then by criterion X U ~ X U CA/CA Also X/A is homeomorphic to X U CA/CA (see picture above), which concludes the result.
□
Exercise 7. Application of the criterion: two types of suspensions, unreduced and reduced.
M8130 Algebraic topology, tutorial 03, 2024
10.10.2024
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 ~ {(ar0,t),t eljcsx —> SX/{x0,t),te 1} = SX
Exercise 8. Given the following diagram, where rows are long exact sequences and m is an iso
->• Ln —-—> Mn —^—>■ Kn_i -> Ln_i
Kn —^ /.„ —^ .U,
u;e ^e^ a long exact sequence
— I Ln © y Ln    y Kn-i
We can denote <9* = h o m 1 o j.
Show exactness in Ln © Kn and also in Ln.
Solution. We have (g — i) o (i, f) = 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(rr) = 0, then there is z G Kn such that i(z) = x. Then, suppose g{x) = a G Zn, then by m being iso we know j(a) = 0, so exists y E Kn such that i(y) = a. Since /(^) and y have the same image, their difference has a preimage, i.e. exists b G Mn+i such that b H> y — f(z). By iso then there exists c H> z, or denote /i(c) = 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 1—^ 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, z H> x — 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. □