M8130 Algebraic topology, tutorial 03, 2017
9.3.2017
Exercise 1. Show dd = 0. Use formula e%n+l o eJn = e£+i o eln, where i < j. The definition for a E Cn{X), a: An -> X, is
9a = ^(-l)Vo4.
Solution. Easily workout
n+l n+1
d(da) = dfei-l)' a o 4+1) = £(-!)< <9(a o 4+i) =
i=0 «=0
= E(-^ ( E("!)V ° 4+1 ° £n + E(-1)'"V ° 4+1 ° ^0 =
i=0 i<« i>«
n+l «
i=l j<i i=l j>i
now, with proper reindex and shift, this yields eln+x o eJn = 4+i ° en = £n+i ° £nT1) both sums are of the same elements but with opposite signs. Hence, dd = 0. □
Exercise 2. Simplicial homology of OA2.
■i
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 d and we know <9ao = v2 — vi, da± = 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
(J	1	0	-1	0	1		0	1	0	-1	0	1
0	(J	1	-1	1		1	Vi	-1	1	0	0	0
therefore ker «9 has a generator
-v0 + v2. We get H0 = —-■-
Z[-v1 + v2,-v0
Hi = ker d = Z[a0 — ax + a2] = Z.
a2 and im<9 has two generators
Z[-«i + w2,^o + «2,^2]
—«i + v2 and Z[t>0] = Z and □
v2]      Z[-v1 + v2,-v0 + v2] Exercise 3. Simplicial complex, model of torus, compute differentials and homology.
M8130 Algebraic topology, tutorial 03, 2017
9.3.2017
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.
= Z,
Z\a,b,a + b — c]     ^r ^ ^
-c]=   [' '   ,-^ = Z[a,b]=Z(BZ
Z\a + o — c
□
M8130 Algebraic topology, tutorial 03, 2017
9.3.2017
First criterion of homotopy equivalence (Hatcher)
Let (X, A) be a pair that satisfies HEP, i.e. A A- X is a cofibration. Let A be contractible in itself. Then q: X —> X/A is a homotopy equivalence.
Exercise 4. 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/B by criterion, therefore X/A ~ X/B and we are done. □
Exercise 5. Let i: A ^ X is a cofibration, show X/A ~ X U 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.
□
M8130 Algebraic topology, tutorial 03, 2017
9.3.2017
Exercise 6. 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
qxidi
X
g
X/A x I —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] 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 = /(M,0), so in this case the map / is homotopy. □
Exercise 7. 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 8. There is a lemma, that says: Given the following diagram, where rows are long exact sequences and m is iso,
->• Lr.
K,
k
we get a long exact sequence
n—l
-> L
n—l
M,
n—l
-^n-1
This might be a homework.