M8130 Algebraic topology, tutorial 04, 2017 16. 3. 2017
Exercise 1. There is a lemma, that says: Given the following diagram, where rows are
long exact sequences and m is an iso
Kn Ln Mn Kn−1 Ln−1 Mn−1
Kn Ln Mn Kn−1 Ln−1 Mn−1
i
f
j
g
h
m
¯i ¯j
we get a long exact sequence
Kn
(i,f)
−→ Ln ⊕ Kn
g−¯i
−→ Ln
i◦iso◦¯j
−→ Kn−1 −→ · · ·
We can denote ∂ = h ◦ m−1
◦ ¯j.
Show exactness in Ln ⊕ ¯Kn and also in ¯Ln.
Solution. We have (g − ¯i) ◦ (i, f) = ¯if − gi = 0 obviously. For x ∈ Ln, y ∈ ¯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 ∈ Kn
such that i(z) = x. Then, suppose g(x) = a ∈ ¯Ln, then by m being iso we know ¯j(a) = 0,
so exists y ∈ ¯Kn such that ¯i(y) = a. Since f(z) and y have the same image, their dierence
has a preimage, i.e. exists b ∈ ¯Mn+1 such that b → y − f(z). By iso then there exists
c → 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 ∂ ◦ (g − ¯i) = 0, so take x ∈ ker ∂ (also, x ∈ ¯Ln).
Now, x → a, by iso there is b in the upper row that maps to zero. Then there exists y
such that y → b. Now we can work with x − g(y). There exists also z such that, obviously,
z → x − g(y) → a − a = 0. Get x = g(y) + ¯i(z) = g(y) − ¯i(−z), that is we needed to
express x as this dierence, hence we are done.
Exercise 2. There is a long exact sequence of the triple (X, A, B), i.e. (B ⊆ A ⊆ X):
· · · → Hn(A, B)
i
−→ Hn(X, B)
jX
−→ Hn(X, A)
D∗
−→ Hn−1(A, B) → · · · ,
with Hn(X, A)
∂∗
−→ Hn−1(A)
jA
−→ Hn−1(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 D∗ = jA ◦ ∂∗.
Solution. The chain complex of a pair is a quotient. We have
0 −→
C∗(A)
C∗(B)
i
−→
C∗(X)
C∗(B)
j
−→
C∗(X)
C∗(A)
→ 0.
Take c ∈ C∗(A), then ji[c] = j[ic] = [jic] = [c], but seen as a dierent class. So, j ◦ i = 0
and inclusion im i ⊆ ker j holds. The other inclusion ker j ⊆ im i is obvious.
M8130 Algebraic topology, tutorial 04, 2017 16. 3. 2017
Analogous: 0 → C∗(A) → C∗(X) →
C∗(X)
C∗(A)
→ 0.
Now, show D∗ equals the composition:
Note, if you are familiar with the denition, 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] ∈ Cn(X, B) and its image [c] ∈ Cn(X, A), same representative,
but dierent equivalence. Take ∂c ∈ Cn−1(X, B), it has preimage [∂c]A,B ∈ Cn−1(A, B).
(Drawing diagram and chase helps.) The equality of D∗, that it is composition, holds,
basically thanks to jA being inclusion.
Exercise 3. Apply previous exercise to the triple (Dk
, Sk−1
, ∗), where ∗ is a point.
Solution.
· · · Hn(Sk−1
, ∗) → Hn(Dk
, ∗) → Hn(Dk
, Sk−1
)
ϕ
→ Hn−1(Sk−1
, ∗) → Hn−1(Dk
, ∗) → · · · ,
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) ⊕
H∗(∗). Since Dk
is contractible, Hn(Dk
, ∗) = ¯Hn(Dk
) = 0 and Hn−1(Dk
, ∗) = 0, so we
have
· · · Hn(Sk−1
, ∗) → 0 → Hn(Dk
, Sk−1
)
ϕ
→ Hn−1(Sk−1
, ∗) → 0 → · · · ,
where ϕ is iso.
Reduced homology for pairs is the same as unreduced:
Hn(Dk
, Sk−1
) = ¯Hn(Dk
, Sk−1
) ∼= ¯Hn−1(Sk−1
).
With this we know that Hn(Dk
, Sk−1
) = ¯Hn(Dk
, Sk−1
) = Z for n = k and it is 0 for n = k.
Note also, that (Dk
, Sk−1
) ∼= (∆k
, ∂∆k
), this might be useful later on.
Exercise 4. Show that the chain in Ck(∆k
, ∂∆k
) given by id: ∆k
→ ∆k
is the representative
of the generator of
Hk(∆k
, ∂∆k
) ∼= Z.
(Use induction and the long exact sequence for triple.)
Solution. First denote ∧k−1
boundary without interior of one face. Then work with the
triple ∆k
, ∂∆k
, ∧k−1
, so the sequence needed is as follows:
0 → Hk(∆k
, ∂∆k
) = Z → Hk−1(∂∆k
, ∧k−1
) → 0,
where we have the zeroes because ∆k
, ∧k−1
are contractible to points. Use excision theorem,
H∗(X − C, A − C) ∼= H∗(X, A), where C is the boundary with bottom cut out (imagine
upper part of the letter ∆, i.e. triangle). We know that Hk−1(∆k−1
, ∂∆k−1
) = Z and this
is isomorphic (by excision theorem) to Hk−1(∂∆k
, ∧k−1
). So, everything is Z. We want to
show that images of idk−1 and idk are the same, this is actually the inductive step.
M8130 Algebraic topology, tutorial 04, 2017 16. 3. 2017
Suppose that the generator in Hk−1(∆k−1
, ∂∆k−1
) is given by idk−1 : ∆k−1
→ ∆k−1
.
Then idk : ∆k
→ ∆k
is cycle again, represents element [idk] ∈ Hk(∆k
, ∂∆k
).
The Beginning of the Induction (coming to theaters this summer):
H1([−1, 1], {−1, 1})
∂∗
→ H0({−1, 1}, {1}),
take id: [−1, 1] → [−1, 1] as a chain complex, 1 − (−1) = [−1], [−1] generator.
Exercise 5. Using the Mayor-Vietoris exact sequence compute the homology groups of the
torus. (note: Vietoris died in 2002, aged 110, remarkable)
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 ∪ 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 ∩ B) → Hn(A) ⊕ Hn(B) → Hn(X) → Hn−1(A ∩ B) → · · · ,
review: X = A ∪ B is torus, A ∩ B = S1
S1
disjoint union, Hn(A) = Hn(B) = Z for
n = 0, 1 and 0 otherwise, Hn(A ∩ B) = Hn(A) ⊕ Hn(B) = Z ⊕ Z for n = 0, 1 and 0
otherwise. We can therefore continue with this sequence:
H2(A) ⊕ H2(B) → H2(X) → H1(A ∩ B)
f
→ H1(A) ⊕ H1(B) → H1(X) →
→ H0(A ∩ B)
g
→ H0(A) ⊕ H0(B) → H0(X) → 0,
and we can rewrite it as
0 → H2(X) → Z ⊕ Z
f
→ Z ⊕ Z → H1(X) →
→ Z ⊕ Z
g
→ Z ⊕ Z → Z → 0,
where we use the fact, that torus is connected, so H0(X) = Z. Now we want to compute
H2(X) and H1(X).
We know H2(X) is ker f, Z ⊕ Z → Z ⊕ Z, (a, b) → (a + b, a + b), then (a, −a) is in the
kernel, so H2(X) = Z[a, −a] = Z.
For the H1(X) group use the fact, that ker g is Z (it has same idea, basically). Now
consider the sequence
0 → Z → H1(X) → ker g = Z → 0,
which splits, so H1(X) = Z ⊕ Z. We are done.
Sphere with two handels might be a homework.