III Algebraic Topology - Four major tools of (co)homology

3Four major tools of (co)homology

III Algebraic Topology

3.6 Repaying the technical debt

Finally, we prove all those theorems we stated without proof.

Long exact sequence of relative homology

We start with the least bad one. In relative homology, we had a short exact

sequence of chain complexes

0 C

(A) C

(X) C

(X, A) 0.

The claim is that when we take the homology groups of this, we get a long exact

sequence of homology groups. This is in fact a completely general theorem.

Theorem

(Snake lemma)

Suppose we have a short exact sequence of complexes

0 A

Then there are maps

∂ : H

) → H

n−1

)

such that there is a long exact sequence

· · · H

(A) H

(B) H

(C)

n−1

(A) H

n−1

(B) H

n−1

∗

∂

∗

The method of proving this is sometimes known as “diagram chasing”, where

we just “chase” around commutative diagrams to find the elements we need.

The idea of the proof is as follows — in the short exact sequence, we can think

as a subgroup of

, and

as the quotient

B/A

, by the first isomorphism

theorem. So any element of

can be represented by an element of

. We

apply the boundary map to this representative, and then exactness shows that

this must come from some element of

. We then check carefully that these is

well-defined, i.e. does not depend on the representatives chosen.

Proof.

The proof of this is in general not hard. It just involves a lot of checking

of the details, such as making sure the homomorphisms are well-defined, are

actually homomorphisms, are exact at all the places etc. The only important

and non-trivial part is just the construction of the map ∂

∗

First we look at the following commutative diagram:

0 A

n−1

To construct

∂

∗

(

)

→ H

n−1

(

), let [

]

∈ H

(

) be a class represented

x ∈ Z

(

). We need to find a cycle

z ∈ A

n−1

. By exactness, we know the

map

→ C

is surjective. So there is a

y ∈ B

such that

(

) =

Since our target is

n−1

, we want to move down to the next level. So consider

(

)

∈ B

n−1

. We would be done if

(

) is in the image of

n−1

. By exactness,

this is equivalent saying

(

) is in the kernel of

n−1

. Since the diagram is

commutative, we know

n−1

◦ d

(y) = d

◦ q

(y) = d

(x) = 0,

using the fact that

is a cycle. So

(

)

∈ ker q

n−1

im i

n−1

. Moreover, by

exactness again,

n−1

is injective. So there is a unique

z ∈ A

n−1

such that

n−1

(z) = d

(y). We have now produced our z.

We are not done. We have

∂

∗

[

] = [

] as our candidate definition, but we

need to check many things:

(i) We need to make sure ∂

∗

is indeed a homomorphism.

(ii) We need d

n−1

(z) = 0 so that [z] ∈ H

n−1

(A);

(iii)

We need to check [

] is well-defined, i.e. it does not depend on our choice

of y and x for the homology class [x].

(iv) We need to check the exactness of the resulting sequence.

We now check them one by one:

(i)

Since everything involved in defining

∂

∗

are homomorphisms, it follows

that ∂

∗

is also a homomorphism.

(ii) We check d

n−1

(z) = 0. To do so, we need to add an additional layer.

0 A

n−1

0 A

n−2

n−1

n−2

We want to check that

n−1

(

) = 0. We will use the commutativity of the

diagram. In particular, we know

n−2

◦ d

n−1

(z) = d

n−1

◦ i

n−1

(z) = d

n−1

◦ d

(y) = 0.

By exactness at

n−2

, we know

n−2

is injective. So we must have

n−1

(z) = 0.

(iii) (a)

First, in the proof, suppose we picked a different

such that

(

) =

(

) =

. Then

(

− y

) = 0. So

− y ∈ ker q

im i

. Let

a ∈ A

be such that i

(a) = y

− y. Then

) = d

− y) + d

(y)

= d

◦ i

(a) + d

(y)

= i

n−1

◦ d

(a) + d

(y).

Hence when we pull back

(

) and

(

) to

n−1

, the results differ

by the boundary

(

), and hence produce the same homology class.

(b)

Suppose [

] = [

]. We want to show that

∂

∗

[

] =

∂

∗

[

]. This time,

we add a layer above.

0 A

n+1

0 A

n−1

n+1

n−1

By definition, since [x

] = [x], there is some c ∈ C

n+1

such that

= x + d

n+1

(c).

By surjectivity of

n+1

, we can write

n+1

(

) for some

b ∈ B

n+1

By commutativity of the squares, we know

= x + q

◦ d

n+1

(b).

The next step of the proof is to find some

such that

(

) =

Then

(y + d

n+1

(b)) = x

So the corresponding

n+1

(

). So

(

) =

(

), and

hence ∂

∗

[x] = ∂

∗

(iv)

This is yet another standard diagram chasing argument. When reading

this, it is helpful to look at a diagram and see how the elements are chased

along. It is even more beneficial to attempt to prove this yourself.

(a) im i

∗

⊆ ker q

∗

: This follows from the assumption that i

◦ q

= 0.

(b) ker q

∗

⊆ im i

∗

: Let [

]

∈ H

(

). Suppose

∗

([

]) = 0. Then there is

some

c ∈ C

n+1

such that

(

) =

n+1

(

). By surjectivity of

n+1

there is some

∈ B

n+1

such that

n+1

(

) =

. By commutativity,

we know q

(b) = q

◦ d

n+1

), i.e.

(b − d

n+1

)) = 0.

By exactness of the sequence, we know there is some

a ∈ A

such

that

(a) = b − d

n+1

Moreover,

n−1

◦ d

(a) = d

◦ i

(a) = d

(b − d

n+1

)) = 0,

using the fact that

is a cycle. Since

n−1

is injective, it follows that

(a) = 0. So [a] ∈ H

(A). Then

∗

([a]) = [b] − [d

n+1

)] = [b].

So [b] ∈ im i

∗

⊆ ker ∂

∗

: Let [

]

∈ H

(

). To compute

∂

∗

(

∗

([

])), we first

pull back

(

) to

b ∈ B

. Then we compute

(

) and then pull it

back to

n+1

. However, we know

(

) = 0 since

is a cycle. So

∂

∗

([b])) = 0, i.e. ∂

∗

◦ q

∗

= 0.

(d) ker ∂

∗

⊆ im q

∗

: Let [

]

∈ H

(

) and suppose

∂

∗

([

]) = 0. Let

b ∈ B

be such that

(

) =

, and

a ∈ A

n−1

such that

n−1

(

) =

(

By assumption,

∂

∗

([

]) = [

] = 0. So we know

is a boundary,

say

(

) for some

∈ A

. Then by commutativity we know

(b) = d

◦ i

). In other words,

(b − i

)) = 0.

So [b − i

)] ∈ H

(B). Moreover,

∗

([b − i

)]) = [q

(b) − q

◦ i

)] = [c].

So [c] ∈ im q

∗

(e) im ∂

∗

⊆ ker i

∗

: Let [

]

∈ H

(

). Let

b ∈ B

be such that

(

) =

and a ∈ A

n−1

be such that i

(a) = d

(b). Then ∂

∗

([c]) = [a]. Then

∗

([a]) = [i

(a)] = [d

(b)] = 0.

So i

∗

◦ ∂

∗

= 0.

(f) ker i

∗

⊆ im ∂

∗

: Let [

]

∈ H

(

) and suppose

∗

([

]) = 0. So we can

find some

b ∈ B

n+1

such that

(

) =

n+1

(

). Let

n+1

(

). Then

n+1

◦ q

n+1

(b) = q

◦ d

n+1

(b) = q

◦ i

(a) = 0.

So [

]

∈ H

(

). Then [

] =

∂

∗

([

]) by definition of

∂

∗

. So [

]

∈

im ∂

∗

Another piece of useful algebra is known as the 5-lemma:

Lemma (Five lemma). Consider the following commutative diagram:

A B C D E

p q

r s t u

If the two rows are exact,

and

are isomorphisms,

is injective and



surjective, then n is also an isomorphism.

Proof. The philosophy is exactly the same as last time.

We first show that

is surjective. Let

∈ C

. Then we obtain

(

)

∈ D

Since

is an isomorphism, we can find

d ∈ D

such that

(

) =

. Then we have

q(j(d)) = u(p(d)) = u(f(c

)) = 0.

Since

is injective, we know

(

) = 0. Since the sequence is exact, there is some

c ∈ C such that h(c) = d.

We are not yet done. We do not know that

(

) =

. All we know is that

(

)) =

(

). So

(

− n

(

)) = 0. By exactness at

, we can find some

such that

(

) =

(

)

− c

. Since

was surjective, we can find

b ∈ B

such that

m(b) = b

. Then we have

n(g(b)) = n(c) − c

So we have

n(c − g(b)) = c

So n is surjective.

Showing that n is injective is similar.

Corollary.

Let

: (

X, A

)

→

(

Y, B

) be a map of pairs, and that any two

∗

(

X, A

)

→ H

∗

(

Y, B

∗

(

)

→ H

∗

(

) and

∗

(

)

→ H

∗

(

) are

isomorphisms. Then the third is also an isomorphism.

Proof. Follows from the long exact sequence and the five lemma.

That wasn’t too bad, as it is just pure algebra.

Proof of homotopy invariance

The next goal is want to show that homotopy of continuous maps does not affect

the induced map on the homology groups. We will do this by showing that

homotopies of maps induce homotopies of chain complexes, and chain homotopic

maps induce the same map on homology groups. To make sense of this, we need

to know what it means to be a homotopy of chain complexes.

Definition

(Chain homotopy)

A chain homotopy between chain maps

, g

→ D

is a collection of homomorphisms F

: C

→ D

n+1

such that

− f

= d

n+1

◦ F

+ F

n−1

◦ d

: C

→ D

for all n.

The idea of the chain homotopy is that

(

) gives us an

+1 simplex whose

boundary is g

− f

, plus some terms arising from the boundary of c itself:

(σ)

+ F

n−1

(dσ)

We will not attempt to justify the signs appearing in the definition; they are

what are needed for it to work.

The relevance of this definition is the following result:

Lemma. If f

and g

are chain homotopic, then f

∗

= g

∗

: H

∗

) → H

∗

Proof. Let [c] ∈ H

). Then we have

n+1

n−1

(c)) = d

n+1

(c),

where the second term dies because

is a cycle. So we have [

(

)] = [

(

)].

That was the easy part. What we need to do now is to show that homotopy

of maps between spaces gives a chain homotopy between the corresponding chain

maps.

We will change notation a bit.

Notation. From now on, we will just write d for d

For

X → Y

, we will write

(

)

→ C

(

) for the map

σ 7→ f ◦ σ

i.e. what we used to call f

Now if

: [0

× X → Y

is a homotopy from

, and

: ∆

→ X

is an

n-chain, then we get a homotopy

[0, 1] × ∆

[0, 1] × X Y

[0,1]×σ

from

(

) to

(

). Note that we write [0

1] for the identity map [0

→

1].

The idea is that we are going to cut up [0

∆

into

+ 1-simplices.

Suppose we can find a collection of chains

∈ C

n+1

([0

∆

) for

n ≥

0 such

that

d(P

) = i

− i

−

j=0

(−1)

([0, 1] × δ

)

n−1

where

: δ

= {0} × ∆

→ [0, 1] × ∆

: δ

= {0} × ∆

→ [0, 1] × ∆

and

: ∆

n−1

→

∆

is the inclusion of the

th face. These are “prisms”

connecting the top and bottom face. Intuitively, the prism P

looks like this:

and the formula tells us its boundary is the top and bottom triangles, plus the

side faces given by the prisms of the edges.

Suppose we managed to find such prisms. We can then define

: C

(X) → C

n+1

(Y )

by sending

(σ : ∆

→ X) 7→ (H ◦ ([0, 1] × σ))

We now calculate.

(σ) = d((H ◦ (1 × ∆

))

= (H ◦ ([0, 1] × σ))

(d(P

))

= (H ◦ ([0, 1] × σ))





− i

−

j=0

(−1)

([0, 1] × δ

)

n−1

)





= H ◦ ([0, 1] × σ) ◦ i

− H ◦ ([0, 1] × σ) ◦ i

−

j=0

(−1)

◦ (([0, 1] × σ) ◦ δ

)

n−1

)

= g ◦ σ − f ◦ σ −

j=0

(−1)

◦ (([0, 1] × σ) ◦ δ

)

n−1

)

= g ◦ σ − f ◦ σ − F

n−1

(dσ)

= g

(σ) − f

(σ) − F

n−1

(dσ).

So we just have to show that

exists. We already had a picture of what it looks

like, so we just need to find a formula that represents it. We view [0

∆

∈

R × R

n+1

. Write

, v

, · · · , v

}

for the vertices of

{

} ×

∆

⊆

∆

, and

, · · · , w

} for the corresponding vertices of {1} × ∆

Now if {x

, x

, · · · , x

n+1

} ⊆ {v

, · · · , v

} ∪ {w

, · · · , w

}, we let

, · · · , x

] : ∆

→ [0, 1] × ∆

, · · · , t

n+1

) =

This is still in the space by convexity. We let

i=0

(−1)

, v

, · · · , v

, w

i+1

, · · · , w

] ∈ C

n+1

([0, 1] × ∆

It is a boring check that this actually works, and we shall not bore the reader

with the details.

Proof of excision and Mayer-Vietoris

Finally, we prove excision and Mayer-Vietoris together. It turns out both follow

easily from what we call the “small simplices theorem”.

Definition

(

) and

(

))

We let

}

α∈I

be a collection of

subspaces of X such that their interiors cover X, i.e.

[

α∈I

= X.

Let

(

)

⊆ C

(

) be the subgroup generated by those singular

-simplices

: ∆

→ X

such that

(∆

)

⊆ U

for some

. It is clear that if

lies in

then so do its faces. So C

(X) is a sub-chain complex of C

(X).

We write H

(X) = H

(X)).

It would be annoying if each choice of open cover gives a different homology

theory, because this would be too many homology theories to think about. The

small simplices theorem says that the natural map

∗

(

)

→ H

∗

(

) is an

isomorphism.

Theorem

(Small simplices theorem)

The natural map

∗

(

)

→ H

∗

(

) is an

isomorphism.

The idea is that we can cut up each simplex into smaller parts by barycentric

subdivision, and if we do it enough, it will eventually lie in on of the open covers.

We then go on to prove that cutting it up does not change homology.

Proving it is not hard, but technically annoying. So we first use this to

deduce our theorems.

Proof of Mayer-Vietoris.

Let

A ∪ B

, with

A, B

open in

. We let

{A, B}, and write C

(A + B) = C

(X). Then we have a natural chain map

(A) ⊕ C

(B) C

(A + B)

−j

that is surjective. The kernel consists of (

x, y

) such that

(

)

− j

(

) = 0,

i.e.

(

) =

(

). But

doesn’t really do anything. It just forgets that the

simplices lie in

. So this means

is a chain in

A ∩ B

. We thus deduce

that we have a short exact sequence of chain complexes

(A ∩ B) C

(A) ⊕ C

(B) C

(A + B).

)

−j

Then the snake lemma shows that we obtain a long exact sequence of homology

groups. So we get a long exact sequence of homology groups

· · · H

(A ∩ B) H

(A) ⊕ H

(B) H

(X) · · ·

)

−j

By the small simplices theorem, we can replace

(

) with

(

). So we

obtain Mayer-Vietoris.

Now what does the boundary map

∂

(

)

→ H

n−1

(

A ∩ B

) do? Suppose

we have

c ∈ H

(

) represented by a cycle

b ∈ C

(

), with

supported in

and

supported in

. By the small simplices theorem, such a representative

always exists. Then the proof of the snake lemma says that

∂

([

]) is given

by tracing through

(A) ⊕ C

(B) C

(A + B)

n−1

(A ∩ B) C

n−1

(A) ⊕ C

n−1

(B)

−j

)

We now pull back

along

− j

to obtain (

a, −b

), then apply

to obtain

(da, −db). Then the required object is [da] = [−db].

We now move on to prove excision.

Proof of excision.

Let

X ⊇ A ⊇ Z

be such that

Z ⊇

. Let

X \ Z

. Then

again take

U = {A, B}.

By assumption, their interiors cover X. We consider the short exact sequences

0 C

(A) C

(A + B) C

(A + B)/C

(A) 0

0 C

(A) C

(X) C

(X, A) 0

Looking at the induced map between long exact sequences on homology, the

middle and left terms induce isomorphisms, so the right term does too by the

5-lemma.

On the other hand, the map

(B)/C

(A ∩ B) C

(A + B)/C

(A)

is an isomorphism of chain complexes. Since their homologies are

(

B, A ∩ B

)

and

(

X, A

), we infer they the two are isomorphic. Recalling that

X \

we have shown that

∗

(X \ Z, A \ Z)

∼

∗

(X, A).

We now provide a sketch proof of the small simplices theorem. As mentioned,

the idea is to cut our simplices up, and one method to do so is barycentric

subdivision.

Given a 0-simplex {v

}, its barycentric subdivision is itself.

If x = {x

, · · · , x

} ⊆ R

spans an n-simplex σ, we let

n + 1

i=0

be its barycenter .

If we have a 1-simplex

Then the barycentric subdivision is obtained as

We can degenerately describe this as first barycentrically subdividing the bound-

ary (which does nothing in this case), and then add the barycenter.

In the case of a 2-simplex:

we first barycentrically subdivide the boundary:

Then add the barycenter

, and for each standard simplex in the boundary, we

“cone it off” towards b

More formally, in the standard

-simplex ∆

⊆ R

n+1

, we let

be its barycenter.

For each singular i-simplex σ : ∆

→ ∆

, we define

Cone

∆

(σ) : ∆

i+1

→ ∆

, t

, · · · , t

i+1

) 7→ t

+ (1 − t

) · σ



, · · · , t

i+1

)

1 − t



We can then extend linearly to get a map Cone

∆

: C

(∆

) → C

i+1

(∆

Example. In the 2-simplex

the cone of the bottom edge is the simplex in orange:

Since this increases the dimension, we might think this is a chain map. Then

for i > 0, we have

dCone

∆

(σ) =

i+1

j=0

Cone

∆

(σ) ◦ δ

= σ +

i+1

j=1

(−1)

Cone

∆

i−1

(σ ◦ δ

j−1

)

= σ − Cone

∆

i−1

(dσ).

For i = 0, we get

dCone

∆

(σ) = σ − ε(σ) · b

In total, we have

dCone

∆

+ Cone

∆

i−1

d = id − c

where c

= 0 for i > 0, and c

(σ) = ε(σ)b

is a map C

(∆

) → C

(∆

We now use this cone map to construct a barycentric subdivision map

(

)

→ C

(

), and insist that it is natural that if

X → Y

is a map,

then f

◦ ρ

= ρ

◦ f

, i.e. the diagram

(X) C

(X)

(Y ) C

(Y )

So if

: ∆

→ X

, we let

: ∆

→

∆

∈ C

(∆

) be the identity map. Then

we must have

(σ) = ρ

(σ

) = σ

∆

(ι

So if know how to do barycentric subdivision for

itself, then by naturality, we

have defined it for all spaces! Naturality makes life easier for us, not harder!

So we define ρ

recursively on n, for all spaces X at once, by

(i) ρ

= id

(X)

(ii) For n > 0, we define the barycentric subdivision of ι

∆

(ι

) = Cone

∆

n−1

(ρ

∆

n−1

(dι

)),

and then extend by naturality.

This has all the expected properties:

Lemma. ρ

is a natural chain map.

Lemma. ρ

is chain homotopic to the identity.

Proof. No one cares.

Lemma.

The diameter of each subdivided simplex in (

∆

)

(

) is bounded by



n+1



diam(∆

Proof. Basic geometry.

Proposition. If c ∈ C

(X), then p

(X).

Moreover, if c ∈ C

(X), then there is some k such that (ρ

)

(X).

Proof.

The first part is clear. For the second part, note that every chain is a

finite sum of simplices. So we only have to check it for single simplices. We let

σ be a simplex, and let

V = {σ

−1

}

be an open cover of ∆

. By the Lebesgue number lemma, there is some

ε >

such that any set of diameter

< ε

is contained in some

−1

. So we can choose

k >

0 such that (

∆

)(

) is a sum of simplices which each has diameter

< ε

So each lies in some σ

−1

. So

(ρ

∆

)

(ι

) = C

(∆

So applying σ tells us

(ρ

∆

)

(σ) ∈ C

(X).

Finally, we get to the theorem.

Theorem

(Small simplices theorem)

The natural map

∗

(

)

→ H

∗

(

)

is an isomorphism.

Proof.

Let [

]

∈ H

(

). By the proposition, there is some

k >

0 such that

(

)

(

)

∈ C

(

). We know that

is chain homotopic to the identity. Thus

so is (ρ

)

. So [(ρ

)

(c)] = [c]. So the map H

(X) → H

(X) is surjective.

To show that it is injective, we suppose

([

]) = 0. Then we can find some

z ∈ H

n+1

(

) such that

. We can then similarly subdivide

enough such

that it lies in C

n+1

(X). So this shows that [c] = 0 ∈ H

(X).

That’s it. We move on to (slightly) more interesting stuff. The next few

sections will all be slightly short, as we touch on various different ideas.