II Algebraic Topology - Simplicial homology

7Simplicial homology

II Algebraic Topology

7.4 Mayer-Vietoris sequence

The Mayer-Vietoris theorem is exactly like the Seifert-van Kampen theorem

for fundamental groups, which tells us what happens when we glue two spaces

together.

Suppose we have a space K = M ∪ N.

M N

We will learn how to compute the homology of the union

M ∪ N

in terms of

those of M, N and M ∩ N.

Recall that to state the Seifert-van Kampen theorem, we needed to learn

some new group-theoretic notions, such as free products with amalgamation.

The situation is somewhat similar here. We will need to learn some algebra in

order to state the Mayer-Vietoris theorem. The objects we need are known as

exact sequences.

Definition (Exact sequence). A pair of homomorphisms of abelian groups

A B C

f g

is exact (at B) if

im f = ker g.

A collection of homomorphisms

· · · A

i+1

i+2

· · ·

i−1

i+1

i+2

is exact at A

ker f

= im f

i−1

We say it is exact if it is exact at every A

Recall that we have seen something similar before. When we defined the

chain complexes, we had

= 0, i.e.

im d ⊆ ker d

. Here we are requiring exact

equivalence, which is something even better.

Algebraically, we can think of an exact sequence as chain complexes with

trivial homology groups. Alternatively, we see the homology groups as measuring

the failure of a sequence to be exact.

There is a particular type of exact sequences that is important.

Definition (Short exact sequence). A short exact sequence is an exact sequence

of the form

0 A B C 0

f g

What does this mean?

–

The kernel of

is equal to the image of the zero map, i.e.

{

}

. So

injective.

–

The image of

is the kernel of the zero map, which is everything. So

surjective.

– im f = ker g.

Since we like chain complexes, we can produce short exact sequences of chain

complexes.

Definition (Short exact sequence of chain complexes). A short exact sequence

of chain complexes is a pair of chain maps i

and j

0 A

such that for each k,

0 A

is exact.

Note that by requiring the maps to be chain maps, we imply that

and

commute with the boundary maps of the chain complexes.

The reason why we care about short exact sequences of chain complexes

(despite them having such a long name) is the following result:

Theorem (Snake lemma). If we have a short exact sequence of complexes

0 A

then a miracle happens to their homology groups. In particular, there is a long

exact sequence (i.e. an exact sequence that is not short)

· · · H

(A) H

(B) H

(C)

n−1

(A) H

n−1

(B) H

n−1

∗

∂

∗

where

∗

and

∗

are induced by

and

, and

∂

∗

is a map we will define in the

proof.

Having an exact sequence is good, since if we know most of the terms in an

exact sequence, we can figure out the remaining ones. To do this, we also need

to understand the maps

∗

and

∂

∗

, but we don’t need to understand all, since

we can deduce some of them with exactness. Yet, we still need to know some of

them, and since they are defined in the proof, you need to remember the proof.

Note, however, if we replace

in the definition of chain groups by a field

(e.g.

), then all the groups become vector spaces. Then everything boils down

to the rank-nullity theorem. Of course, this does not get us the right answer in

exams, since we want to have homology groups over

, and not

, but this helps

us to understand the exact sequences somewhat. If, at any point, homology

groups confuse you, then you can try to work with homology groups over

and

get a feel for what homology groups are like, since this is easier.

We are first going to apply this result to obtain the Mayer-Vietoris theorem.

Theorem (Mayer-Vietoris theorem). Let

K, L, M, N

be simplicial complexes

with K = M ∪ N and L = M ∩ N. We have the following inclusion maps:

L M

N K.

Then there exists some natural homomorphism

∂

∗

(

)

→ H

n−1

(

) that

gives the following long exact sequence:

· · · H

(L) H

(M) ⊕ H

(N) H

(K)

n−1

(L) H

n−1

(M) ⊕ H

n−1

(N) H

n−1

(K) · · ·

· · · H

(M) ⊕ H

(N) H

(K) 0

∂

∗

−`

∗

∂

∗

−`

∗

−`

∗

Here

A ⊕ B

is the direct sum of the two (abelian) groups, which may also be

known as the Cartesian product.

Note that unlike the Seifert-van Kampen theorem, this does not require the

intersection

M ∩ N

to be (path) connected. This restriction was needed

for Seifert-van Kampen since the fundamental group is unable to see things

outside the path component of the basepoint, and hence it does not like non-path

connected spaces well. However, homology groups don’t have these problems.

Proof.

All we have to do is to produce a short exact sequence of complexes. We

have

0 C

(L) C

(M) ⊕ C

(N) C

(K) 0

−`

Here

(

)

→ C

(

)

⊕ C

(

) is the map

x 7→

(

x, x

), while

− `

(

)

⊕ C

(

)

→ C

(

) is the map (

a, b

)

7→ a − b

(after applying the

appropriate inclusion maps).

It is easy to see that this is a short exact sequence of chain complexes. The

image of

is the set of all elements of the form (

x, x

), and the kernel of

− `

is also these. It is also easy to see that

is injective and

− `

surjective.

At first sight, the Mayer-Vietoris theorem might look a bit scary to use, since

it involves all homology groups of all orders at once. However, this is actually

often a good thing, since we can often use this to deduce the higher homology

groups from the lower homology groups.

Yet, to properly apply the Mayer-Vietoris sequence, we need to understand

the map ∂

∗

. To do so, we need to prove the snake lemma.

Theorem (Snake lemma). If we have a short exact sequence of complexes

0 A

then there is a long exact sequence

· · · H

(A) H

(B) H

(C)

n−1

(A) H

n−1

(B) H

n−1

∗

∂

∗

where

∗

and

∗

are induced by

and

, and

∂

∗

is a map we will define in the

proof.

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 this 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 to saying

(

) is in the kernel of

n−1

. Since the diagram is

commutative, we know

n−1

◦ d

(y) = d

◦ j

(y) = d

(x) = 0,

using the fact that

is a cycle. So

(

)

∈ ker j

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 j

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 + j

◦ 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 j

∗

: This follows from the assumption that i

◦ j

= 0.

(b) ker j

∗

⊆ 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 j

(b) = j

◦ 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. ∂

∗

◦ j

∗

= 0.

(d) ker ∂

∗

⊆ im j

∗

: 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

)]) = [j

(b) − j

◦ i

)] = [c].

So [c] ∈ im j

∗

(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

◦ j

n+1

(b) = j

◦ d

n+1

(b) = j

◦ i

(a) = 0.

So [

]

∈ H

(

). Then [

] =

∂

∗

([

]) by definition of

∂

∗

. So [

]

∈

im ∂

∗