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

3Four major tools of (co)homology

III Algebraic Topology

3.5 Applications

We now use all these tools to do lots of computations. The first thing we compute

will be the homology of spheres.

Theorem. We have

) =

(

Z i = 0, 1

0 otherwise

Proof. We can split S

up as

We want to apply Mayer-Vietoris. We have

∼

R ' ∗, A ∩ B

∼

R ' {p}

{q}.

We obtain

0 0

· · · H

(A ∩ B) H

(A) ⊕ H

(B) H

)

(A ∩ B) H

(A) ⊕ H

(B) H

) 0

Z ⊕ Z Z ⊕ Z Z

∂

A∗

⊕i

B∗

Notice that the map into

(

) is zero. So the kernel of

∂

is trivial, i.e.

∂

is an

injection. So

(

) is isomorphic to the image of

∂

, which is, by exactness, the

kernel of i

A∗

⊕ i

B∗

. So we want to know what this map does.

We know that

(

A ∩ B

)

∼

Z ⊕ Z

is generated by

and

, and the inclusion

map sends each of

and

to the unique connected components of

and

. So

the homology classes are both sent to (1

∈ H

(

)

⊕ H

(

)

∼

Z ⊕ Z

. We then

see that the kernel of

A∗

⊕ i

B∗

is generated by (

p − q

), and is thus isomorphic

to Z. So H

)

∼

By looking higher up the exact sequence, we see that all other homology

groups vanish.

We can do the sphere in general in the same way.

Theorem. For any n ≥ 1, we have

) =

(

Z i = 0, n

0 otherwise

Proof. We again cut up S

A = S

\ {N}

∼

' ∗,

B = S

\ {S}

∼

' ∗,

where N and S are the north and south poles. Moreover, we have

A ∩ B

∼

R × S

n−1

' S

n−1

So we can “induct up” using the Mayer-Vietoris sequence:

· · · H

n−1

) H

(∗) ⊕ H

(∗) H

)

i−1

n−1

) H

i−1

(∗) ⊕ H

i−1

(∗) H

i−1

) · · ·

∂

Now suppose

n ≥

2, as we already did

already. If

i >

1, then

(

∗

) = 0 =

i−1

(∗). So the Mayer-Vietoris map

) H

i−1

n−1

)

∂

is an isomorphism.

All that remains is to look at

= 0

1. The

= 0 case is trivial. For

= 1,

we look at

· · · H

(∗) ⊕ H

(∗) H

)

n−1

) H

(∗) ⊕ H

(∗) H

) 0

Z Z ⊕ Z Z

∂

To conclude that

(

) is trivial, it suffices to show that the map

is injective.

By picking the canonical generators, it is given by 1

7→

1). So we are done.

Corollary.

n 6

, then

n−1

6' S

m−1

, since they have different homology

groups.

Corollary. If n 6= m, then R

∼

Now suppose we have a map

→ S

. As always, it induces a map

∗

(

)

→ H

(

). Since

(

)

∼

, we know the map

∗

is given by

multiplication by some integer. We call this the degree of the map f.

Definition

(Degree of a map)

Let

→ S

be a map. The degree

deg

(

)

is the unique integer such that under the identification

(

)

∼

, the map

∗

is given by multiplication by deg(f).

In particular, for n = 1, the degree is the winding number.

Note that there are two ways we can identify

(

) and

, which differ by

a sign. For the degree to be well-defined, and not just defined up to a sign, we

must ensure that we identify both H

) in the same way.

Using the degree, we can show that certain maps are not homotopic to each

other. We first note the following elementary properties:

Proposition.

(i) deg(id

) = 1.

(ii) If f is not surjective, then deg(f) = 0.

(iii) We have deg(f ◦ g) = (deg f)(deg g).

(iv) Homotopic maps have equal degrees.

Proof.

(i) Obvious.

(ii) If f is not surjective, then f can be factored as

\ {p} S

where

is some point not in the image of

. But

\ {p}

is contractible.

So f

∗

factors as

∗

: H

) H

(∗) = 0 H

) .

So f

∗

is the zero homomorphism, and is thus multiplication by 0.

(iii) This follows from the functoriality of H

(iv) Obvious as well.

As a corollary, we obtain the renowned Brouwer’s fixed point theorem, a

highly non-constructive fixed point existence theorem proved by a constructivist.

Corollary

(Brouwer’s fixed point theorem)

Any map

→ D

has a fixed

point.

Proof.

Suppose

has no fixed point. Define

→ S

n−1

∂D

by taking

the intersection of the ray from f(x) through x with ∂D

. This is continuous.

f(x)

r(x)

Now if x ∈ ∂D

, then r(x) = x. So we have a map

n−1

= ∂D

∂D

= S

n−1

i r

and the composition is the identity. This is a contradiction — contracting

to a point, this gives a homotopy from the identity map

n−1

→ S

n−1

to the

constant map at a point. This is impossible, as the two maps have different

degrees.

A more manual argument to show this would be to apply

n−1

to the chain

of maps above to obtain a contradiction.

We know there is a map of degree 1, namely the identity, and a map of degree

0, namely the constant map. How about other degrees?

Proposition.

A reflection

→ S

about a hyperplane has degree

−

1. As

before, we cover S

A = S

\ {N}

∼

' ∗,

B = S

\ {S}

∼

' ∗,

where we suppose the north and south poles lie in the hyperplane of reflection.

Then both A and B are invariant under the reflection. Consider the diagram

) H

n−1

(A ∩ B) H

n−1

)

) H

n−1

(A ∩ B) H

n−1

)

∂

∼

∗

∼

∗

∂

∼ ∼

where the S

n−1

on the right most column is given by contracting A ∩ B to the

equator. Note that

restricts to a reflection on the equator. By tracing through

the isomorphisms, we see that

deg

(

) =

deg

(

equator

). So by induction, we only

have to consider the case when n = 1. Then we have maps

0 H

) H

(A ∩ B) H

(A) ⊕ H

(B)

0 H

) H

(A ∩ B) H

(A) ⊕ H

(B)

∂

∼

∗

⊕r

∗

∂

∼

Now the middle vertical map sends

p 7→ q

and

q 7→ p

. Since

(

) is given by

the kernel of

(

A ∩ B

)

→ H

(

)

⊕ H

(

), and is generated by

p − q

, we see

that this sends the generator to its negation. So this is given by multiplication

by −1. So the degree is −1.

Corollary. The antipodal map a : S

→ S

given by

a(x

, · · · , x

n+1

) = (−x

, · · · , −x

n+1

)

has degree (−1)

n+1

because it is a composition of (n + 1) reflections.

Corollary

(Hairy ball theorem)

. S

has a nowhere 0 vector field iff

is odd.

More precisely, viewing

⊆ R

n+1

, a vector field on

is a map

→ R

n+1

such that hv(x), xi = 0, i.e. v(x) is perpendicular to x.

Proof. If n is odd, say n = 2k − 1, then

v(x

, y

, x

, y

, · · · , x

, y

) = (y

, −x

, y

, −x

, · · · , y

, −x

)

works.

Conversely, if

→ R

n+1

\ {

}

is a vector field, we let

|v|

→

. We can construct a homotopy from

to the antipodal map by “linear

interpolation”, but in a way so that it stays on the sphere. We let

H : [0, π] × S

→ S

(t, x) 7→ cos(t)x + sin(t)w(x)

Since

(

) and

are perpendicular, it follows that this always has norm 1, so

indeed it stays on the sphere.

Now we have

H(0, x) = x, H(π, x) = −x.

So this gives a homotopy from

, which is a contradiction since they have

different degrees.

So far, we’ve only been talking about spheres all the time. We now move on

to something different.

Example. Let K be the Klein bottle. We cut them up by

Then both A and B are cylinders, and their intersection is two cylinders:

A B

A ∩ B

We have a long exact sequence

0 H

(K) H

(A ∩ B) H

(A) ⊕ H

(B)

(K) H

(A ∩ B) H

(A) ⊕ H

(B)

(K) 0

)

−j

)

−j

We plug in the numbers to get

0 H

(K) Z ⊕ Z Z ⊕ Z

(K) Z ⊕ Z Z ⊕ Z Z 0

)

−j

)

−j

Now let’s look at the first (

, i

) map. We suppose each

is generated by

the left-to-right loops. Then we have

(s) = −1

(t) = 1

(s) = 1

(t) = 1

So the matrix of (i

, i

) is



−1 1

1 1



This has determinant 2, so it is injective and has trivial kernel. So we must have

(K) = 0. We also have

(A) ⊕ H

(B)

im(i

, i

)

ha, bi

h−a + b, a + bi

hai

h2ai

∼

Z/2Z.

Now the second (i

, i

) is given by



1 1



whose kernel is isomorphic to Z. So we have

0 Z/2Z H

(K) Z 0 .

So we have H

(K)

∼

Z ⊕ Z/2Z.

Finally, we look at some applications of relative homology.

Lemma.

Let

be a

-dimensional manifold (i.e. a Hausdorff, second-countable

space locally homeomorphic to R

). Then

(M, M \ {x})

∼

(

Z n = d

0 otherwise

This is known as the local homology.

This gives us a homological way to define the dimension of a space.

Proof.

Let

be an open neighbourhood isomorphic to

that maps

x 7→

We let Z = M \ U . Then

Z = Z ⊆ M \ {x} =

(M \ {x}).

So we can apply excision, and get

(U, U \ {x}) = H

(M \ Z, (M \ {x}) \ Z)

∼

(M, M \ {x}).

So it suffices to do this in the case

∼

and

= 0. The long exact sequence

for relative homology gives

) H

, R

\ {0}) H

n−1

\ {0}) H

n−1

) .

Since H

) = H

n−1

) = 0 for n ≥ 2 large enough, it follows that

, R

\ {0})

∼

n−1

\ {0})

∼

n−1

d−1

and the result follows from our previous computation of the homology of

d−1

We will have to check the lower-degree terms manually, but we shall not.

These relative homology groups are exactly the homology group of spheres.

So we might be able to define a degree-like object for maps between manifolds.

However, there is the subtlety that the identification with

is not necessarily

canonical, and this involves the problem of orientation. For now, we will just

work with S

Definition

(Local degree)

Let

→ S

be a map, and

x ∈ S

. Then

induces a map

∗

: H

, S

\ {x}) → H

, S

\ {f(x)}).

We identify

(

, S

\{x}

)

∼

(

)

∼

for any

via the inclusion

(

)

→

(

, S

\ {x}

), which is an isomorphism by the long exact sequence. Then

∗

is given by multiplication by a constant

deg

(

)

, called the local degree of

What makes the local degree useful is that we can in fact compute the degree

of a map via local degrees.

Theorem. Let f : S

→ S

be a map. Suppose there is a y ∈ S

such that

−1

(y) = {x

, · · · , x

}

is finite. Then

deg(f) =

i=1

deg(f)

Proof.

Note that by excision, instead of computing the local degree at

via

(

, S

\ {x}

), we can pick a neighbourhood

and a neighbourhood

of x

such that f(U

) ⊆ V , and then look at the map

∗

: H

, U

\ {x

}) → H

(V, V \ {y})

instead. Moreover, since

is Hausdorff, we can pick the

such that they are

disjoint. Consider the huge commutative diagram:

) H

)

, S

\ {x

, · · · , x

}) H

, S

\ {y})

(

\ x

))

i=1

, U

\ x

) H

(V, V \ {y})

∗

∼

∗

excision

∗

∼

Consider the generator 1 of

(

). By definition, its image in

(

) is

deg

(

Also, its image in

(

, U

\ {x

}

) is (1

, · · · ,

1). The bottom horizontal map

sends this to

deg(f)

. So by the isomorphisms, it follows that

deg(f) =

i=1

deg(f)