III Algebraic Topology - Cell complexes

5Cell complexes

III Algebraic Topology

5 Cell complexes

So far, everything we’ve said is true for arbitrary spaces. This includes, for

example, the topological space with three points

a, b, c

, whose topology is

{∅, {a}, {a, b, c}}

. However, these spaces are horrible. We want to restrict

our attention to nice spaces. Spaces that do feel like actual, genuine spaces.

The best kinds of space we can imagine would be manifolds, but that is a

bit too strong a condition. For example, the union of the two axes in R

is not

a manifold, but it is still a sensible space to talk about. Perhaps we can just

impose conditions like Hausdorffness and maybe second countability, but we can

still produce nasty spaces that satisfy these properties.

So the idea is to provide a method to build spaces, and then say we only

consider spaces built this way. These are known as cell complexes, or CW

complexes

Definition

(Cell complex)

A cell complex is any space built out of the following

procedure:

(i) Start with a discrete space X

. The set of points in X

are called I

(ii)

n−1

has been constructed, then we may choose a family of maps

{ϕ

: S

n−1

→ X

n−1

}

α∈I

, and set

n−1

α∈I

/{x ∈ ∂D

∼ ϕ

(x) ∈ X

n−1

We call

the

-skeleton of

We call the image of

\ ∂D

the

open cell e

(iii) Finally, we define

X =

[

n≥0

with the weak topology, namely that

A ⊆ X

is open if

A ∩ X

is open in

for all n.

We write Φ

→ X

for the obvious inclusion map. This is called the

characteristic map for the cell e

Definition

(Finite-dimensional cell complex)

for some

, we say

X is finite-dimensional.

Definition

(Finite cell complex)

is finite-dimensional and

are all finite,

then we say X is finite.

Definition

(Subcomplex)

A subcomplex

is a simplex obtained by using

a subset I

⊆ I

Note that we cannot simply throw away some cells to get a subcomplex, as

the higher cells might want to map into the cells you have thrown away, and

you need to remove them as well.

We note the following technical result without proof:

Lemma. If A ⊆ X is a subcomplex, then the pair (X, A) is good.

Proof. See Hatcher 0.16.

Corollary. If A ⊆ X is a subcomplex, then

(X, A)

(X/A)

∼

is an isomorphism.

We are next going to show that we can directly compute the cohomology of

a cell complex by looking at the cell structures, instead of going through all the

previous rather ad-hoc mess we’ve been through. We start with the following

lemma:

Lemma. Let X be a cell complex. Then

(i)

, X

n−1

) =

(

0 i 6= n

i∈I

Z i = n

(ii) H

) = 0 for all i > n.

(iii) H

) → H

(X) is an isomorphism for i < n.

Proof.

(i) As (X

, X

n−1

) is good, we have an isomorphism

, X

n−1

)

n−1

)

∼

But we have

n−1

∼

α∈I

the space obtained from

α∈I

by collapsing down the subspace

α ∈ I

}

, where each

is the south pole of the sphere. To

compute the homology of the wedge

n−1

, we then note that (

Y, Z

)

is good, and so we have a long exact sequence

(Z) H

(Y )

(Y/Z) H

i−1

(Z) H

i−1

(Y ) .

Since

(

) vanishes for

i ≥

1, the result then follows from the homology

of the spheres plus the fact that H

(

) =

(ii) This follows by induction on n. We have (part of) a long exact sequence

n−1

) H

, X

n−1

)

We know the first term vanishes by induction, and the third term vanishes

for i > n. So it follows that H

) vanishes.

(iii)

To avoid doing too much point-set topology, we suppose

is finite-

dimensional, so

for some

. Then we have a long exact sequence

i+1

n+1

, X

) H

n+1

) H

n+1

, X

)

Now if

i < n

, we know the first and last groups vanish. So we have

)

∼

n+1

). By continuing, we know that

)

∼

n+1

)

∼

n+2

)

∼

· · ·

∼

) = H

(X).

To prove this for the general case, we need to use the fact that any map

from a compact space to a cell complex hits only finitely many cells, and

then the result would follow from this special case.

For a cell complex X, let

cell

(X) = H

, X

n−1

)

∼

α∈I

We define d

cell

: C

cell

(X) → C

cell

n−1

(X) by the composition

, X

n−1

) H

n−1

) H

n−1

, X

n−2

)

∂

We consider

0 H

n+1

)

n+1

, X

) H

, X

n−1

) H

n−1

, X

n−2

)

n−1

)

0 H

n−1

)

∂

cell

n+1

cell

∂

n−1

Referring to the above diagram, we see that

cell

◦ d

cell

n+1

= q

n−1

◦ ∂ ◦ q

◦ ∂ = 0,

since the middle

∂ ◦ q

is part of an exact sequence. So (

cell

(

)

, d

cell

) is a

chain complex, and the corresponding homology groups are known as the cellular

homology of X, written H

cell

(X).

Theorem.

cell

(X)

∼

(X).

Proof. We have

(X)

∼

n+1

)

= H

)/ im(∂ : H

n+1

, X

) → H

))

Since q

is injective, we apply it to and bottom to get

= q

))/ im(d

cell

n+1

: H

n+1

, X

) → H

, X

n−1

))

By exactness, the image of q

is the kernel of ∂. So we have

= ker(∂ : H

, X

n−1

) → H

n−1

))/ im(d

cell

n+1

)

= ker(d

cell

)/ im(d

cell

n+1

)

= H

cell

(X).

Corollary.

is a finite cell complex, then

(

) is a finitely-generated

abelian group for all

, generated by at most

elements. In particular, if

there are no n-cells, then H

(X) vanishes.

has a cell-structure with cells in even-dimensional cells only, then

∗

(

)

are all free.

We can similarly define cellular cohomology.

Definition (Cellular cohomology). We define cellular cohomology by

cell

(X) = H

, X

n−1

)

and let d

cell

be the composition

, X

n−1

) H

n+1

, X

∗

∂

This defines a cochain complex

cell

(

) with cohomology

∗

cell

(

), and we have

∗

cell

(X)

∼

∗

(X).

One can directly check that

cell

(X)

∼

Hom(C

cell

(X), Z).

This is all very good, because cellular homology is very simple and concrete.

However, to actually use it, we need to understand what the map

cell

: C

cell

(X) =

α∈I

Z{e

} → C

cell

n−1

(X) =

β∈I

n−1

Z{e

}

is. In particular, we want to find the coefficients d

αβ

such that

cell

) =

αβ

It turn out this is pretty easy

Lemma. The coefficients d

αβ

are given by the degree of the map

n−1

= ∂D

n−1

n−2

γ∈I

n−1

αβ

where the final map is obtained by collapsing the other spheres in the wedge.

In the case of cohomology, the maps are given by the transposes of these.

This is easier in practice that in sounds. In practice, the map is given by

“the obvious one”.

Proof. Consider the diagram

, ∂D

) H

n−1

(∂D

)

n−1

)

, X

n−1

) H

n−1

)

n−1



n−1



n−1

, X

n−2

)

n−1

n−2

)

(Φ

)

∗

∂

∼

(ϕ

)

∗

∂

cell

collapse

excision

∼

By the long exact sequence, the top left horizontal map is an isomorphism.

Now let’s try to trace through the diagram. We can find

1 1 d

αβ

αγ

isomorphism

αβ

So the degree of f

αβ

is indeed d

αβ

Example. Let K be the Klein bottle.

We give it a cell complex structure by

– K

= {v}. Note that all four vertices in the diagram are identified.

– K

= {a, b}.

– K

is the unique 2-cell

we see in the picture, where

→ K

given

by aba

−1

The cellular chain complex is given by

0 C

cell

(K) C

cell

(K) C

cell

(K)

Zπ Za ⊕ Zb Z

cell

We can now compute the maps d

cell

. The d

map is easy. We have

(a) = d

(b) = v − v = 0.

In the

map, we can figure it out by using local degrees. Locally, the attaching

map is just like the identity map, up to an orientation flip, so the local degrees

are

1. Moreover, the inverse image of each point has two elements. If we think

hard enough, we realize that for the attaching map of

, the two elements have

opposite degree and cancel each other out; while in the case of

they have the

same sign and give a degree of 2. So we have

(π) = 0a + 2b.

So we have

(K) = Z

(K) =

Z ⊕ Z

h2bi

= Z ⊕ Z/2Z

(K) = 0

We can similarly compute the cohomology. By dualizing, we have

cell

(K)(K) C

cell

(K) C

cell

(K)

Z Z ⊕ Z Z

(0 2) (0)

So we have

(K) = Z

(K) = Z/2Z.

Note that the second cohomology is not the dual of the second homology!

However, if we forget where each factor is, and just add all the homology

groups together, we get

Z⊕Z⊕Z/

. Also, if we forget all the torsion components

Z/2Z, then they are the same!

This is a general phenomenon. For a cell complex, if we know what all the

homology groups are, we can find the cohomologies by keeping the

’s unchanged

and moving the torsion components up. The general statement will be given by

the universal coefficient theorem.

Example.

Consider

(

x ∼ −x

). We notice that for any point in

if it is not in the equator, then it is represented by a unique element in the

northern hemisphere. Otherwise, it is represented by two points. So we have

∼

(

x ∼ −x for x ∈ ∂D

). This is a nice description, since if we throw

out the interior of the disk, then we are left with an

n−1

with antipodal points

identified, i.e. an RP

n−1

! So we can immediately see that

= RP

n−1

∪

for f : S

n−1

→ RP

n−1

given by

f(x) = [x].

has a cell structure with one cell in every degree up to

. What are the

boundary maps?

We write

for the

-th degree cell. We know that

is attached along the

map f described above. More concretely, we have

f : S

i−1

/RP

i−2

= S

i−1

The open upper hemisphere and lower hemisphere of

i−1

are mapped homeo-

morphically to S

i−1

\ {∗}. Furthermore,

upper

= f|

lower

◦ a,

where

is the antipodal map. But we know that

deg

(

) = (

−

. So we have a

zero map if i is odd, and a 2 map if i is even. Then we have

· · · Ze

2 0 2 0

What happens on the left end depends on whether

is even or odd. So we have

(RP

) =











Z i = 0

Z/2Z i odd, i < n

0 i even, 0 < i < n

Z i = n is odd

0 otherwise

We can immediately work out the cohomology too. We will just write out the

answer:

(RP

) =











Z i = 0

0 i odd, i < n

Z/2Z i even, 0 < i ≤ n

Z i = n is odd

0 otherwise