11Manifolds and Poincare duality
III Algebraic Topology
11.1 Compactly supported cohomology
Definition
(Support of cochain)
.
Let
ϕ ∈ C
n
(
X
) be a cochain. We say
ϕ
has
support in
S ⊆ X
if whenever
σ
: ∆
n
→ X \ S ⊆ X
, then
ϕ
(
σ
) = 0. In this case,
dϕ also has support in S.
Note that this has a slight subtlety. The definition only requires that if
σ
lies
completely outside S, then ϕ(σ) vanishes. However, we can have simplices that
extends very far and only touches
S
slightly, and the support does not tell us
anything about the value of
σ
. Later, we will get around this problem by doing
sufficient barycentric subdivision.
Definition
(Compactly-supported cochain)
.
Let
C
·
c
(
X
)
⊆ C
·
(
X
) be the sub-
chain complex consisting of these
ϕ
which has support in some compact
K ⊆ X
.
Note that this makes sense — we have seen that if
ϕ
has support in
K
,
then d
ϕ
has support in
K
. To see it is indeed a sub-chain complex, we need to
show that
C
·
c
(
X
) is a subgroup! Fortunately, if
ϕ
has support on
K
, and
ψ
has
support in L, then ϕ + ψ has support in K ∪ L, which is compact.
Definition
(Compactly-supported cohomology)
.
The compactly supported co-
homology of X is
H
∗
c
(X) = H
∗
(C
·
c
(X)).
Note that we can write
C
·
c
(X) =
[
K compact
C
·
(X, X \ K) ⊆ C
·
(X).
We would like to say that the compactly supported cohomology is also “built out
of” those relative cohomology, but we cannot just take the union, because the
relative cohomology is not a subgroup of
H
∗
(
X
). To do that, we need something
more fancy.
Definition
(Directed set)
.
A directed set is a partial order (
I, ≤
) such that for
all i, j ∈ I, there is some k ∈ I such that i ≤ k and j ≤ k.
Example. Any total order is a directed system.
Example. N with divisibility | as the partial order is a directed system.
Definition
(Direct limit)
.
Let
I
be a directed set. An direct system of abelian
groups indexed by
I
is a collection of abelian groups
G
i
for each
i ∈ I
and
homomorphisms
ρ
ij
: G
i
→ G
j
for all i, j ∈ I such that i ≤ j, such that
ρ
ii
= id
G
i
and
ρ
ik
= ρ
jk
◦ ρ
ij
whenever i ≤ j ≤ k.
We define the direct limit on the system (G
i
, ρ
ij
) to be
lim
−→
i∈I
G
i
=
M
i∈I
G
i
!
/hx − ρ
ij
(x) : x ∈ G
i
i.
The underlying set of it is
a
i∈I
G
i
!
/{x ∼ ρ
ij
(x) : x ∈ G
i
}.
In terms of the second description, the group operation is given as follows:
given
x ∈ G
i
and
y ∈ G
j
, we find some
k
such that
i, j ≤ k
. Then we can view
x, y
as elements as
G
k
and do the operation there. It is an exercise to show that
these two descriptions are indeed the same.
Now observe that if
J ⊆ I
is a sub-directed set such that for all
a ∈ I
, there
is some b ∈ J such that a ≤ b. Then we have
lim
−→
i∈J
G
i
∼
=
lim
−→
i∈I
G
i
.
So our claim is now
Theorem. For any space X, we let
K(X) = {K ⊆ X : K is compact}.
This is a directed set under inclusion, and the map
K 7→ H
n
(X, X \ K)
gives a direct system of abelian groups indexed by
K
(
X
), where the maps
ρ
are
given by restriction.
Then we have
H
∗
c
(X)
∼
=
lim
−→
K(X)
H
n
(X, X \ K).
Proof. We have
C
n
c
(X)
∼
=
lim
−→
K(X)
C
n
(X, X \ K),
where we have a map
lim
−→
K(α)
C
n
(X, X \ K) → C
n
c
(X)
given in each component of the direct limit by inclusion, and it is easy to see
that this is well-defined and bijective.
It is then a general algebraic fact that
H
∗
commutes with inverse limits, and
we will not prove it.
Lemma. We have
H
i
c
(R
d
; R)
∼
=
(
R i = d
0 otherwise
.
Proof. Let B ∈ K(R
d
) be the balls, namely
B = {nD
d
, n = 0, 1, 2, · · · }.
Then since every compact set is contained in one of them, we have
H
n
c
(X)
∼
=
lim
−→
K∈K(R
d
)
H
n
(R
d
, R
d
\ K; R)
∼
=
lim
−→
nD
d
∈B
H
n
(R
d
, R
d
\ nD
d
; R)
We can compute that directly. Since R
d
is contractible, the connecting map
H
i
(R
d
, R
d
\ nD
d
; R) → H
i−1
(R
d
\ nD
d
; R)
in the long exact sequence is an isomorphism. Moreover, the following diagram
commutes:
H
i
(R
d
, R
d
\ nD
n
; R) H
i
(R
d
, R
d
\ (n + 1)D
d
; R)
H
i−1
(R
d
\ nD
d
; R) H
i−1
(R
d
\ (n + 1)D
d
; R)
∂
ρ
n,n+1
∂
But all maps here are isomorphisms because the horizontal maps are homotopy
equivalences. So we know
lim
−→
H
i
(R
d
, R
d
\ nD
d
; R)
∼
=
H
i
(R
d
, R
d
\ {0}; R)
∼
=
H
i−1
(R
d
\ {0}; R).
So it follows that
H
i
(R
d
, R
d
\ {0}; R) =
(
R i = d
0 otherwise
.
In general, this is how we always compute compactly-supported cohomology
— we pick a suitable subset of K(X) and compute the limit of that instead.
Note that compactly-supported cohomology is not homotopy-invariant! It
knows about the dimension of
R
d
, since the notion of compactness is not ho-
motopy invariant. Even worse, in general, a map
f
:
X → Y
does not induce a
map
f
∗
:
H
∗
c
(
Y
)
→ H
∗
c
(
X
). Indeed, the usual map does not work because the
preimage of a compact set of a compact set is not necessarily compact.
Definition
(Proper map)
.
A map
f
:
X → Y
of spaces is proper if the preimage
of a compact space is compact.
Now if
f
is proper, then it does induce a map
H
∗
c
(
·
) by the usual construction.
From now on, we will assume all spaces are Hausdorff, so that all compact
subsets are closed. This isn’t too bad a restriction since we are ultimately
interested in manifolds, which are by definition Hausdorff.
Let
i
:
U → X
be the inclusion of an open subspace. We let
K ⊆ U
be
compact. Then by excision, we have an isomorphism
H
∗
(U, U \ K)
∼
=
H
∗
(X, X \ K),
since the thing we cut out, namely
X \ U
, is already closed, and
U \ K
is open,
since K is closed.
So a compactly supported cohomology class on
U
gives one on
X
. So we get
a map
i
∗
: H
∗
c
(U) → H
∗
c
(X).
We call this “extension by zero”. Indeed, this is how the cohomology class works
— if you have a cohomology class
φ
on
U
supported on
K ⊆ U
, then given any
simplex in
X
, if it lies inside
U
, we know how to evaluate it. If it lies outside
K
,
then we just send it to zero. Then by barycentric subdivision, we can assume
every simplex is either inside U or outside K, so we are done.
Example. If i : U → R
d
is an open ball, then the map
i
∗
: H
∗
c
→ H
∗
c
(R
d
)
is an isomorphism. So each cohomology class is equivalent to something with a
support as small as we like.
Since it is annoying to write H
n
(X, X \ K) all the time, we write
H
n
(X | K; R) = H
n
(X, X \ K; R).
By excision, this depends only on a neighbourhood of
K
in
X
. In the case where
K
is a point, this is local cohomology at a point. So it makes sense to call this
local cohomology near K ⊆ X.
Our end goal is to produce a Mayer-Vietoris for compactly-supported coho-
mology. But before we do that, we first do it for local cohomology.
Proposition. Let K, L ⊆ X be compact. Then there is a long exact sequence
H
n
(X | K ∩ L) H
n
(X | K) ⊕ H
n
(X | L) H
n
(X | K ∪ L)
H
n+1
(X | K ∩ L) H
n+1
(X | K) ⊕ H
n+1
(X | L) · · ·
∂
,
where the unlabelled maps are those induced by inclusion.
We are going to prove this by relating it to a Mayer-Vietoris sequence of
some sort.
Proof. We cover X \ K ∩ L by
U = {X \ K, X \ L}.
We then draw a huge diagram (here
∗
denotes the dual, i.e.
X
∗
=
Hom
(
X
;
R
),
and C
·
(X | K) = C
·
(X, X \ K)):
0 0 0
0
C
·
(X)
C
U
·
(X\K∩L)
∗
C
·
(X | K) ⊕ C
·
(X | L) C
·
(X | K ∪ L) 0
0 C
·
(X) C
·
(X) ⊕ C
·
(X) C
·
(X) 0
0 C
·
U
(X \ K ∩ L) C
·
(X \ K) ⊕ C
·
(X \ L) C
·
(X \ K ∪ L) 0
0 0 0
(id,− id)
id + id
(j
∗
1
,−j
∗
2
) i
∗
1
+i
∗
2
This is a diagram. Certainly.
The bottom two rows and all columns are exact. By a diagram chase (the
nine lemma), we know the top row is exact. Taking the long exact sequence
almost gives what we want, except the first term is a funny thing.
We now analyze that object. We look at the left vertical column:
0 Hom
C
·
(x)
C
U
·
(X\K∩L)
, R
C
·
(X) Hom(C
U
·
(X \ K ∩ L), R) 0
Now by the small simplices theorem, the right hand object gives the same
(co)homology as
C
·
(
X \ K ∩ L
;
R
). So we can produce another short exact
sequence:
0 Hom
C
·
(x)
C
U
·
(X\(K∩L))
, R
C
·
(X) Hom(C
U
·
(X \ K ∩ L), R) 0
0 C
·
(X, X \ K ∩ L) C
·
(X) Hom(C
·
(X \ K ∩ L), R) 0
Now the two right vertical arrows induce isomorphisms when we pass on to
homology. So by taking the long exact sequence, the five lemma tells us the left
hand map is an isomorphism on homology. So we know
H
∗
Hom
C
·
(x)
C
U
·
(X \ (K ∩ L))
, R
!!
∼
=
H
∗
(X | K ∩ L).
So the long exact of the top row gives what we want.
Corollary.
Let
X
be a manifold, and
X
=
A ∪ B
, where
A, B
are open sets.
Then there is a long exact sequence
H
n
c
(A ∩ B) H
n
c
(A) ⊕ H
n
c
(B) H
n
c
(X)
H
n+1
c
(A ∩ B) H
n+1
c
(A) ⊕ H
n+1
c
(B) · · ·
∂
Note that the arrows go in funny directions, which is different from both
homology and cohomology versions!
Proof.
Let
K ⊆ A
and
L ⊆ B
be compact sets. Then by excision, we have
isomorphisms
H
n
(X | K)
∼
=
H
n
(A | K)
H
n
(X | L)
∼
=
H
n
(B | L)
H
n
(X | K ∩ L)
∼
=
H
n
(A ∩ B | K ∩ L).
So the long exact sequence from the previous proposition gives us
H
n
(A ∩ B | K ∩ L) H
n
(A | K) ⊕ H
n
(B | L) H
n
(X | K ∪ L)
H
n+1
(A ∩ B | K ∩ L) H
n+1
(A | K) ⊕ H
n+1
(B | L) · · ·
∂
The next step is to take the direct limit over
K ∈ K
(
A
) and
L ∈ K
(
B
). We
need to make sure that these do indeed give the right compactly supported
cohomology. The terms
H
n
(
A | K
)
⊕ H
n
(
B | L
) are exactly right, and the one
for
H
n
(
A ∩ B | K ∩ L
) also works because every compact set in
A ∩ B
is a
compact set in
A
intersect a compact set in
B
(take those to be both the original
compact set).
So we get a long exact sequence
H
n
c
(A ∩ B) H
n
c
(A) ⊕ H
n
c
(B) lim
−→
K∈K(A)
L∈K(B)
H
n
(X | K ∪ L) H
n+1
c
(A ∩ B)
∂
To show that that funny direct limit is really what we want, we have to show
that every compact set
C ∈ X
lies inside some
K ∪ L
, where
K ⊆ A
and
L ⊆ B
are compact.
Indeed, as
X
is a manifold, and
C
is compact, we can find a finite set of
closed balls in
X
, each in
A
or in
B
, such that their interiors cover
C
. So done.
(In general, this will work for any locally compact space)
This is all we have got to say about compactly supported cohomology. We
now start to talk about manifolds.