6(Co)homology with coefficients
III Algebraic Topology
6 (Co)homology with coefficients
Recall that when we defined (co)homology, we constructed these free
Z
-modules
from our spaces. However, we did not actually use the fact that it was
Z
, we
might as well replace it with any abelian group A.
Definition
((Co)homology with coefficients)
.
Let
A
be an abelian group, and
X be a topological space. We let
C
·
(X; A) = C
·
(X) ⊗ A
with differentials
d ⊗id
A
. In other words
C
·
(
X
;
A
) is the abelian group obtained
by taking the direct sum of many copies of A, one for each singular simplex.
We let
H
n
(X; A) = H
n
(C
·
(X; A), d ⊗ id
A
).
We can also define
H
cell
n
(X; A) = H
n
(C
cell
·
(X) ⊗ A),
and the same proof shows that H
cell
n
(X; A) = H
n
(X; A).
Similarly, we let
C
·
(X; A) = Hom(C
·
(X), A),
with the usual (dual) differential. We again set
H
n
(X; A) = H
n
(C
·
(X; A)).
We similarly define cellular cohomology.
If A is in fact a commutative ring, then these are in fact R-modules.
We call A the “coefficients”, since a general member of C
·
(X; A) looks like
X
n
σ
s, where n
σ
∈ A, σ : ∆
n
→ X.
We will usually take
A
=
Z, Z/nZ
or
Q
. Everything we’ve proved for homology
holds for these with exactly the same proof.
Example.
In the case of
C
cell
·
(
RP
n
), the differentials are all 0 or 2. So in
C
cell
·
(RP
n
, Z/2), all the differentials are 0. So we have
H
i
(RP
n
, Z/2) =
(
Z/2 0 ≤ i ≤ n
0 i > n
Similarly, the cohomology groups are the same.
On the other hand, if we take the coefficients to be
Q
, then multiplication
by 2 is now an isomorphism. Then we get
C
cell
·
(RP
n
, Q)
(
Q n odd
0 n even
for n not too large.