9Kunneth theorem and universal coefficients theorem

III Algebraic Topology



9 K¨unneth theorem and universal coefficients
theorem
We are going to prove two theorems of similar flavour unneth’s theorem
and the universal coefficients theorem. They are both fairly algebraic results
that relate different homology and cohomology groups. They will be very useful
when we prove things about (co)homologies in general.
In both cases, we will not prove the “full” theorem, as they require knowledge
of certain objects known as
Tor
and
Ext
. Instead, we will focus on a particular
case where the
Tor
and
Ext
vanish, so that we can avoid mentioning them at all.
We start with K¨unneth’s theorem.
Theorem
(K¨unneth’s theorem)
.
Let
R
be a commutative ring, and suppose
that H
n
(Y ; R) is a free R-module for each n. Then the cross product map
M
k+`=n
H
k
(X; R) H
`
(Y ; R) H
n
(X × Y ; R)
×
is an isomorphism for every n, for every finite cell complex X.
It follows from the five lemma that the same holds if we have a relative
complex (Y, A) instead of just Y .
For convenience, we will write
H
(
X
;
R
)
H
(
Y
;
R
) for the graded
R
-module
which in grade n is given by
M
k+`=n
H
k
(X; R) H
`
(Y ; R).
Then K¨unneth says the map given by
H
(X; R) H
(Y ; R) H
(X × Y ; R)
×
is an isomorphism of graded rings.
Proof. Let
F
n
() =
M
k+`=n
H
k
(; R) H
`
(Y ; R).
We similarly define
G
n
() = H
n
( × Y ; R).
We observe that for each
X
, the cross product gives a map
×
:
F
n
(
X
)
G
n
(
X
),
and, crucially, we know
×
:
F
n
(
)
G
n
(
) is an isomorphism, since
F
n
(
)
=
G
n
()
=
H
n
(Y ; R).
The strategy is to show that
F
n
(
) and
G
n
(
) have the same formal structure
as cohomology and agree on a point, and so must agree on all finite cell complexes.
It is clear that both
F
n
and
G
n
are homotopy invariant, because they are
built out of homotopy invariant things.
We now want to define the cohomology of pairs. This is easy. We define
F
n
(X, A) =
M
i+j=n
H
i
(X, A; R) H
j
(Y ; R)
G
n
(X, A) = H
n
(X × Y, A × Y ; R).
Again, the relative cup product gives us a relative cross product, which gives us
a map F
n
(X, A) G
n
(X, A).
It is immediate
G
n
has a long exact sequence associated to (
X, A
) given by
the usual long exact sequence of (
X × Y, A × Y
). We would like to say
F
has a
long exact sequence as well, and this is where our hypothesis comes in.
If
H
(
Y
;
R
) is a free
R
-module, then we can take the long exact sequence of
(X, A)
· · · H
n
(A; R) H
n
(X, A; R) H
n
(X; R) H
n
(A; R) · · ·
,
and then tensor with
H
j
(
Y
;
R
). This preserves exactness, since
H
j
(
Y
;
R
)
=
R
k
for some
k
, so tensoring with
H
j
(
Y
;
R
) just takes
k
copies of this long exact
sequence. By adding the different long exact sequences for different
j
(with
appropriate translations), we get a long exact sequence for F .
We now want to prove unneth by induction on the number of cells and the
dimension at the same time. We are going to prove that if
X
=
X
0
f
D
n
for some
S
n1
X
0
, and
×
:
F
(
X
0
)
G
(
X
0
) is an isomorphism, then
×
:
F
(
X
)
G
(
X
)
is also an isomorphism. In doing so, we will assume that the result is true for
attaching any cells of dimension less than n.
Suppose
X
=
X
0
f
D
n
for some
f
:
S
n1
X
0
. We get long exact sequences
F
∗−1
(X
0
) F
(X, X
0
) F
(X) F
(X
0
) F
+1
(X, X
0
)
G
∗−1
(X
0
) G
(X, X
0
) G
(X) G
(X
0
) G
+1
(X, X
0
)
×
×
×
×
×
Note that we need to manually check that the boundary maps
commute with
the cross product, since this is not induced by maps of spaces, but we will not
do it here.
Now by the five lemma, it suffices to show that the maps on the relative
cohomology × : F
n
(X, X
0
) G
n
(X, X
0
) is an isomorphism.
We now notice that
F
(
) and
G
(
) have excision. Since (
X, X
0
) is a good
pair, we have a commutative square
F
(D
n
, D
n
) F
(X, X
0
)
G
(D
n
, D
n
) G
(X, X
0
)
×
×
So we now only need the left-hand map to be an isomorphism. We look at the
long exact sequence for (D
n
, D
n
)!
F
∗−1
(D
n
) F
(D
n
, D
n
) F
(D
n
) F
(D
n
) F
+1
(D
n
, D
n
)
G
∗−1
(D
n
) G
(D
n
, D
n
) G
(D
n
) G
(D
n
) G
+1
(D
n
, D
n
)
×
×
× ×
××
But now we know the vertical maps for
D
n
and
D
n
are isomorphisms the
ones for
D
n
are because they are contractible, and we have seen the result of
already; whereas the result for D
n
follows by induction.
So we are done.
The conditions of the theorem require that
H
n
(
Y
;
R
) is free. When will this
hold? One important example is when
R
is actually a field, in which case all
modules are free.
Example.
Consider
H
(
S
1
, Z
). We know it is
Z
in
= 0
,
1, and 0 elsewhere.
Let’s call the generator of
H
0
(
S
1
, Z
) “1”, and the generator of
H
1
(
S
1
, Z
) as
x
.
Then we know that
x ^ x
= 0 since there isn’t anything in degree 2. So we
know
H
(S
1
, Z) = Z[x]/(x
2
).
Then K¨unneth’s theorem tells us that
H
(T
n
, Z)
=
H
(S
1
; Z)
n
,
where
T
n
= (
S
1
)
n
is the
n
-torus, and this is an isomorphism of rings. So this is
H
(T
n
, Z)
=
Z[x
1
, · · · , x
n
]/(x
2
i
, x
i
x
j
+ x
j
x
i
),
using the fact that
x
i
, x
j
have degree 1 so anti-commute. Note that this has
an interesting cup product! This ring is known as the exterior algebra in
n
generators.
Example.
Let
f
:
S
n
T
n
be a map for
n >
1. We claim that this induces
the zero map on the nth cohomology.
We look at the induced map on cohomology:
f
: H
n
(T
n
; Z) H
n
(S
n
, Z) .
Looking at the presentation above, we know that
H
n
(
T
n
, Z
) is generated by
x
1
^
· · · ^ x
n
, and
f
sends it to (
f
x
1
)
^ · · · ^
(
f
x
n
). But
f
x
i
H
1
(
S
n
, Z
) = 0
for all n > 1. So f
(x
1
· · · x
n
) = 0.
Note that the statement does not involve cup products at all, but it would
be much more difficult to prove this without using cup products!
We are now going to prove another useful result.
Theorem
(Universal coefficients theorem for (co)homology)
.
Let
R
be a PID
and M an R-module. Then there is a natural map
H
(X; R) M H
(X; M).
If
H
(
X
;
R
) is a free module for each
n
, then this is an isomorphism. Similarly,
there is a natural map
H
(X; M) Hom
R
(H
(X; R), M), ,
which is an isomorphism again if H
(X; R) is free.
In particular, when R = Z, then an R-module is just an abelian group, and
this tells us how homology and cohomology with coefficients in an abelian group
relate to the usual homology and cohomology theory.
Proof.
Let
C
n
be
C
n
(
X
;
R
) and
Z
n
C
n
be the cycles and
B
n
Z
n
the
boundaries. Then there is a short exact sequence
0 Z
n
C
n
B
n1
0
i
g
,
and
B
n1
C
n1
is a submodule of a free
R
-module, and is free, since
R
is
a PID. So by picking a basis, we can find a map
s
:
B
n1
C
n
such that
g s = id
B
n1
. This induces an isomorphism
i s : Z
n
B
n1
C
n
.
Now tensoring with M, we obtain
0 Z
n
M C
n
M B
n1
M 0 ,
which is exact because we have
C
n
M
=
(Z
n
B
n1
) M
=
(Z
n
M) (B
n1
M).
So we obtain a short exact sequence of chain complexes
0 (Z
n
M, 0) (C
n
M, d id) (B
n1
M, 0) 0 ,
which gives a long exact sequence in homology:
· · · B
n
M Z
n
M H
n
(X; M) B
n1
M · · ·
We’ll leave this for a while, and look at another short exact sequence. By
definition of homology, we have a long exact sequence
0 B
n
Z
n
H
n
(X; R) 0 .
As
H
n
(
X
;
R
) is free, we have a splitting
t
:
H
n
(
X
;
R
)
Z
n
, so as above,
tensoring with M preserves exactness, so we have
0 B
n
M Z
n
M H
n
(X; R) M 0 .
Hence we know that
B
n
M Z
n
M
is injective. So our previous long exact
sequence breaks up to
0 B
n
M Z
n
M H
n
(X; M) 0.
Since we have two short exact sequence with first two terms equal, the last terms
have to be equal as well.
The cohomology version is similar.