3Four major tools of (co)homology

III Algebraic Topology



3.2 Mayer-Vietoris
The next tool is something that allows us to compute (co)homology by computing
the (co)homology of smaller subspaces. Unfortunately, the result doesn’t just
say we can directly add the cohomologies, or anything like that. Instead, the
information is encoded in an exact sequence.
Definition (Exact sequence). We say a pair of homomorphisms
A B C
f g
is exact at B if im(f) = ker(g).
We say a sequence
· · · X
0
X
1
X
2
· · ·
is exact if it is exact at each X
n
.
Example. If
0 A B 0
f
.
is exact, then f is an isomorphism.
Definition
(Short exact sequence)
.
A short exact sequence is a sequence of the
form
0 A B C 0 .
It is an easy consequence of the first isomorphism theorem that
Lemma. In a short exact sequence
0 A B C 0
f g
,
the map f is injective; g is surjective, and C
=
B/A.
Example. Consider the sequence
0 Z/nZ A Z/mZ 0
There are many possible values of
A
. A few obvious examples are
A
=
Z/nmZ
and
A
=
Z/nZ Z/mZ
. If
n
and
m
are not coprime, then these are different
groups.
Many of the important theorems for computing homology groups tell us that
the homology groups of certain spaces fall into some exact sequence, and we
have to try figure out what the groups are.
Theorem
(Mayer-Vietoris theorem)
.
Let
X
=
A B
be the union of two open
subsets. We have inclusions
A B A
B X
i
A
i
B
j
A
j
B
.
Then there are homomorphisms
MV
:
H
n
(
X
)
H
n1
(
A B
) such that the
following sequence is exact:
H
n
(A B) H
n
(A) H
n
(B) H
n
(X)
H
n1
(A B) H
n1
(A) H
n1
(B) H
n1
(X) · · ·
· · · H
0
(A) H
0
(B) H
0
(X) 0
MV
i
A
i
B
j
A
j
B
MV
i
A
i
B
j
A
j
B
j
A
j
B
Furthermore, the Mayer-Vietoris sequence is natural, i.e. if
f
:
X
=
A B
Y = U V satisfies f(A) U and f(B) V , then the diagram
H
n+1
(X) H
n
(A B) H
n
(A) H
n
(B) H
n
(X)
H
n+1
(Y ) H
n
(U V ) H
n
(U) H
n
(V ) H
n
(Y )
MV
f
i
A
i
B
f|
AB
j
A
j
B
f|
A
f|
B
f
MV
i
U
i
V
j
U
j
V
commutes.
For certain elements of
H
n
(
X
), we can easily specify what
MV
does to it.
The meat of the proof is to show that every element of
H
n
(
X
) can be made to
look like that. If [
a
+
b
]
H
n
(
X
) is such that
a C
n
(
A
) and
b C
n
(
B
), then
the map
MV
is specified by
MV
([a + b]) = [d
n
(a)] = [d
n
(b)] H
n1
(A B).
To see this makes sense, note that we have
d
n
(
a
+
b
) = 0. So
d
n
(
a
) =
d
n
(
b
).
Moreover, since
a C
n
(
A
), we have
d
n
(
a
)
C
n1
(
A
). Similarly,
d
n
(
b
)
C
n1
(B). So d
n
(a) = d
n
(b) C
n1
(A) C
n1
(B) = C
n1
(A B).