Graphics Reference
In-Depth Information
Solution.
We shall only sketch the computation of H
1
(K) and H
2
(K) in this example
and leave the details and the computation of the other groups as an exercise to the
reader.
Claim 1.
Z
2
(K) = 0.
Orient the 2-simplices in K as shown in Figure 7.6. If z is any nonzero 2-cycle,
then z must contain all the oriented 2-simplices of K with equal multiplicity. For
example, if
[
]
+
za
=
vvv
...,
679
then all the oriented 2-simplices adjacent to
v
6
v
7
v
9
must appear with the same mul-
tiplicity because that is the only way that the boundary of z can vanish. In other words,
z must be of the form aS where S is the sum of all the oriented 2-simplices. But
∂
()
= 2u,
where
u =
[
]
+
[
]
+
[
]
+
[
]
+
[
]
+
[
]
vv
vv
vv
vv
vv
vv
50
.
01
12
23
34
45
It follows that a must be zero and Claim 1 is proved. This also shows that
()
=
HK
2
0
.
Claim 2.
If z is a 1-cycle, then z is homologous to ku for some integer k.
Claim 2 is proved by first showing that z is homologous to a 1-cycle z
1
of the form
[
]
+
[
]
+
[
]
+
[
]
+
[
]
za
=
vv
a
vv
a
vv
a
vv
a
vv
1
1
0
1
2
1
2
3
2
3
4
3
4
5
4
5
[
]
+
[
]
+
[
]
+
[
]
+
[
]
+
a
vv
a
vv
a
vv
a
vv
a
vv
,
for some a
i
Œ
Z
.
650
747
876
959
098
But since ∂
1
(z
1
) = 0 we must have a
7
= a
8
= a
9
= a
10
= 0. Finally, one shows that all the
remaining 1-simplices in z
1
must appear with the same multiplicity, proving Claim 2.
Claim 3.
[u] π 0.
If u =∂
2
(c) for some 2-chain c, then one can show by an argument similar to the one
in the proof of Claim 1 that all the oriented 2-simplices must appear in c with multi-
plicity 1, that is, c =S. This contradicts the fact that ∂(S) = 2u, and proves Claim 3.
Claims 2 and 3 and the fact that 2[u] = 0 (since ∂(S) = 2u) clearly prove that
()
ª
Z
.
HK
1
2
This finishes what we have to say about Example 7.2.1.8.
