Graphics Reference
In-Depth Information
Assume that
[]
+
[]
= 0
au
bv
in H
1
(K) for some integers a and b. It follows that the cycle au + bv is a boundary,
that is,
+=
()
au
bv
∂
2
x
for some 2-chain x. Let [s] be any oriented 2-simplex that appears in x. The simplex
s will have at least one edge
v
i
v
j
that is different from those appearing in u and v.
Since every edge belongs to precisely two 2-simplices, let s¢ be the other 2-simplex
that has
v
i
v
j
for an edge. The only way that the coefficient of [
v
i
v
j
] will vanish in ∂
2
(x)
is the coefficients of [s] and [s¢] in x are equal. A simple extension of this argument
shows that every oriented 2-simplex [s] of K must appear in x with the same coeffi-
cient. Therefore, x = kS, for some integer k and where S is the 2-chain defined earlier,
and
()
=
()
=
()
=◊
∂
x
∂
k
k
∂
k
00
=.
2
2
2
This means that a = b = 0 and Claim 4 is proved.
Claim 3 and 4 prove that the map that sends (a,b) Œ
Z
≈
Z
to a[u] + b[v] Œ H
1
(K)
is an isomorphism and
1
()
ª
Z
.
HK
Finally, one can show that
0
()
ª
Z
HK
using the same argument as in Examples 7.2.1.5 and 7.2.1.6. This finishes Example
7.2.1.7.
7.2.1.8. Example.
The simplicial complex K in Figure 7.6 triangulates the projec-
tive plane. We want to compute the homology groups of K.
u
v
4
v
5
v
3
v
0
v
1
v
2
v
7
v
9
v
1
v
2
v
6
v
8
v
0
v
5
v
4
v
3
u
Figure 7.6.
A triangulating complex for the
projective plane.
|K| = P
2