Simplicial Homology Homology Groups
ϕ
3
ϕ
1
ϕ
1
ker ϕ
2
im im ϕ
1
ϕ
2
ϕ A
B C
=
Fig. 4.11. Groups in Lemma 4.1. ϕ
2
is injective and ϕ
1
is surjective.
The operators between the homology groups are induced by the boundary op- erators: We map a homology class to the class of the boundary of one of its
members. The Euler characteristic of
H
∗
C
∗
, according to the new definition, is simply ∑
i
−1
i
rank
H
i
= ∑
i
−1
i
β
i
. Surprisingly, the homology functor preserves the Euler characteristic of a chain complex.
Theorem 4.5 Euler-Poincaré χ
C
∗
= χ H
∗
C
∗
. The theorem states that ∑
i
−1
i
s
i
= ∑
i
−1
i
β
i
for a simplicial complex K, deriving the invariance of the Euler characteristic from the invariance of ho-
mology. To prove the theorem, we need a lemma.
Lemma 4.1 Let A , B,C be finitely generated Abelian groups related by the
sequence of maps ϕ
i
:
ϕ
3
−→ A
ϕ
2
−→ B
ϕ
1
−→ C
ϕ
−→ 0, 4.8
where im
ϕ
i
= ker ϕ
i
−
1
. Then, rank B
= rank A + rankC. Proof
The sequence is shown in Figure 4.11. First, we establish two facts. a
ϕ
1
is surjective: im ϕ
1
= ker ϕ = C.
b ϕ
2
is injective: ker ϕ
2
= im ϕ
3
= {e}, so by Corollary 3.1, ϕ
2
is 1-1. By the fundamental homomorphism theorem Theorem 3.10,
B ker ϕ
1
∼ =
im ϕ
1
. By fact a, B ker ϕ
1
∼ = C. Corollary 3.3 gives rank B ker ϕ
1
= rank B
− rank ker ϕ
1
, so rankC = rank B − rank ker ϕ
1
. By fact b, A ∼ =
im ϕ
2
and rank A = rank im ϕ
2
. But im ϕ
2
= ker ϕ
1
, so rank A = rank ker ϕ
1
. Substituting, we get the desired result.
The sequence in the lemma has a name.
Definition 4.16 short exact sequence The sequence in Lemma 4.1 is a short exact sequence
.
We use the lemma to prove the Euler-Poincaré relation. Proof
[Euler-Poincaré] Consider the following sequences: −→
Z
n i
−→
C
n ∂
n
−→
B
n
−
1
−→ −→
B
n i
−→
Z
n ϕ
−→
H
n
−→ 0, where 0 is the zero map, i is the inclusion map, and
ϕ assigns to a cycle z ∈
Z
n
its homology class [z] ∈
H
n
. Both sequences are short exact. Applying Lemma 4.1, we get:
rank
C
n
= rank
Z
n
+ rank
B
n
−
1
, 4.9
rank
Z
n
= rank
B
n
+ rank
H
n
. 4.10
Substituting the second equation into the first, multiplying by −1
n
, and sum- ming over n gives the theorem.