478 APPENDIX A. HOMOLOGICAL METHODS∗
∂dn+1. is in Im f . Let en+1 ≡ g(dn+1). Then ∂en+1 = g(∂dn+1) = 0. Thus this en+1 isa cycle. Then f−1∂g−1en+1 = f−1∂g−1g(dn+1) = f−1∂dn+1 = cn. Therefore, ∆ [en+1] ≡[
f−1∂g−1en+1]= [cn] . Therefore, ker( f )⊆ Im∆. This completes the proof. ■
I think it may be useful to have a description of this connecting homomorphism in termsof a sequence of steps.
en+1 = g(dn+1) since g is onto En+1
0 = g(∂dn+1) since en+1 is cyclef (cn) = ∂dn+1 by Im( f ) = ker(g)f (∂cn) = ∂∂dn+1 = 0⇒ ∂cn = 0∆([en+1])≡ [cn] ∈ HCn
(1.13)
Definition A.6.4 One of the columns in 1.9 is called a chain complex. The homomorphismsf ,g are called chain maps. Recall that the diagram commutes so that f (∂c) = ∂ f (c). Inthe above theorem about the connecting homomorphism, what happens to the bottom row?It is of the form
0→ C0f→ D0
g→ E0 → 0
and ∂ will map these Abelian groups to 0. Thus the connecting homomorphism ∆ will justbe the zero map so everything just disappears after this. To save on notation, we write 1.9as
0→ Cf→ D
g→ E → 0
where C,D,E symbolize the entire column. Thus, in words, a short exact sequence of chaincomplexes yields a long exact sequence of homology groups. This is completely algebraic.∂ is just a mapping from the nth to the (n−1)st level in one of these chain complexes suchthat ∂ 2 = 0 which enables the definition of the homology groups. Now suppose you havetwo of these short exact sequences.
0→ Cf→ D
g→ E → 0↓ α ↓ β ↓ γ
0→ C′f ′→ D′
g′→ E ′ → 0
(1.14)
where the mappings α,β ,γ are homomorphisms which act between the groups Cn,Dn,Enand C′n,D
′n,E
′n in such a way that the squares in the above diagram commute. That is,
for c ∈ Cn, β f (c) = f ′α (c) with a similar relation holding for the next square involvingDn,En,D′n,E
′n. Such mappings α,β ,γ are called “chain homomorphisms”. They are said
to have degree 0 because they act on the same level.Also we insist that for
c ∈ Cn,∂′ (α (c)) = α (∂c) ,d ∈ Dn,∂
′ (β (d)) = β (∂ (d)) , (1.15)e ∈ En,∂
′ (γ (e)) = β (∂ (e))
Thus we can consider the following in terms of homology groups. For c a cycle in C, wecan say the following is well defined.
[αc]′ = α [c] , [βd]′ = β [d] , [γe]′ = γ [e]
This is because of 1.15. Cycles go to cycles and boundaries go to boundaries. Thus thisdefinition yields a homomorphism of homology groups also.