476 APPENDIX A. HOMOLOGICAL METHODS∗

Thus ∂dn+1 ∈ ker(g) and so ∂dn+1 = f (cn+1) for some unique cn+1 ∈ Cn+1. Same withĉn+1. Is cn+1 a cycle?

f (∂cn+1) = ∂ f (cn+1) = ∂∂dn+1 = 0

so ∂cn+1 = 0 since f is one to one and so this is indeed a cycle and the result on the rightin 1.10 is [cn+1] . Same with ĉn+1. Will [cn+1− ĉn+1] = 0? If so, this will prove that theexpression on the right in 1.10 is well defined.

f (cn+1− ĉn+1) =(∂dn+1−∂ d̂n+1

), (cn+1− ĉn+1) = f−1 (

∂(dn+1− d̂n+1

))(1.11)

Now f(

f−1 (∂d))= ∂d and f

(∂ f−1 (d)

)= ∂ f

(f−1 (d)

)= ∂d so f−1 (∂d)= ∂ f−1 (d) . It

follows from this observation and 1.11 that cn+1− ĉn+1 is a boundary and so [cn+1− ĉn+1] =0. Therefore, ∆ is well defined.

Is ∆ a homomorphism?

∆([en+1]+ [ên+1])≡ ∆ [en+1 + ên+1]≡[

f−1∂g−1 (en+1 + ên+1)

]Letting dn+1 ∈ g−1 (en+1) , d̂n+1 ∈ g−1 (ên+1), it follows that

dn+1 + d̂n+1 ∈ g−1 (en+1 + ên+1) .

Similar considerations will now apply to f−1. We can have

f (cn+1) = ∂dn+1, f (ĉn+1) = ∂ d̂n+1

and the result will be [cn+1]+ [ĉn+1] = ∆ [en+1]+∆ [ên+1] so this is a homomorphism. ■

Theorem A.6.3 Suppose the situation of 1.9 described in Definition A.6.1. Here ∂∂ = 0and ∂ f = f ∂ , same with g. Let the homology groups be defined as before HCn ≡ ZCn/BCn

where ZCn ≡ {c ∈C : ∂c = 0} and BCn ≡ ∂c for some c ∈Cn+1. Then we can consider f ,gacting on the homology groups in the natural way

f ([c]) = [ f (c)] ,g([d]) = [g(d)] (1.12)

and also there is a connecting homomorphism ∆ such that

· · · → HCn

f→ HDng→ HEn

∆→ HCn−1

f→ HDn−1

g→ ···

is an exact sequence.

Before proving this, consider that just because g is onto En does not mean that g is ontoHEn . This is because the things in HDn are equivalence classes of cycles.

Proof: The definitions in 1.12 are clearly true and this was shown earlier in LemmaA.2.3. Recall

0→...Cn+1

f→...Dn+1

g→...En+1 → 0

↓ ∂ ↓ ∂ ↓ ∂

0→ Cn...

f→ Dn...

g→ En...

→ 0