Bourbaki

Exercise 9, § 2

Algebra, X, p. 173

Let u:C′→Cu : C' \to C be a morphism of A-complexes. We denote by v:C→Coker (u)v : C \to \mathrm{Coker}\,(u) the passing map to the quotient. We consider the graded A-homomorphisms

α:(Ker u) (−1)→Con (u)andφ:Con (u)→Coker (u) \alpha : (\mathrm{Ker}\,u)\ (-1) \to \mathrm{Con}\,(u) \quad \text{and} \quad \varphi : \mathrm{Con}\,(u) \to \mathrm{Coker}\,(u)

defined by α(x′)=(x′,0)\alpha(x') = (x', 0), φ(x′,x)=v(x)\varphi(x', x) = v(x) for x′∈C′x' \in C', x∈Cx \in C. Now prove that α,φ\alpha, \varphi are morphisms of complexes and that they give rise to an exact sequence

⋯→Hn−1(Ker u)→H(α)Hn(Con (u))→H(φ)Hn(Coker u)→δHn−2(Ker u)→⋯ \cdots \to H_{n-1}(\mathrm{Ker}\,u) \xrightarrow{H(\alpha)} H_n(\mathrm{Con}\,(u)) \xrightarrow{H(\varphi)} H_n(\mathrm{Coker}\,u) \xrightarrow{\delta} H_{n-2}(\mathrm{Ker}\,u) \to \cdots

where δ\delta is the composite of the connecting homomorphisms relating to the two exact sequences

0→Ker u→C′→Im u→0and0→Im u→C→Coker u→0. 0 \to \mathrm{Ker}\,u \to C' \to \mathrm{Im}\,u \to 0 \quad \text{and} \quad 0 \to \mathrm{Im}\,u \to C \to \mathrm{Coker}\,u \to 0 .
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