Let (C,d) be a complex. The following conditions are equivalent: (i) there exists a homotopy equivalence from (C,d) to (H(C),0); (ii) there exists an A-endomorphism s of C, graded of degree 1, such that d=d∘s∘d; (iii) B(C) and Z(C) are direct summands of C; (iv) (C,d) is a direct sum of subcomplexes which are either of length 0, or of length 1 and have zero homology.
(i) ⇒ (ii): let φ:C→H(C) be a homotopy equivalence; then there exists a morphism of complexes ψ:H(C)→C and an endomorphism s of C, graded of degree 1, such that ψ∘φ=1C−s∘d−d∘s. We have d∘ψ=ψ∘0=0, hence
0=d∘ψ∘φ=d−d∘s∘d−d∘d∘s=d−d∘s∘d, whence (ii).
(ii)⇒(iii): let s be as in (ii). Then d∘(1C−s∘d)=0, so 1C−s∘d is a projector of C onto Z(C), and (d∘s)∘d=d, so d∘s is a projector of C onto B(C).
(iii)⇒(iv): for each n∈Z, let Zn=Zn(C), Bn=Bn(C) and choose submodules Kn and Bn′ of Cn such that Cn=Bn′⊕Zn, Zn=Kn⊕Bn. Then
E(n)=Kn(−n)andF(n)=Bn′(−n)⊕Bn−1(1−n)
are subcomplexes of (C,d); one has
(C,d)=n∈Z⨁(E(n)⊕F(n))
; each E(n) is either zero or of length 0, each F(n) is either zero or of length 1 and of zero homology, whence (iv).
(iv)⇒(i): it suffices to remark that (i) is satisfied when C is of length zero, or of zero homology and of length 1.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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