Bourbaki

Proposition 6

Tag 0DRY, permanent. Cite it as [Bourbaki, Tag 0DRY].

Algebra, chapter X, § 2. Complexes de A-modules

Let (C,d)(C, d) be a complex. The following conditions are equivalent: (i) there exists a homotopy equivalence from (C,d)(C, d) to (H(C),0)(\mathrm{H}(C), 0); (ii) there exists an AA-endomorphism ss of CC, graded of degree 1, such that d=d∘s∘dd = d \circ s \circ d; (iii) B(C)B(C) and Z(C)Z(C) are direct summands of CC; (iv) (C,d)(C, d) is a direct sum of subcomplexes which are either of length 0, or of length 1 and have zero homology.

(i) ⇒\Rightarrow (ii): let φ:C→H(C)\varphi : C \to \mathrm{H}(C) be a homotopy equivalence; then there exists a morphism of complexes ψ:H(C)→C\psi : \mathrm{H}(C) \to C and an endomorphism ss of CC, graded of degree 1, such that ψ∘φ=1C−s∘d−d∘s\psi \circ \varphi = 1_C - s \circ d - d \circ s. We have d∘ψ=ψ∘0=0d \circ \psi = \psi \circ 0 = 0, hence

0=d∘ψ∘φ=d−d∘s∘d−d∘d∘s=d−d∘s∘d, 0 = d \circ \psi \circ \varphi = d - d \circ s \circ d - d \circ d \circ s = d - d \circ s \circ d ,
whence (ii).

(ii)⇒(iii)(ii) \Rightarrow (iii): let ss be as in (ii). Then d∘(1C−s∘d)=0d \circ (1_C-s\circ d)=0, so 1C−s∘d1_C-s\circ d is a projector of CC onto Z(C)Z(C), and (d∘s)∘d=d(d\circ s)\circ d=d, so d∘sd\circ s is a projector of CC onto B(C)B(C).

(iii)⇒(iv)(iii) \Rightarrow (iv): for each n∈Zn\in\mathbf{Z}, let Zn=Zn(C)Z_n=Z_n(C), Bn=Bn(C)B_n=B_n(C) and choose submodules KnK_n and Bn′B'_n of CnC_n such that Cn=Bn′⊕ZnC_n=B'_n\oplus Z_n, Zn=Kn⊕BnZ_n=K_n\oplus B_n. Then

E(n)=Kn(−n)andF(n)=Bn′(−n)⊕Bn−1(1−n) E_{(n)}=K_n(-n)\quad\text{and}\quad F_{(n)}=B'_n(-n)\oplus B_{n-1}(1-n)

are subcomplexes of (C,d)(C,d); one has

(C,d)=⨁n∈Z(E(n)⊕F(n)) (C,d)=\bigoplus_{n\in\mathbf{Z}}(E_{(n)}\oplus F_{(n)})

; each E(n)E_{(n)} is either zero or of length 00, each F(n)F_{(n)} is either zero or of length 11 and of zero homology, whence (iv).

(iv)⇒(i)(iv) \Rightarrow (i): it suffices to remark that (i) is satisfied when CC 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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Cites

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Translated with machine help, not yet reviewed by a person.