Bourbaki

Remark

Tag 0DS5, permanent. Cite it as [Bourbaki, Tag 0DS5].

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

a) Suppose that the canonical exact sequence of graded A-modules

0⟶B(C)⟶Z(C)→πH(C)⟶0(II) \tag{II} 0\longrightarrow B(C)\longrightarrow Z(C)\xrightarrow{\pi}H(C)\longrightarrow0

be split (this occurs for example if H(C)H(C) is projective, or Bn(C)B_n(C) injective for each nn); let σ:H(C)→Z(C)\sigma:H(C)\to Z(C) be a graded A-linear section of π\pi, and let ψ\psi be the homomorphism x↦σ(x)x\mapsto\sigma(x) from H(C)H(C) into CC. Then ψ\psi is a homologism from (H(C),0)(H(C),0) into CC, such that

H(ψ)=1H(C). H(\psi)=1_{H(C)}.

b) Suppose that the canonical exact sequence of graded A-modules

0⟶H(C)→iC/B(C)⟶B(C)(−1)⟶0(IV) \tag{IV} 0\longrightarrow H(C)\xrightarrow{i}C/B(C)\longrightarrow B(C)(-1)\longrightarrow0

is split (this occurs for example if B(C)B(C) is projective, or Hn(C)H_n(C) injective for each nn); let τ:C/B(C)→H(C)\tau : C/B(C) \to H(C) be a graded A-linear retraction of ii and let φ\varphi be the homomorphism from CC into H(C)H(C) which associates to each element of CC the image by τ\tau of its class modulo B(C)B(C). Then φ\varphi is a homologism from CC into (H(C),0)(H(C), 0) such that H(φ)=1H(C)H(\varphi) = 1_{H(C)}.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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