Bourbaki

Lemma 1

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

Algebra, chapter X, § 5. Modules d’extensions

a) If CC or C′C' is bounded on the right, if CC is projective and if H(C′)=0H(C') = 0, then H(HomgrA(C,C′))=0H(\mathrm{Homgr}_A(C, C')) = 0. b) If CC or C′C' is bounded on the left, if C′C' is injective and if H(C)=0H(C) = 0, then H(HomgrA(C,C′))=0H(\mathrm{Homgr}_A(C, C')) = 0.

Let f∈Zn(HomgrA(C,C′))f \in Z_n(\mathrm{Homgr}_A(C, C')); ff is then a morphism of complexes from CC to C′(n)C'(n); in case a) (resp. b)), fmf_m vanishes for mm sufficiently small (resp. sufficiently large). By X, p. 47, Proposition 1, ff is then homotopic to zero, hence belongs to Bn(HomgrA(C,C′))B_n(\mathrm{Homgr}_A(C, C')), which gives the conclusion.

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

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