Bourbaki

Proposition 3

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

Algebra, chapter X, § 3. Résolutions

Let f:M′→Mf : M' \to M be a homomorphism of A-modules, p′:P′→M′p' : P' \to M' a morphism into M′M' of a complex null on the right and projective P′P', and p:P→Mp : P \to M a left resolution of M. There exists a morphism of complexes f~:P′→P\tilde{f} : P' \to P, and only one up to homotopy, such that p∘f~=f∘p′p \circ \tilde{f} = f \circ p'.

Let us consider the complex P‾\overline{P} defined as follows: P‾n=Pn\overline{P}_n = P_n for n≠−1n \neq -1, P‾−1=M\overline{P}_{-1} = M, dP‾,n=dP,nd_{\overline{P},n} = d_{P,n} for n≠0,−1n \neq 0, -1, dP‾,0=p0d_{\overline{P},0} = p_0, dP‾,−1=0d_{\overline{P},-1} = 0, and the complex P‾′\overline{P}' defined analogously. Applying to the complexes P‾\overline{P} and P‾′\overline{P}' Prop. 1 a) with r=0r = 0, ui=0u_i = 0 for i<−1i < -1 and u−1=fu_{-1} = f, we obtain Prop. 3.

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

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