Bourbaki

Corollary

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

Algebra, chapter X, § 3. Résolutions

Let (P,p)(P, p) and (P′,p′)(P', p') be two projective resolutions of M. There exists a morphism, and only one up to homotopy, α:P′→P\alpha : P' \to P such that p∘α=p′p \circ \alpha = p'.

In fact, there exists a morphism α:P′→P\alpha : P' \to P (resp. β:P→P′\beta : P \to P') such that p∘α=p′p \circ \alpha = p' (resp. p′∘β=pp' \circ \beta = p). Since p∘α∘β=pp \circ \alpha \circ \beta = p (resp. p′∘β∘α=p′p' \circ \beta \circ \alpha = p'), α∘β\alpha \circ \beta is homotopic to 1P1_P (resp. β∘α\beta \circ \alpha is homotopic to 1P′1_{P'}).

PROPOSITION 3 bis. — Let g:N→N′g : N \to N' be a homomorphism of a-modules, e′:N′→E′e' : N' \to E' a morphism from N′N' into a left zero and injective complex E′E', and e:N→Ee : N \to E a right resolution of N. There exists a morphism of complexes g~:E→E′\tilde{g} : E \to E', and only one up to homotopy, such that g~∘e=e′∘g\tilde{g} \circ e = e' \circ g.

This is proved like Prop. 3 with the aid of Prop. 1 b).

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

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