Bourbaki

Proposition 1

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

Algebra, chapter X, § 3. Résolutions

Let (P,dP)(P, d_P) and (E,dE)(E, d_E) be two complexes of AA-modules and rr an integer.

a) Let (ui:Pi→Ei)i≤r(u_i : P_i \to E_i)_{i \leq r} be a family of homomorphisms such that dE∘ui=ui−1∘dPd_E \circ u_i = u_{i-1} \circ d_P for i≤ri \leq r. Suppose that PiP_i is projective for i>ri > r and that Hi(E)=0H_i(E) = 0 for i≥ri \geq r. Then the family of the uiu_i extends to a morphism of complexes from PP into EE; two such extensions are homotopic.

b) Let (ui:Pi→Ei)i≤r(u^i : P^i \to E^i)_{i \leq r} be a family of homomorphisms such that ui∘dP=dE∘ui−1u^i \circ d_P = d_E \circ u^{i-1} for i≤ri \leq r. Suppose that EiE^i is injective for i>ri > r and that Hi(P)=0H^i(P) = 0 for i≥ri \geq r. Then the family of the uiu^i extends to a morphism of complexes from PP into EE; two such extensions are homotopic.

Let us prove a). The existence of an extension vv of the family (ui)i≤r(u_i)_{i \leq r} follows at once from Lemma 2 by induction. Let v′v' be another extension; put f=v′−vf = v' - v, and construct by induction on the integer nn a homomorphism kn:Pn→En+1k_n : P_n \to E_{n+1} such that fn=dE∘kn+kn−1∘dPf_n = d_E \circ k_n + k_{n-1} \circ d_P. For i≤ri \leq r, take ki=0k_i = 0. Let n≥rn \geq r and suppose that the kik_i have been constructed for i≤ni \leq n. Consider then the diagram

Pn+1→dPPn→dPPn−1fn+1↓kn−1↓En+2→dEEn+1→dEEn. \begin{array}{cccccc} P_{n+1} & \xrightarrow{d_P} & P_n & \xrightarrow{d_P} & P_{n-1} \\ f_{n+1} \downarrow & & & & k_{n-1} \downarrow \\ E_{n+2} & \xrightarrow{d_E} & E_{n+1} & \xrightarrow{d_E} & E_n . \end{array}

The hypotheses of Lemma 1 are satisfied; there therefore exists an AA-homomorphism kn+1:Pn+1→En+2k_{n+1} : P_{n+1} \to E_{n+2} such that fn+1=dE∘kn+1+kn∘dPf_{n+1} = d_E \circ k_{n+1} + k_n \circ d_P, whence a).

The proof of b) is analogous, via Lemmas 1 bis and 2 bis.

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

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