Bourbaki

Definition 2

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

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

Let (C,d)(C, d) and (C′,d′)(C', d') be two complexes. A morphism 1^1 from (C,d)(C, d) to (C′,d′)(C', d') is a graded A-homomorphism uu of degree 0 from CC to C′C' such that

d′∘u=u∘d. d' \circ u = u \circ d .

¹ Or morphism of degree 0 (cf. X, p. 81).

For every nn, we therefore have dn′∘un=un−1∘dnd'_n \circ u_n = u_{n-1} \circ d_n and d′′n∘un=un+1∘dn{d''}^n \circ u^n = u^{n+1} \circ d^n. We have

u(Z(C))⊂Z(C′),u(B(C))⊂B(C′), u(Z(C)) \subset Z(C') , \quad u(B(C)) \subset B(C') ,
and we denote by Z(u):Z(C)→Z(C′)Z(u) : Z(C) \to Z(C'), B(u):B(C)→B(C′)B(u) : B(C) \to B(C'), H(u):H(C)→H(C′)H(u) : H(C) \to H(C') the homomorphisms of A-modules deduced from this; the homogeneous components of these morphisms are denoted Zn(u)Z_n(u), Zn(u)Z^n(u), ...

If vv is another morphism from (C,d)(C, d) to (C′,d′)(C', d'), then u+vu + v is a morphism from (C,d)(C, d) to (C′,d′)(C', d'), and we have

Z(u+v)=Z(u)+Z(v),B(u+v)=B(u)+B(v),H(u+v)=H(u)+H(v). Z(u + v) = Z(u) + Z(v) , \quad B(u + v) = B(u) + B(v) , \quad H(u + v) = H(u) + H(v) .
Analogously, if AA is an algebra over a commutative ring kk, and if λ∈k\lambda \in k, then λu\lambda u is a morphism from (C,d)(C, d) to (C′,d′)(C', d') and we have
Z(λu)=λZ(u),B(λu)=λB(u),H(λu)=λH(u). Z(\lambda u) = \lambda Z(u) , \quad B(\lambda u) = \lambda B(u) , \quad H(\lambda u) = \lambda H(u) .
If u′:(C′,d′)→(C′′,d′′)u' : (C', d') \to (C'', d'') is another morphism of complexes, then u′∘uu' \circ u is a morphism from (C,d)(C, d) to (C′′,d′′)(C'', d'') and we have
Z(u′∘u)=Z(u′)∘Z(u),B(u′∘u)=B(u′)∘B(u),H(u′∘u)=H(u′)∘H(u). Z(u' \circ u) = Z(u') \circ Z(u) , \quad B(u' \circ u) = B(u') \circ B(u) , \quad H(u' \circ u) = H(u') \circ H(u) .
It is clear that a bijective morphism is an isomorphism.

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

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Translated with machine help, not yet reviewed by a person.