Let (C,d) and (C′,d′) be two complexes. A morphism 1 from (C,d) to (C′,d′) is a graded A-homomorphism u of degree 0 from C to C′ such that
d′∘u=u∘d.
¹ Or morphism of degree 0 (cf. X, p. 81).
For every n, we therefore have dn′∘un=un−1∘dn and d′′n∘un=un+1∘dn. We have
u(Z(C))⊂Z(C′),u(B(C))⊂B(C′), and we denote by
Z(u):Z(C)→Z(C′),
B(u):B(C)→B(C′),
H(u):H(C)→H(C′) the homomorphisms of A-modules deduced from this; the homogeneous components of these morphisms are denoted
Zn(u),
Zn(u), ...
If v is another morphism from (C,d) to (C′,d′), then u+v is a morphism from (C,d) to (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). Analogously, if
A is an algebra over a commutative ring
k, and if
λ∈k, then
λu is a morphism from
(C,d) to
(C′,d′) and we have
Z(λu)=λZ(u),B(λu)=λB(u),H(λu)=λH(u). If
u′:(C′,d′)→(C′′,d′′) is another morphism of complexes, then
u′∘u is a morphism from
(C,d) to
(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). 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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