Let (C,d) and (C′,d′) be two complexes. A homological isomorphism (or quasi-isomorphism) from (C,d) to (C′,d′) is a morphism u from (C,d) to (C′,d′) such that H(u) is bijective.
Every isomorphism is a homological isomorphism, every morphism composed of homological isomorphisms is a homological isomorphism.
We say that (C,d) is of null homology if H(C)=0, that is to say if the unique morphism of complexes 0→C (resp. C→0) is a homological isomorphism. We say that (C,d) is acyclic in descending degree n (resp. in ascending degree n) if Hn(C)=0 (resp. Hn(C)=0).
Let (C,d) be a complex and p∈Z. The p-th translate of (C,d) is the complex (C(p),d(p)) obtained as follows: C(p) is the a-module obtained by shifting the grading of C by p (II, p. 163; Example 3), so that
C(p)n=Cn+p,C(p)n=Cn−p; in particular
C(p)0=Cp; note also that
C is the direct sum of its graded submodules
Cp(−p),
p∈Z (resp.
Cp(p),
p∈Z). We set
d(p)=(−1)pd. We have
Z(C(p))=Z(C)(p),
B(C(p))=B(C)(p) and
H(C(p))=H(C)(p).
For example, d is a morphism of complexes from C to C(−1) and
H(d):H(C)→H(C)(−1) is zero.
For every morphism of complexes u:(C,d)→(C′,d′), and every p∈Z, u is also a morphism from (C(p),d(p)) to (C′(p),d′(p)); it is sometimes denoted u(p) and we have
u(p)n=un+p,u(p)n=un−p.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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