Bourbaki

Definition 3

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

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

Let (C,d)(C, d) and (C′,d′)(C', d') be two complexes. A homological isomorphism (or quasi-isomorphism) from (C,d)(C, d) to (C′,d′)(C', d') is a morphism uu from (C,d)(C, d) to (C′,d′)(C', d') such that H(u)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)(C, d) is of null homology if H(C)=0H(C) = 0, that is to say if the unique morphism of complexes 0→C0 \to C (resp. C→0C \to 0) is a homological isomorphism. We say that (C,d)(C, d) is acyclic in descending degree nn (resp. in ascending degree nn) if Hn(C)=0H_n(C) = 0 (resp. Hn(C)=0H^n(C) = 0).

Let (C,d)(C, d) be a complex and p∈Zp \in \mathbf{Z}. The pp-th translate of (C,d)(C, d) is the complex (C(p),d(p))(C(p), d(p)) obtained as follows: C(p)C(p) is the a-module obtained by shifting the grading of CC by pp (II, p. 163; Example 3), so that

C(p)n=Cn+p,C(p)n=Cn−p; C(p)_n = C_{n+p} , \quad C(p)^n = C^{n-p} ;
in particular C(p)0=CpC(p)_0 = C_p; note also that CC is the direct sum of its graded submodules Cp(−p)C_p(-p), p∈Zp \in \mathbf{Z} (resp. Cp(p)C^p(p), p∈Zp \in \mathbf{Z}). We set d(p)=(−1)pdd(p) = (-1)^p d. We have Z(C(p))=Z(C)(p)Z(C(p)) = Z(C)(p), B(C(p))=B(C)(p)B(C(p)) = B(C)(p) and H(C(p))=H(C)(p)H(C(p)) = H(C)(p).

For example, dd is a morphism of complexes from CC to C(−1)C(-1) and

H(d):H(C)→H(C)(−1) H(d) : H(C) \to H(C)(-1)
is zero.

For every morphism of complexes u:(C,d)→(C′,d′)u : (C,d) \to (C',d'), and every p∈Zp \in \mathbf{Z}, uu is also a morphism from (C(p),d(p))(C(p),d(p)) to (C′(p),d′(p))(C'(p),d'(p)); it is sometimes denoted u(p)u(p) and we have

u(p)n=un+p,u(p)n=un−p. u(p)_n = u_{n+p}, \qquad u(p)^n = u^{n-p}.

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

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