Bourbaki

Example 3

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

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

If M is bounded of type C\mathscr{C}, it is analogously the case for M(p)M(p) for every p∈Zp \in \mathbf{Z}, and one has χ(M(p))=(−1)pχ(M)\chi(M(p)) = (-1)^p \chi(M).

4) Let 0→M′→M→M′′→00 \to M' \to M \to M'' \to 0 be an exact sequence of graded A-modules and graded homomorphisms of degree 0. If M, M' and M'' are bounded of type C\mathscr{C}, one has

χ(M)=χ(M′)+χ(M′′). \chi(M) = \chi(M') + \chi(M'').
If M and M'' are bounded of type C\mathscr{C}, it is analogously the case for M'; if C\mathscr{C} is stable and if two of the three modules are bounded of type C\mathscr{C}, it is analogously the case for the third.

5) Let u:C′→Cu : C' \to C be a morphism of bounded complexes of type C\mathscr{C}. Then Con (uu) is bounded of type C\mathscr{C}, and one has:

χ(Con⁡(u))=χ(C)−χ(C′). \chi(\operatorname{Con}(u)) = \chi(C) - \chi(C').

6) One may take for G the group K(C)K(\mathscr{C}) itself, and for φ\varphi the identity; in this case one denotes by χC(M)\chi_{\mathscr{C}}(M) the element χφ(M)=∑(−1)n[Mn]\chi_\varphi(M) = \sum (-1)^n[M_n] of K(C)K(\mathscr{C}).

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.