Bourbaki

Proposition 11

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

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

Let C be a bounded complex of type C\mathscr{C}. a) If C\mathscr{C} is stable, H(C)H(C) is of type C\mathscr{C}. b) If H(C)H(C) is of type C\mathscr{C}, it is analogously the case for B(C)B(C) and Z(C)Z(C), and one has χ(H(C))=χ(C)\chi(H(C)) = \chi(C).

a) If C\mathscr{C} is stable, for every nn the module Zn(C)Z_n(C) is of type C\mathscr{C} as kernel of dn:Cn→Cn−1d_n : C_n \to C_{n-1}, and Hn(C)H_n(C) is of type C\mathscr{C} as cokernel of Cn+1→ZnC_{n+1} \to Z_n. On the other hand, Hn(C)=0H_n(C) = 0 as soon as Cn=0C_n = 0.

b) Suppose H(C)H(C) of type C\mathscr{C}. The canonical exact sequences:

0→Zn(C)→Cn→Bn−1(C)→0 0 \to Z_n(C) \to C_n \to B_{n-1}(C) \to 0
0→Bn(C)→Zn(C)→Hn(C)→0 0 \to B_n(C) \to Z_n(C) \to H_n(C) \to 0

show by induction on nn starting from the right bound of CC that Zn(C)Z_n(C) and Bn(C)B_n(C) are of type C\mathscr{C} for all nn. One then has

χ(C)=χ(Z(C))+χ(B(C)(−1))=χ(Z(C))−χ(B(C))=χ(H(C)). \chi(C) = \chi(Z(C)) + \chi(B(C) (-1)) = \chi(Z(C)) - \chi(B(C)) = \chi(H(C)) .

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

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