Proposition 10

Let C and C' be two bounded complexes of type $\mathscr{C}$. If there exists a homologism $u : C' \to C$, one has $\chi(C) = \chi(C')$. Indeed, Con ($u$) is bounded of type $\mathscr{C}$ and one has $\chi(\operatorname{Con}(u)) = \chi(C) - \chi(C')$; on the other hand, $H(\operatorname{Con}(u)) = 0$ by X, p. 38, cor., hence $\chi(\operatorname{Con}(u)) = 0$ (lemma 4).