Lemma 36.9.3. In Situation 36.9.1. Let M^\bullet be a complex of A-modules and denote \mathcal{F}^\bullet the associated complex of \mathcal{O}_ X-modules. Then there is a canonical isomorphism of complexes
functorial in M^\bullet .
Lemma 36.9.3. In Situation 36.9.1. Let M^\bullet be a complex of A-modules and denote \mathcal{F}^\bullet the associated complex of \mathcal{O}_ X-modules. Then there is a canonical isomorphism of complexes
functorial in M^\bullet .
Proof. Consider the double complex F^{\bullet , \bullet } with terms F^{p, q} = \mathcal{C}_{alt}^ p(\mathcal{U}, \mathcal{F}^ q) discussed in Cohomology, Section 20.25. Consider the double complex G^{\bullet , \bullet } with terms G^{p, q} = \mathop{\mathrm{colim}}\nolimits _ e \mathop{\mathrm{Hom}}\nolimits _ A(I^{-p}(f_1^ e, \ldots , f_ r^ e), M^ q) and differentials given by functoriality (without the intervention of signs). The maps \psi ^{p, q} : G^{p, q} \to F^{p, q} constructed in the proof of Lemma 36.9.2 are isomorphisms and compatible with the differentials d_1 (by the lemma) and d_2 (this is clear). However, the differentials d on the complexes on the left and right hand side of the arrow in the lemma have different signs. Namely, for g \in G^{p, q} is given by
(see More on Algebra, Section 15.71) and the differential for f \in F^{p, q} is given by
Thus we can fix the signs by multiplying \psi ^{p, q} by (-1)^{pq + p(p - 1)/2}. \square
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