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

$\mathop{\mathrm{colim}}\nolimits _ e \mathop{\mathrm{Hom}}\nolimits ^\bullet (I^\bullet (f_1^ e, \ldots , f_ r^ e), M^\bullet ) \longrightarrow \text{Tot}(\check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \mathcal{F}^\bullet ))$

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

$d(g) = d_2(g) - (-1)^{p + q} d_1(g)$

(see More on Algebra, Section 15.71) and the differential for $f \in F^{p, q}$ is given by

$d(f) = d_1(f) + (-1)^ p d_2(f)$

Thus we can fix the signs by multiplying $\psi ^{p, q}$ by $(-1)^{pq + p(p - 1)/2}$. $\square$

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