The Stacks project

Proof. Recall that we have the quasi-isomorphism $K^\bullet \to \underline{\mathbf{Z}}[0]$, see (69.6.1.1). Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet$ in $\textit{Ab}(X_{\acute{e}tale})$. Consider the double complex $\mathop{\mathrm{Hom}}\nolimits (K^\bullet , \mathcal{I}^\bullet )$ with terms $\mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{I}^ q)$. The differential $d_1^{p, q} : A^{p, q} \to A^{p + 1, q}$ is the one coming from the differential $K^{p + 1} \to K^ p$ and the differential $d_2^{p, q} : A^{p, q} \to A^{p, q + 1}$ is the one coming from the differential $\mathcal{I}^ q \to \mathcal{I}^{q + 1}$. Denote $\text{Tot}(\mathop{\mathrm{Hom}}\nolimits (K^\bullet , \mathcal{I}^\bullet ))$ the associated total complex, see Homology, Section 12.18. We will use the two spectral sequences $({}'E_ r, {}'d_ r)$ and $({}''E_ r, {}''d_ r)$ associated to this double complex, see Homology, Section 12.25.

Because $K^\bullet$ is a resolution of $\underline{\mathbf{Z}}$ we see that the complexes

are acyclic in positive degrees and have $H^0$ equal to $\Gamma (X, \mathcal{I}^ q)$. Hence by Homology, Lemma 12.25.4 the natural map

is a quasi-isomorphism of complexes of abelian groups. In particular we conclude that $H^ n(\text{Tot}(\mathop{\mathrm{Hom}}\nolimits (K^\bullet , \mathcal{I}^\bullet ))) = H^ n(X, \mathcal{F})$.

The map $\check{\mathcal{C}}^\bullet _{alt}(f, \mathcal{F}) \to R\Gamma (X, \mathcal{F})$ of the lemma is the composition of $\check{\mathcal{C}}^\bullet _{alt}(f, \mathcal{F}) \to \text{Tot}(\mathop{\mathrm{Hom}}\nolimits (K^\bullet , \mathcal{I}^\bullet ))$ with the inverse of the displayed quasi-isomorphism.

Finally, consider the spectral sequence $({}'E_ r, {}'d_ r)$. We have

This proves the lemma. $\square$