The Stacks project

Proof. First, let $\mathcal{I}$ be an injective abelian sheaf on $\mathcal{C}$. Then the spectral sequence of Lemma 85.9.3 for the sheaf $a^{-1}\mathcal{I}$ degenerates as $(a^{-1}\mathcal{I})_ p = a_ p^{-1}\mathcal{I}$ is injective by Lemma 85.15.4. Thus the complex

computes $Ra_*a^{-1}\mathcal{I}$. By Lemma 85.17.2 this is equal to the complex $\mathop{\mathcal{H}\! \mathit{om}}\nolimits (s(\mathbf{Z}_{F(K)}^\# ), \mathcal{I})$. Because $K$ is a hypercovering, we see that $s(\mathbf{Z}_{F(K)}^\# )$ is exact in degrees $> 0$ by Hypercoverings, Lemma 25.4.4 applied to the simplicial presheaf $F(K)$. Since $\mathcal{I}$ is injective, the functor $\mathop{\mathcal{H}\! \mathit{om}}\nolimits (-, \mathcal{I})$ is exact and we conclude that $\mathop{\mathcal{H}\! \mathit{om}}\nolimits (s(\mathbf{Z}_{F(K)}^\# ), \mathcal{I})$ is exact in positive degrees. We conclude that $R^ pa_*a^{-1}\mathcal{I} = 0$ for $p > 0$. On the other hand, we have $\mathcal{I} = a_*a^{-1}\mathcal{I}$ by Lemma 85.17.1.

Bounded case. Let $E \in D^+(\mathcal{C})$. Choose a bounded below complex $\mathcal{I}^\bullet $ of injectives representing $E$. By the result of the first paragraph and Leray's acyclicity lemma (Derived Categories, Lemma 13.16.7) $Ra_*a^{-1}\mathcal{I}^\bullet $ is computed by the complex $a_*a^{-1}\mathcal{I}^\bullet = \mathcal{I}^\bullet $ and we conclude the lemma is true in this case.

Unbounded case. We urge the reader to skip this, since the argument is the same as above, except that we use explicit representation by double complexes to get around convergence issues. Let $E \in D(\mathcal{C})$. To show the map $E \to Ra_*a^{-1}E$ is an isomorphism, it suffices to show for every object $U$ of $\mathcal{C}$ that

We will compute both sides and show the map $E \to Ra_*a^{-1}E$ induces an isomorphism. Choose a K-injective complex $\mathcal{I}^\bullet $ representing $E$. Choose a quasi-isomorphism $a^{-1}\mathcal{I}^\bullet \to \mathcal{J}^\bullet $ for some K-injective complex $\mathcal{J}^\bullet $ on $(\mathcal{C}/K)_{total}$. We have

and

By Lemma 85.9.1 we have a quasi-isomorphism

Hence $R\mathop{\mathrm{Hom}}\nolimits (a^{-1}\mathbf{Z}_ U^\# , a^{-1}E)$ is equal to

By the construction in Cohomology on Sites, Section 21.35 and since $\mathcal{J}^\bullet $ is K-injective, we see that this is represented by the complex of abelian groups with terms

See Cohomology on Sites, Lemmas 21.34.6 and 21.35.1 for more information. Thus we find that $R\Gamma (U, Ra_*a^{-1}E)$ is computed by the product total complex $\text{Tot}_\pi (B^{\bullet , \bullet })$ with $B^{p, q} = \mathop{\mathrm{Hom}}\nolimits (a_ p^{-1}\mathbf{Z}_ U^\# , g_ p^{-1}\mathcal{J}^ q)$. For the other side we argue similarly. First we note that

is a quasi-isomorphism of complexes on $\mathcal{C}$ by Hypercoverings, Lemma 25.4.4. Since $\mathbf{Z}_ U^\# $ is a flat sheaf of $\mathbf{Z}$-modules we see that

is a quasi-isomorphism. Therefore $R\mathop{\mathrm{Hom}}\nolimits (\mathbf{Z}_ U^\# , E)$ is equal to

By the construction of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits $ and since $\mathcal{I}^\bullet $ is K-injective, this is represented by the complex of abelian groups with terms

The equality of terms follows from the fact that $\mathbf{Z}^\# _{K_ p} \otimes _\mathbf {Z} \mathbf{Z}_ U^\# = a_{p!}a_ p^{-1}\mathbf{Z}_ U^\# $ by Modules on Sites, Remark 18.27.10. Thus we find that $R\Gamma (U, E)$ is computed by the product total complex $\text{Tot}_\pi (A^{\bullet , \bullet })$ with $A^{p, q} = \mathop{\mathrm{Hom}}\nolimits (a_ p^{-1}\mathbf{Z}_ U^\# , a_ p^{-1}\mathcal{I}^ q)$.

Since $\mathcal{I}^\bullet $ is K-injective we see that $a_ p^{-1}\mathcal{I}^\bullet $ is K-injective, see Lemma 85.15.4. Since $\mathcal{J}^\bullet $ is K-injective we see that $g_ p^{-1}\mathcal{J}^\bullet $ is K-injective, see Lemma 85.3.6. Both represent the object $a_ p^{-1}E$. Hence for every $p \geq 0$ the map of complexes

induced by $g_ p^{-1}$ applied to the given map $a^{-1}\mathcal{I}^\bullet \to \mathcal{J}^\bullet $ is a quasi-isomorphisms as these complexes both compute

By More on Algebra, Lemma 15.103.2 we conclude that the right vertical arrow in the commutative diagram

is a quasi-isomorphism. Since we saw above that the horizontal arrows are quasi-isomorphisms, so is the left vertical arrow. $\square$