Lemma 21.37.4. Let $u : \mathcal{C} \to \mathcal{D}$ be a continuous and cocontinuous functor of sites. Let $g : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{D})$ be the corresponding morphism of topoi. Let $\mathcal{O}_\mathcal {D}$ be a sheaf of rings and let $\mathcal{I}$ be an injective $\mathcal{O}_\mathcal {D}$-module. If $g_!^{Sh} : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{D})$ commutes with fibre products1, then $g^{-1}\mathcal{I}$ is totally acyclic.

Proof. We will use the criterion of Lemma 21.13.5. Condition (1) holds by Lemma 21.37.1. Let $K' \to K$ be a surjective map of sheaves of sets on $\mathcal{C}$. Since $g_!^{Sh}$ is a left adjoint, we see that $g_!^{Sh}K' \to g_!^{Sh}K$ is surjective. Observe that

\begin{align*} H^0(K' \times _ K \ldots \times _ K K', g^{-1}\mathcal{I}) & = H^0(g_!^{Sh}(K' \times _ K \ldots \times _ K K'), \mathcal{I}) \\ & = H^0(g_!^{Sh}K' \times _{g_!^{Sh}K} \ldots \times _{g_!^{Sh}K} g_!^{Sh}K', \mathcal{I}) \end{align*}

by our assumption on $g_!^{Sh}$. Since $\mathcal{I}$ is an injective module it is totally acyclic by Lemma 21.14.1 (applied to the identity). Hence we can use the converse of Lemma 21.13.5 to see that the complex

$0 \to H^0(K, g^{-1}\mathcal{I}) \to H^0(K', g^{-1}\mathcal{I}) \to H^0(K' \times _ K K', g^{-1}\mathcal{I}) \to \ldots$

is exact as desired. $\square$

 Holds if $\mathcal{C}$ has finite connected limits and $u$ commutes with them, see Sites, Lemma 7.21.6.

There are also:

• 3 comment(s) on Section 21.37: Derived lower shriek

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).