**Proof.**
Proof of (1). Let $E \subset \mathop{\mathrm{Ob}}\nolimits (\textit{Ab}(Y_{\acute{e}tale}))$ be the class consisting of products of skyscraper sheaves. We claim that

every $\mathcal{G}$ in $\textit{Ab}(Y_{\acute{e}tale})$ is a subsheaf of an element of $E$, and

for every $\mathcal{G} \in E$ there exists an object $\mathcal{H}$ of $\textit{Ab}(X_{\acute{e}tale})$ such that $\mathop{\mathrm{Hom}}\nolimits (f_!\mathcal{F}, \mathcal{G}) = \mathop{\mathrm{Hom}}\nolimits (\mathcal{F}, \mathcal{H})$ functorially in $\mathcal{F}$.

Once the claim has been verified, the dual of Homology, Lemma 12.29.6 produces the adjoint functor $f^!$.

Part (a) is true because we can map $\mathcal{G}$ to the sheaf $\prod \overline{y}_*\mathcal{G}_{\overline{y}}$ where the product is over all geometric points of $Y$. This is an injection by Étale Cohomology, Theorem 59.29.10. (This is the first step in the Godement resolution when done in the setting of abelian sheaves on topological spaces.)

Part (b) and part (2) of the lemma can be seen as follows. Suppose that $\mathcal{G} = \prod \overline{y}_*A_{\overline{y}}$ for some abelian groups $A_{\overline{y}}$. Then

\[ \mathop{\mathrm{Hom}}\nolimits (f_!\mathcal{F}, \mathcal{G}) = \prod \mathop{\mathrm{Hom}}\nolimits (f_!\mathcal{F}, \overline{y}_*A_{\overline{y}}) \]

Thus it suffices to find abelian sheaves $\mathcal{H}_{\overline{y}}$ on $X_{\acute{e}tale}$ representing the functors $\mathcal{F} \mapsto \mathop{\mathrm{Hom}}\nolimits (f_!\mathcal{F}, \overline{y}_*A_{\overline{y}})$ and to take $\mathcal{H} = \prod \mathcal{H}_{\overline{y}}$. This reduces us to the case $\mathcal{H} = \overline{y}_*A$ for some fixed geometric point $\overline{y} : \mathop{\mathrm{Spec}}(k) \to Y$ and some fixed abelian group $A$. We claim that in this case $\mathcal{H} = \prod _{f(\overline{x}) = \overline{y}} \overline{x}_*A$ works. This will finish the proof of parts (1) and (2) of the lemma. Namely, we have

\[ \mathop{\mathrm{Hom}}\nolimits (f_!\mathcal{F}, \overline{y}_*A) = \mathop{\mathrm{Hom}}\nolimits _{\textit{Ab}}((f_!\mathcal{F})_{\overline{y}}, A) = \mathop{\mathrm{Hom}}\nolimits _{\textit{Ab}}(\bigoplus \nolimits _{f(\overline{x}) = \overline{y}} \mathcal{F}_{\overline{x}}, A) \]

by the description of stalks in Lemma 63.4.5 on the one hand and on the other hand we have

\[ \mathop{\mathrm{Hom}}\nolimits (\mathcal{F}, \mathcal{H}) = \prod \nolimits _{f(\overline{x}) = \overline{y}} \mathop{\mathrm{Hom}}\nolimits (\mathcal{F}, \overline{x}_*A) = \prod \nolimits _{f(\overline{x}) = \overline{y}} \mathop{\mathrm{Hom}}\nolimits _{\textit{Ab}}(\mathcal{F}_{\overline{x}}, A) \]

We leave it to the reader to identify these as functors of $\mathcal{F}$.

Proof of part (3). Observe that an object $\textit{Mod}(X_{\acute{e}tale}, \Lambda )$ is the same thing as an object $\mathcal{F}$ of $\textit{Ab}(X_{\acute{e}tale})$ together with a map $\Lambda \to \text{End}(\mathcal{F})$. Hence the functors $f_!$ and $f^!$ in (1) define functors $f_!$ and $f^!$ as in (3). A straightforward computation shows that they are adjoints.
$\square$

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