
Lemma 54.69.5. Let $j : U \to X$ be finite and étale. Then $j_! = j_*$ on abelian sheaves and sheaves of $\Lambda$-modules.

Proof. We prove this in the case of abelian sheaves. We claim there is a natural transformation $j_! \to j_*$. We will construct a canonical map

$j_!^{PSh}\mathcal{F} \to j_*\mathcal{F}$

of functors $X_{\acute{e}tale}\to \textit{Ab}$ for any abelian sheaf $\mathcal{F}$ on $U_{\acute{e}tale}$. Sheafification of this map will be the desired map $j_!\mathcal{F} \to j_*\mathcal{F}$. Namely, given $V \to X$ étale we have

$j_!^{PSh}\mathcal{F}(V) = \bigoplus \nolimits _{\varphi : V \to U} \mathcal{F}(V \xrightarrow {\varphi } U) \quad \text{and}\quad j_*\mathcal{F}(V) = \mathcal{F}(V \times _ X U)$

For each $\varphi$ we have an open and closed immersion

$\Gamma _\varphi = (1, \varphi ) : V \longrightarrow V \times _ X U$

over $U$. (It is open as it is a morphism between schemes étale over $U$ and it is closed as it is a section of a scheme finite over $V$.) Thus for a section $s_\varphi \in \mathcal{F}(V \xrightarrow {\varphi } U)$ there exists a unique section $s'_\varphi$ in $\mathcal{F}(V \times _ X U)$ which pulls back to $s_\varphi$ by $\Gamma _\varphi$ and which restricts to zero on the complement of the image of $\Gamma _\varphi$. Then we map $(s_\varphi )$ in $j_!^{PSh}\mathcal{F}(V)$ to $\sum _\varphi s'_\varphi$ in $j_*\mathcal{F}(V) = \mathcal{F}(V \times _ X U)$. We leave it to the reader to see that this construction is compatible with restriction mappings.

It suffices to check $j_!\mathcal{F} \to j_*\mathcal{F}$ is an isomorphism étale locally on $X$. Thus we may assume $U \to X$ is a finite disjoint union of isomorphisms, see Étale Morphisms, Lemma 40.18.3. We omit the proof in this case. $\square$

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