Lemma 59.70.6. Let $j : U \to X$ be separated and étale. Then there is a functorial injective map $j_!\mathcal{F} \to j_*\mathcal{F}$ on abelian sheaves and sheaves of $\Lambda$-modules.

Proof. We prove this in the case of abelian sheaves. Let us construct a canonical map

$j_{p!}\mathcal{F} \to j_*\mathcal{F}$

of abelian presheaves on $X_{\acute{e}tale}$ for any abelian sheaf $\mathcal{F}$ on $U_{\acute{e}tale}$ where $j_{p!}$ is as in (59.70.1.1). Sheafification of this map will be the desired map $j_!\mathcal{F} \to j_*\mathcal{F}$. Evaluating both sides on $V \to X$ étale we obtain

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

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 separated over $V$ (Schemes, Lemma 26.21.11). 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 $\Gamma _\varphi$.

To show that our map is injective suppose that $\sum _{i = 1, \ldots , n} s_{\varphi _ i}$ is an element of $j_{p!}\mathcal{F}(V)$ in the formula above maps to zero in $j_*\mathcal{F}(V)$. Our task is to show that $\sum _{i = 1, \ldots , n} s_{\varphi _ i}$ restricts to zero on the members of an étale covering of $V$. Looking at all pairwise equalizers (which are open and closed in $V$) of the morphisms $\varphi _ i : V \to U$ and working locally on $V$, we may assume the images of the morphisms $\Gamma _{\varphi _1}, \ldots , \Gamma _{\varphi _ n}$ are pairwise disjoint. Since our assumption is that $\sum _{i = 1, \ldots , n} s'_{\varphi _ i} = 0$ we then immediately conclude that $s'_{\varphi _ i} = 0$ for each $i$ (by the disjointness of the supports of these sections), whence $s_{\varphi _ i} = 0$ for all $i$ as desired. $\square$

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