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
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
For each \varphi we have an open and closed immersion
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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