The Stacks project

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0F4L. Beware of the difference between the letter 'O' and the digit '0'.