The Stacks project

Lemma 85.6.3. With assumptions as in Lemma 85.6.1 the functor $g_{n!} : \textit{Mod}(\mathcal{O}_ n) \to \textit{Mod}(\mathcal{O})$ is exact if the maps $f_\varphi ^{-1}\mathcal{O}_ n \to \mathcal{O}_ m$ are flat for all $\varphi : [n] \to [m]$.

Proof. Recall that $g_{n!}\mathcal{G}$ is the $\mathcal{O}$-module whose degree $m$ part is the $\mathcal{O}_ m$-module

\[ \bigoplus \nolimits _{\varphi : [n] \to [m]} f_\varphi ^*\mathcal{G} \]

Here the morphism of ringed topoi $f_\varphi : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ m), \mathcal{O}_ m) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ n), \mathcal{O}_ n)$ uses the map $f_\varphi ^{-1}\mathcal{O}_ n \to \mathcal{O}_ m$ of the statement of the lemma. If these maps are flat, then $f_\varphi ^*$ is exact (Modules on Sites, Lemma 18.31.2). By definition of the site $\mathcal{C}_{total}$ we see that these functors have the desired exactness properties and we conclude. $\square$

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Your email address will not be published.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 0D74. Beware of the difference between the letter 'O' and the digit '0'.