Processing math: 100%

The Stacks project

Lemma 18.12.4. Let \mathcal{C}, \mathcal{D} be sites. Let f : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) be a morphism of topoi. Let \mathcal{O} be a sheaf of rings on \mathcal{C}. Let \mathcal{F} be a sheaf of \mathcal{O}-modules. Let \mathcal{G} be a sheaf of f_*\mathcal{O}-modules. Then

\mathop{\mathrm{Mor}}\nolimits _{\textit{Mod}(\mathcal{O})}( \mathcal{O} \otimes _{f^{-1}f_*\mathcal{O}} f^{-1}\mathcal{G}, \mathcal{F}) = \mathop{\mathrm{Mor}}\nolimits _{\textit{Mod}(f_*\mathcal{O})}(\mathcal{G}, f_*\mathcal{F}).

Here we use Lemmas 18.12.2 and 18.12.1, and we use the canonical map f^{-1}f_*\mathcal{O} \to \mathcal{O} in the definition of the tensor product.

Proof. Note that we have

\mathop{\mathrm{Mor}}\nolimits _{\textit{Mod}(\mathcal{O})}( \mathcal{O} \otimes _{f^{-1}f_*\mathcal{O}} f^{-1}\mathcal{G}, \mathcal{F}) = \mathop{\mathrm{Mor}}\nolimits _{\textit{Mod}(f^{-1}f_*\mathcal{O})}( f^{-1}\mathcal{G}, \mathcal{F}_{f^{-1}f_*\mathcal{O}})

by Lemma 18.11.3. Hence the result follows from Lemma 18.12.3. \square


Comments (0)


Your email address will not be published.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.