The Stacks project

Lemma 21.18.5. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}')$ be a morphism of ringed topoi. There is a canonical bifunctorial isomorphism

\[ \mathcal{F}^\bullet \otimes _\mathcal {O}^{\mathbf{L}} Lf^*\mathcal{G}^\bullet = \mathcal{F}^\bullet \otimes _{f^{-1}\mathcal{O}_ Y}^{\mathbf{L}} f^{-1}\mathcal{G}^\bullet \]

for $\mathcal{F}^\bullet $ in $D(\mathcal{O})$ and $\mathcal{G}^\bullet $ in $D(\mathcal{O}')$.

Proof. Let $\mathcal{F}$ be an $\mathcal{O}$-module and let $\mathcal{G}$ be an $\mathcal{O}'$-module. Then $\mathcal{F} \otimes _{\mathcal{O}} f^*\mathcal{G} = \mathcal{F} \otimes _{f^{-1}\mathcal{O}'} f^{-1}\mathcal{G}$ because $f^*\mathcal{G} = \mathcal{O} \otimes _{f^{-1}\mathcal{O}'} f^{-1}\mathcal{G}$. The lemma follows from this and the definitions. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 21.18: Derived pullback

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Your email address will not be published.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 08I6. Beware of the difference between the letter 'O' and the digit '0'.