The Stacks project

Lemma 61.12.18. Let $\mathit{Sch}_{pro\text{-}\acute{e}tale}$ be a big pro-étale site. Consider a cartesian diagram

\[ \xymatrix{ T' \ar[r]_{g'} \ar[d]_{f'} & T \ar[d]^ f \\ S' \ar[r]^ g & S } \]

in $\mathit{Sch}_{pro\text{-}\acute{e}tale}$. Then $i_ g^{-1} \circ f_{big, *} = f'_{small, *} \circ (i_{g'})^{-1}$ and $g_{big}^{-1} \circ f_{big, *} = f'_{big, *} \circ (g'_{big})^{-1}$.

Proof. Since the diagram is cartesian, we have for $U'/S'$ that $U' \times _{S'} T' = U' \times _ S T$. Hence both $i_ g^{-1} \circ f_{big, *}$ and $f'_{small, *} \circ (i_{g'})^{-1}$ send a sheaf $\mathcal{F}$ on $(\mathit{Sch}/T)_{pro\text{-}\acute{e}tale}$ to the sheaf $U' \mapsto \mathcal{F}(U' \times _{S'} T')$ on $S'_{pro\text{-}\acute{e}tale}$ (use Lemmas 61.12.12 and 61.12.15). The second equality can be proved in the same manner or can be deduced from the very general Sites, Lemma 7.28.1. $\square$


Comments (0)

There are also:

  • 6 comment(s) on Section 61.12: The pro-étale site

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Your email address will not be published.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 0F61. Beware of the difference between the letter 'O' and the digit '0'.