Lemma 63.3.2. Let $X$ be a scheme and $g : X \to X$ a morphism. Assume that for all $\varphi : U \to X$ étale, there is an isomorphism

$\xymatrix{ U \ar[rd]_\varphi \ar[rr]^-\sim & & {U \times _{\varphi , X, g} X} \ar[ld]^{\text{pr}_2} \\ & X }$

functorial in $U$. Then $g$ induces the identity on cohomology (for any sheaf).

Proof. The proof is formal and without difficulty. $\square$

## Comments (2)

Comment #5116 by Laurent Moret-Bailly on

The meaning of the statement is not completely formal. To make sense of "the identity on cohomology" we need to show that we can identify $\mathcal{F}$ with $g_*(\mathcal{F})$ and/or $g^{-1}(\mathcal{F})$, for any $\mathcal{F}$. This is of course the case in subsequent lemmas where $\mathcal{F}$ is a constant sheaf.

Comment #5323 by on

Hmm... yes. OK, well, I think Exercise 63.3.1 about topological spaces just above the lemma explains why this would be so. To be precise $g_*\mathcal{F}(U) = \mathcal{F}(U \times_{\varphi, X, g} X)$ which is functorially equal to $\mathcal{F}(U)$ by the assumption of the lemma. So indeed $\mathcal{F} = g_*\mathcal{F}$ for all sheaves $\mathcal{F}$ and hence also $g^{-1}\mathcal{F} = \mathcal{F}$ for all sheaves $\mathcal{F}$. Then of course this agrees with what you would do on constant sheaves. Sigh!

There are also:

• 12 comment(s) on Section 63.3: Frobenii

## 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Your email address will not be published.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 03SP. Beware of the difference between the letter 'O' and the digit '0'.