The Stacks project

Lemma 59.39.3. Let $S$ be a scheme. Let $f : T \to S$ be a morphism such that

  1. $f$ is flat and quasi-compact, and

  2. the geometric fibres of $f$ are connected.

Let $\mathcal{F}$ be a sheaf on $S_{\acute{e}tale}$. Then $\Gamma (S, \mathcal{F}) = \Gamma (T, f^{-1}_{small}\mathcal{F})$.

Proof. There is a canonical map $\Gamma (S, \mathcal{F}) \to \Gamma (T, f_{small}^{-1}\mathcal{F})$. Since $f$ is surjective (because its fibres are connected) we see that this map is injective.

To show that the map is surjective, let $\alpha \in \Gamma (T, f_{small}^{-1}\mathcal{F})$. Since $\{ T \to S\} $ is an fpqc covering we can use Lemma 59.39.2 to see that suffices to prove that $\alpha $ pulls back to the same section over $T \times _ S T$ by the two projections. Let $\overline{s} \to S$ be a geometric point. It suffices to show the agreement holds over $(T \times _ S T)_{\overline{s}}$ as every geometric point of $T \times _ S T$ is contained in one of these geometric fibres. In other words, we are trying to show that $\alpha |_{T_{\overline{s}}}$ pulls back to the same section over

\[ (T \times _ S T)_{\overline{s}} = T_{\overline{s}} \times _{\overline{s}} T_{\overline{s}} \]

by the two projections to $T_{\overline{s}}$. However, since $\mathcal{F}|_{T_{\overline{s}}}$ is the pullback of $\mathcal{F}|_{\overline{s}}$ it is a constant sheaf with value $\mathcal{F}_{\overline{s}}$. Since $T_{\overline{s}}$ is connected by assumption, any section of a constant sheaf is constant. Hence $\alpha |_{T_{\overline{s}}}$ corresponds to an element of $\mathcal{F}_{\overline{s}}$. Thus the two pullbacks to $(T \times _ S T)_{\overline{s}}$ both correspond to this same element and we conclude. $\square$

Comments (2)

Comment #3243 by Dario Weißmann on

Typo: should be . Also in the sentence ' the two projections .' I'd expect the projections, not the fibre product. Or don't mention them explicitly as done earlier.

The last sentence confuses me a bit. Firstly we want that is connected as well (Lemma 32.7.4). Secondly the claim is that the and right? Don't we need locally Noetherian as well as connected for this? Something like is locally of finite type?

Comment #3342 by on

OK, thanks very much. I've clarified the argument (you don't need to prove connectedness of the product at all to do the argument). See here.

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Your email address will not be published.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 0A3H. Beware of the difference between the letter 'O' and the digit '0'.