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

$f$ is flat and quasi-compact, and

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

Comment #3342 by Johan on