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. $\square$

Comment #3243 by Dario Weißmann on

Typo: $\alpha|_{X_{\overline{s}}}$ should be $\alpha|_{T_{\overline{s}}}$. Also in the sentence '...by the two projections $T_{\overline{s}}\times_{\overline{s}}T_{\overline{s}}$.' 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 $(T\times_S T)_{\overline{s}}$ is connected as well (Lemma 32.7.4). Secondly the claim is that the $\mathcal{F}(pr_i)=\text{id}$ and $\mathcal{F}(T_{\overline{s}})=\mathcal{F}_{\overline{s}}= \mathcal{F}((T\times_S T)_{\overline{s}})$ right? Don't we need locally Noetherian as well as connected for this? Something like $T\to S$ 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.

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