The Stacks project

Lemma 76.11.2. Let $S$ be a scheme. Let $F : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$ be a functor. Let $\{ S_ i \to S\} _{i \in I}$ be a covering of $(\mathit{Sch}/S)_{fppf}$. Assume that

  1. $F$ is a sheaf,

  2. each $F_ i = h_{S_ i} \times F$ is an algebraic space, and

  3. the morphisms $F_ i \to S_ i$ are of finite type.

Then $F$ is an algebraic space.

Proof. We will use Lemma 76.11.1 above. To do this we will show that the assumption that $F_ i$ is of finite type over $S_ i$ to prove that the set theoretic condition in the lemma is satisfied (after perhaps refining the given covering of $S$ a bit). We suggest the reader skip the rest of the proof.

If $S'_ i \to S_ i$ is a morphism of schemes then

\[ h_{S'_ i} \times F = h_{S'_ i} \times _{h_{S_ i}} h_{S_ i} \times F = h_{S'_ i} \times _{h_{S_ i}} F_ i \]

is an algebraic space of finite type over $S'_ i$, see Spaces, Lemma 61.7.3 and Morphisms of Spaces, Lemma 63.23.3. Thus we may refine the given covering. After doing this we may assume: (a) each $S_ i$ is affine, and (b) the cardinality of $I$ is at most the cardinality of the set of points of $S$. (Since to cover all of $S$ it is enough that each point is in the image of $S_ i \to S$ for some $i$.)

Since each $S_ i$ is affine and each $F_ i$ of finite type over $S_ i$ we conclude that $F_ i$ is quasi-compact. Hence by Properties of Spaces, Lemma 62.6.3 we can find an affine $U_ i \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and a surjective ├ętale morphism $U_ i \to F_ i$. The fact that $F_ i \to S_ i$ is locally of finite type then implies that $U_ i \to S_ i$ is locally of finite type, and in particular $U_ i \to S$ is locally of finite type. By Sets, Lemma 3.9.7 we conclude that $\text{size}(U_ i) \leq \text{size}(S)$. Since also $|I| \leq \text{size}(S)$ we conclude that $\coprod _{i \in I} U_ i$ is isomorphic to an object of $(\mathit{Sch}/S)_{fppf}$ by Sets, Lemma 3.9.5 and the construction of $\mathit{Sch}$. This implies that $\coprod F_ i$ is an algebraic space by Spaces, Lemma 61.8.4 and we win. $\square$


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