The definition of an algebraic space is fppf local.

Lemma 76.11.1. 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. $\coprod _{i \in I} F_ i$ is an algebraic space (see Spaces, Lemma 61.8.4).

Then $F$ is an algebraic space.

Proof. Consider the morphism $\coprod F_ i \to F$. This is the base change of $\coprod S_ i \to S$ via $F \to S$. Hence it is representable, locally of finite presentation, flat and surjective by our definition of an fppf covering and Lemma 76.4.2. Thus Theorem 76.10.1 applies to show that $F$ is an algebraic space. $\square$

Comment #1225 by David Corwin on

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