Theorem 76.6.1. Let $S$ be a scheme. Let $F : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$ be a functor. Assume that

the presheaf $F$ is a sheaf,

the diagonal morphism $F \to F \times F$ is representable by algebraic spaces, and

there exists an algebraic space $X$ and a map $X \to F$ which is surjective, and étale.

Then $F$ is an algebraic space.

**Proof.**
We will use the remarks directly below Definition 76.4.1 without further mention. In the situation of the theorem, let $U \to X$ be a surjective étale morphism from a scheme towards $X$. By Lemma 76.3.8 $U \to F$ is surjective and étale also. Hence the theorem boils down to proving that $\Delta _ F$ is representable. This follows immediately from Lemma 76.5.3. On the other hand we can circumvent this lemma and show directly $F$ is an algebraic space as in the next paragraph.

Let $U$ be a scheme, and let $U \to F$ be surjective and étale. Set $R = U \times _ F U$, which is an algebraic space (see Lemma 76.5.1). The morphism of algebraic spaces $R \to U \times _ S U$ is a monomorphism, hence separated (as the diagonal of a monomorphism is an isomorphism). Moreover, since $U \to F$ is étale, we see that $R \to U$ is étale, by Lemma 76.4.2. In particular, we see that $R \to U$ is locally quasi-finite, see Morphisms of Spaces, Lemma 63.39.5. We conclude that also $R \to U \times _ S U$ is locally quasi-finite by Morphisms of Spaces, Lemma 63.27.8. Hence Morphisms of Spaces, Proposition 63.50.2 applies and $R$ is a scheme. Hence $F = U/R$ is an algebraic space according to Spaces, Theorem 61.10.5.
$\square$

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