Theorem 80.6.1. Let $S$ be a scheme. Let $F : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$ be a functor. Assume that

1. the presheaf $F$ is a sheaf,

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

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

or assume that

1. the presheaf $F$ is a sheaf, and

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

Then $F$ is an algebraic space.

Proof. We will use the remarks directly below Definition 80.4.1 without further mention.

Assume (1), (2), and (3) and let $X \to F$ be as in (3). By Lemma 80.5.1 the morphism $X \to F$ is representable by algebraic spaces. Thus we see that (a) and (b) hold.

Assume (a) and (b) and let $X \to F$ be as in (b). Let $U \to X$ be a surjective étale morphism from a scheme towards $X$. By Lemma 80.3.8 the transformation $U \to F$ is representable by algebraic spaces, surjective, and étale. Hence to prove that $F$ is an algebraic space boils down to proving that $\Delta _ F$ is representable (Spaces, Definition 65.6.1). This follows immediately from Lemma 80.5.3. On the other hand we can circumvent this lemma and show directly $F$ is an algebraic space as in the next paragraph.

Namely, let $U$ be a scheme and let $U \to F$ be representable by algebraic spaces, surjective, and étale. Consider the fibre product $R = U \times _ F U$. Both projections $R \to U$ are representable by algebraic spaces, surjective, and étale (Lemma 80.4.2). In particular $R$ is an algebraic space by Lemma 80.3.6. 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). Since $R \to U$ is étale, we see that $R \to U$ is locally quasi-finite, see Morphisms of Spaces, Lemma 67.39.5. We conclude that also $R \to U \times _ S U$ is locally quasi-finite by Morphisms of Spaces, Lemma 67.27.8. Hence Morphisms of Spaces, Proposition 67.50.2 applies and $R$ is a scheme. By Lemma 80.4.6 the map $U \to F$ is a surjection of sheaves. Thus $F = U/R$. We conclude that $F$ is an algebraic space by Spaces, Theorem 65.10.5. $\square$

Comment #6069 by DatPham on

I think in the first paragraph where we use Lemma 046K to deduce that $\Delta_F$ is representable by schemes, we don't need the assumption (2). I suppose you put it here just because you want to emphasize the similar with the original definition of an algebraic space. Is my understanding correct?

Comment #6199 by on

Thanks very much for this question/remark. I have clarified the theorem by stating exactly what is needed for the argument as you suggested. See this commit. Thanks again!

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