Theorem 80.6.1. 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.
or assume that
the presheaf F is a sheaf, and
there exists an algebraic space X and a map X \to F which is representable by algebraic spaces, 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
Comments (2)
Comment #6069 by DatPham on
Comment #6199 by Johan on
There are also: