The Stacks project

A functor that admits a representable morphism to an algebraic space is an algebraic space.

Lemma 64.11.3. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F$ be an algebraic space over $S$. Let $G \to F$ be a representable transformation of functors. Then $G$ is an algebraic space.

Proof. By Lemma 64.3.5 we see that $G$ is a sheaf. The diagram

\[ \xymatrix{ G \times _ F G \ar[r] \ar[d] & F \ar[d]^{\Delta _ F} \\ G \times G \ar[r] & F \times F } \]

is cartesian. Hence we see that $G \times _ F G \to G \times G$ is representable by Lemma 64.3.3. By Lemma 64.3.6 we see that $G \to G \times _ F G$ is representable. Hence $\Delta _ G : G \to G \times G$ is representable as a composition of representable transformations, see Lemma 64.3.2. Finally, let $U$ be an object of $(\mathit{Sch}/S)_{fppf}$ and let $U \to F$ be surjective and ├ętale. By assumption $U \times _ F G$ is representable by a scheme $U'$. By Lemma 64.5.5 the morphism $U' \to G$ is surjective and ├ętale. This verifies the final condition of Definition 64.6.1 and we win. $\square$


Comments (1)

Comment #893 by Kestutis Cesnavicius on

Suggested slogan: A functor that admits a representable morphism to an algebraic space is representable by an algebraic space


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