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

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02WY. Beware of the difference between the letter 'O' and the digit '0'.