Lemma 80.5.1 (03Y2)—The Stacks project

The diagonal of a presheaf is representable by algebraic spaces if and only if every map from a scheme to the presheaf is representable by algebraic spaces.

Lemma 80.5.1. Let $S$ be a scheme. If $F$ is a presheaf on $(\mathit{Sch}/S)_{fppf}$. The following are equivalent:

$\Delta _ F : F \to F \times F$ is representable by algebraic spaces,
for every scheme $T$ any map $T \to F$ is representable by algebraic spaces, and
for every algebraic space $X$ any map $X \to F$ is representable by algebraic spaces.

Proof. Assume (1). Let $X \to F$ be as in (3). Let $T$ be a scheme, and let $T \to F$ be a morphism. Then we have

\[ T \times _ F X = (T \times _ S X) \times _{F \times F, \Delta } F \]

which is an algebraic space by Lemma 80.3.7 and (1). Hence $X \to F$ is representable, i.e., (3) holds. The implication (3) $\Rightarrow $ (2) is trivial. Assume (2). Let $T$ be a scheme, and let $(a, b) : T \to F \times F$ be a morphism. Then

\[ F \times _{\Delta _ F, F \times F} T = (T \times _{a, F, b} T) \times _{T \times T, \Delta _ T} T \]

which is an algebraic space by assumption. Hence $\Delta _ F$ is representable by algebraic spaces, i.e., (1) holds. $\square$

Comments (3)

Comment #1283 by Johan Commelin on January 29, 2015 at 12:09

Suggested slogan: The diagonal of an fppf-presheaf is representable if and only if every map from an algebraic space into the presheaf is representable.

Comment #4926 by Sean Cotner on February 11, 2020 at 23:16

The second displayed equation is not correct: the LHS is functorially the set of $t \in T$ with $a(t) = b(t)$ , whereas the RHS is functorially the set of pairs $(t, t') \in T \times T$ with $a(t) = b(t')$ . However, it is true that $F \times_{\Delta_F, F\times F} T = (T \times_{a, F, b} T) \times_{T \times T, \Delta_T} T,$ which is enough.

Comment #5194 by Johan on June 10, 2020 at 13:10

Thanks very much! Fixed here.

Comments (3)

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