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

1. $\Delta _ F : F \to F \times F$ is representable by algebraic spaces,

2. for every scheme $T$ any map $T \to F$ is representable by algebraic spaces, and

3. 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 79.3.7 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. $\square$

Comment #1283 by on

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

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 which is enough.

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