Lemma 96.4.4. Let $\mathcal{X}$ be a category fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. The following are equivalent

1. $\Delta _\Delta : \mathcal{X} \to \mathcal{X} \times _{\mathcal{X} \times \mathcal{X}} \mathcal{X}$ is representable by algebraic spaces,

2. for every $1$-morphism $\mathcal{V} \to \mathcal{X} \times \mathcal{X}$ with $\mathcal{V}$ representable (by a scheme) the fibre product $\mathcal{Y} = \mathcal{X} \times _{\Delta , \mathcal{X} \times \mathcal{X}} \mathcal{V}$ has diagonal representable by algebraic spaces.

Proof. Although this is a bit of a brain twister, it is completely formal. Namely, recall that $\mathcal{X} \times _{\mathcal{X} \times \mathcal{X}} \mathcal{X} = \mathcal{I}_\mathcal {X}$ is the inertia of $\mathcal{X}$ and that $\Delta _\Delta$ is the identity section of $\mathcal{I}_\mathcal {X}$, see Categories, Section 4.34. Thus condition (1) says the following: Given a scheme $V$, an object $x$ of $\mathcal{X}$ over $V$, and a morphism $\alpha : x \to x$ of $\mathcal{X}_ V$ the condition “$\alpha = \text{id}_ x$” defines an algebraic space over $V$. (In other words, there exists a monomorphism of algebraic spaces $W \to V$ such that a morphism of schemes $f : T \to V$ factors through $W$ if and only if $f^*\alpha = \text{id}_{f^*x}$.)

On the other hand, let $V$ be a scheme and let $x, y$ be objects of $\mathcal{X}$ over $V$. Then $(x, y)$ define a morphism $\mathcal{V} = (\mathit{Sch}/V)_{fppf} \to \mathcal{X} \times \mathcal{X}$. Next, let $h : V' \to V$ be a morphism of schemes and let $\alpha : h^*x \to h^*y$ and $\beta : h^*x \to h^*y$ be morphisms of $\mathcal{X}_{V'}$. Then $(\alpha , \beta )$ define a morphism $\mathcal{V}' = (\mathit{Sch}/V)_{fppf} \to \mathcal{Y} \times \mathcal{Y}$. Condition (2) now says that (with any choices as above) the condition “$\alpha = \beta$” defines an algebraic space over $V$.

To see the equivalence, given $(\alpha , \beta )$ as in (2) we see that (1) implies that “$\alpha ^{-1} \circ \beta = \text{id}_{h^*x}$” defines an algebraic space. The implication (2) $\Rightarrow$ (1) follows by taking $h = \text{id}_ V$ and $\beta = \text{id}_ x$. $\square$

Comment #4922 by Robot0079 on

We can replace $X\times X$ with any category fibred in groupoid, say $Z$.

Note that morphism $Y\times_V Y \to Y\times Y$ is pull back of $V \to V\times V$, hence it's both representable by algebraic spaces and diagonal representable by algebraic spaces. Therefore condition 2 is equivalent to requiring diagonal of $Y \to V$ is representable for all V.

1 implies 2 is easy, we will show the inverse below. Fix scheme T, let $u: S \to T$ be a pull back of $\Delta_{X/Z}$. According to previous paragraph, u becomes representable after fibred producting with any scheme V over Z. Just take $V=T$ and further pull back by $\Delta_{T/Z}$, we see that u is representable and we win.

Comment #5190 by on

@#4922: OK, I think the proof as given is pretty perspicuous. But I did notice a typo which I fixed here.

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