**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 $\beta = \text{id}_ x$.
$\square$

## Comments (2)

Comment #4922 by Robot0079 on

Comment #5190 by Johan on