The Stacks project

Proof. We have seen in Lemma 94.9.2 that $j$ is faithful on fibre categories. Consider a scheme $U$, two objects $u, v$ of $\mathcal{X}_ U$, and an isomorphism $t : j(u) \to j(v)$ in $\mathcal{Y}_ U$. We have to construct an isomorphism in $\mathcal{X}_ U$ between $u$ and $v$. By the $2$-Yoneda lemma (see Section 94.5) we think of $u$, $v$ as $1$-morphisms $u, v : (\mathit{Sch}/U)_{fppf} \to \mathcal{X}$ and we consider the $2$-fibre product

By assumption this is representable by an algebraic space $F_{j \circ v}$, over $U$ and the morphism $F_{j \circ v} \to U$ is a monomorphism. But since $(1_ U, v, 1_{j(v)})$ gives a $1$-morphism of $(\mathit{Sch}/U)_{fppf}$ into the displayed $2$-fibre product, we see that $F_{j \circ v} = U$ (here we use that if $V \to U$ is a monomorphism of algebraic spaces which has a section, then $V = U$). Therefore the $1$-morphism projecting to the first coordinate

is an equivalence of fibre categories. Since $(1_ U, u, t)$ and $(1_ U, v, 1_{j(v)})$ give two objects in $((\mathit{Sch}/U)_{fppf} \times _{j \circ v, \mathcal{Y}} \mathcal{X})_ U$ which have the same first coordinate, there must be a $2$-morphism between them in the $2$-fibre product. This is by definition a morphism $\tilde t : u \to v$ such that $j(\tilde t) = t$. $\square$