The Stacks project

Proof. By Lemmas 100.9.7 and 100.9.8 we find that the map is a bijection. If $\mathcal{Y} \to \mathcal{X}$ factors through $\mathcal{Z}$ then of course the base change $\mathcal{Y} \times _\mathcal {X} U \to U$ factors through $Z$. Converse, suppose that $\mathcal{Y} \to \mathcal{X}$ is a morphism such that $\mathcal{Y} \times _\mathcal {X} U \to U$ factors through $Z$. We will show that for every scheme $T$ and morphism $T \to \mathcal{Y}$, given by an object $y$ of the fibre category of $\mathcal{Y}$ over $T$, the object $y$ is in fact in the fibre category of $\mathcal{Z}$ over $T$. Namely, the fibre product $T \times _\mathcal {X} U$ is an algebraic space and $T \times _\mathcal {X} U \to T$ is a surjective smooth morphism. Hence there is an fppf covering $\{ T_ i \to T\} $ such that $T_ i \to T$ factors through $T \times _\mathcal {X} U \to T$ for all $i$. Then $T_ i \to \mathcal{X}$ factors through $\mathcal{Y} \times _\mathcal {X} U$ and hence through $Z \subset U$. Thus $y|_{T_ i}$ is an object of $\mathcal{Z}$ (as $Z$ is the fibre product of $U$ with $\mathcal{Z}$ over $\mathcal{X}$). Since $\mathcal{Z}$ is a strictly full substack, we conclude that $y$ is an object of $\mathcal{Z}$ as desired. $\square$