The Stacks project

Proof. Let $U$ be a scheme and let $\xi = (U, Z, p, x, 1)$ be an object of $\mathcal{H}_ d(\mathcal{X}) = \mathcal{H}_ d(\mathcal{X}/S)$ over $U$. Here $p$ is just the structure morphism of $U$. The fifth component $1$ exists and is unique since everything is over $S$. Also, let $y$ be an object of $\mathcal{Y}$ over $U$. We have to show the $2$-fibre product

is representable by an algebraic space. To explain why this is so we introduce

which is an algebraic space over $Z$ by assumption. Let $a : U' \to U$ be a scheme over $U$. What does it mean to give an object of the fibre category of (97.12.5.1) over $U'$? Well, it means that we have an object $\xi ' = (U', Z', y', x', \alpha ')$ of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ over $U'$ and isomorphisms $(U', Z', p', x', 1) \cong (U, Z, p, x, 1)|_{U'}$ and $y' \cong y|_{U'}$. Thus $\xi '$ is isomorphic to $(U', U' \times _{a, U} Z, a^*y, x|_{U' \times _{a, U} Z}, \alpha )$ for some morphism

in the fibre category of $\mathcal{Y}$ over $U' \times _{a, U} Z$. Hence we can view $\alpha $ as a morphism $b : U' \times _{a, U} Z \to I$. In this way we see that (97.12.5.1) is representable by $\text{Res}_{Z/U}(I)$ which is an algebraic space by Proposition 97.11.5. $\square$