**Proof.**
This is formal. Proof of (1). Let $T = \mathop{\mathrm{lim}}\nolimits _{i \in I} T_ i$ be the directed limit of affine schemes $T_ i$ over $S$. We will prove that the functor $\mathop{\mathrm{colim}}\nolimits \mathcal{X}_{T_ i} \to \mathcal{X}_ T$ is essentially surjective. Recall that an object of the fibre product over $T$ is a quadruple $(T, x, z, \alpha )$ where $x$ is an object of $\mathcal{X}$ lying over $T$, $z$ is an object of $\mathcal{Z}$ lying over $T$, and $\alpha : p(x) \to q(z)$ is a morphism in the fibre category of $\mathcal{Y}$ over $T$. By assumption on $\mathcal{X}$ and $\mathcal{Z}$ we can find an $i$ and objects $x_ i$ and $z_ i$ over $T_ i$ such that $x_ i|_ T \cong T$ and $z_ i|_ T \cong z$. Then $\alpha $ corresponds to an isomorphism $p(x_ i)|_ T \to q(z_ i)|_ T$ which comes from an isomorphism $\alpha _{i'} : p(x_ i)|_{T_{i'}} \to q(z_ i)|_{T_{i'}}$ by our assumption on $\mathcal{Y}$. After replacing $i$ by $i'$, $x_ i$ by $x_ i|_{T_{i'}}$, and $z_ i$ by $z_ i|_{T_{i'}}$ we see that $(T_ i, x_ i, z_ i, \alpha _ i)$ is an object of the fibre product over $T_ i$ which restricts to an object isomorphic to $(T, x, z, \alpha )$ over $T$ as desired.

We omit the arguments showing that $\mathop{\mathrm{colim}}\nolimits \mathcal{X}_{T_ i} \to \mathcal{X}_ T$ is fully faithful in (2).
$\square$

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