The Stacks project

Proof. First we use exactly the same procedure as in the proof of Lemma 102.6.1 (and we borrow its notation) to construct the embedding $\mathcal{X} \to \mathcal{X}'$ as a morphism $\mathcal{X} \to \mathcal{X}' = Y \times _{Y_ i} \mathcal{X}_ i$ with $\mathcal{X}_ i = [U_ i/R_ i]$. Thus it is enough to show that $\mathcal{X}_ i \to Y_ i$ is separated for sufficiently large $i$. In other words, it is enough to show that $\mathcal{X}_ i \to \mathcal{X}_ i \times _{Y_ i} \mathcal{X}_ i$ is proper for $i$ sufficiently large. Since the morphism $U_ i \times _{Y_ i} U_ i \to \mathcal{X}_ i \times _{Y_ i} \mathcal{X}_ i$ is surjective and smooth and since $R_ i = \mathcal{X}_ i \times _{\mathcal{X}_ i \times _{Y_ i} \mathcal{X}_ i} U_ i \times _{Y_ i} U_ i$ it is enough to show that the morphism $(s_ i, t_ i) : R_ i \to U_ i \times _{Y_ i} U_ i$ is proper for $i$ sufficiently large, see Properties of Stacks, Lemma 100.3.3. We prove this in the next paragraph.

Observe that $U \times _ Y U \to Y$ is quasi-separated and of finite type. Hence we can use the construction of Limits of Spaces, Remark 70.23.5 to find an $i_1 \in I$ and an inverse system $(V_ i)_{i \geq i_1}$ with $U \times _ Y U = \mathop{\mathrm{lim}}\nolimits _{i \geq i_1} V_ i$. By Limits of Spaces, Lemma 70.23.9 for $i$ sufficiently large the functoriality of the construction applied to the projections $U \times _ Y U \to U$ gives closed immersions

(There is a small mismatch here because in truth we should replace $Y_ i$ by the scheme theoretic image of $Y \to Y_ i$, but clearly this does not change the fibre product.) On the other hand, by Limits of Spaces, Lemma 70.23.8 the functoriality applied to the proper morphism $(s, t) : R \to U \times _ Y U$ (here we use that $\mathcal{X}$ is separated) leads to morphisms $R_ i \to V_ i$ which are proper for large enough $i$. Composing these morphisms we obtain a proper morphisms $R_ i \to U_ i \times _{Y_ i} U_ i$ for all $i$ large enough. The functoriality of the construction of Limits of Spaces, Remark 70.23.5 shows that this is the morphism is the same as $(s_ i, t_ i)$ for large enough $i$ and the proof is complete. $\square$