Lemma 102.4.2. In Situation 102.4.1 assume that \mathcal{X}_0 \to Y_0 is a morphism from algebraic stack to Y_0. Assume \mathcal{X}_0 is quasi-compact and quasi-separated. If Y \times _{Y_0} \mathcal{X}_0 \to Y is separated, then Y_ i \times _{Y_0} \mathcal{X}_0 \to Y_ i is separated for all sufficiently large i \in I.
Proof. Write \mathcal{X} = Y \times _{Y_0} \mathcal{X}_0 and \mathcal{X}_ i = Y_ i \times _{Y_0} \mathcal{X}_0. Choose an affine scheme U_0 and a surjective smooth morphism U_0 \to \mathcal{X}_0. Set U = Y \times _{Y_0} U_0 and U_ i = Y_ i \times _{Y_0} U_0. Then U and U_ i are affine and U \to \mathcal{X} and U_ i \to \mathcal{X}_ i are smooth and surjective. Set R_0 = U_0 \times _{\mathcal{X}_0} U_0. Set R = Y \times _{Y_0} R_0 and R_ i = Y_ i \times _{Y_0} R_0. Then R = U \times _\mathcal {X} U and R_ i = U_ i \times _{\mathcal{X}_ i} U_ i.
With this notation note that \mathcal{X} \to Y is separated implies that R \to U \times _ Y U is proper as the base change of \mathcal{X} \to \mathcal{X} \times _ Y \mathcal{X} by U \times _ Y U \to \mathcal{X} \times _ Y \mathcal{X}. Conversely, we see that \mathcal{X}_ i \to Y_ i is separated if R_ i \to U_ i \times _{Y_ i} U_ i is proper because U_ i \times _{Y_ i} U_ i \to \mathcal{X}_ i \times _{Y_ i} \mathcal{X}_ i is surjective and smooth, see Properties of Stacks, Lemma 100.3.3. Observe that R_0 \to U_0 \times _{Y_0} U_0 is locally of finite type and that R_0 is quasi-compact and quasi-separated. By Limits of Spaces, Lemma 70.6.13 we see that R_ i \to U_ i \times _{Y_ i} U_ i is proper for large enough i which finishes the proof. \square
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