Lemma 102.6.2. Let f : \mathcal{X} \to Y be a morphism from an algebraic stack to an algebraic space. Assume:
f is of finite type and separated,
Y is quasi-compact and quasi-separated.
Then there exists a separated morphism of finite presentation f' : \mathcal{X}' \to Y and a closed immersion \mathcal{X} \to \mathcal{X}' of algebraic stacks over Y.
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
V_ i \to U_ i \times _{Y_ i} U_ i
(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
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