The Stacks project

Proof. The rough idea is to use that $\mathcal{X}$ has a dense open which is a gerbe (Morphisms of Stacks, Proposition 101.29.1) and appeal to Lemma 106.10.2. The reason this does not work is that the open may not be quasi-compact and one runs into technical problems. Thus we first do a (standard) reduction to the Noetherian case.

First we choose a closed immersion $\mathcal{X} \to \mathcal{X}'$ where $\mathcal{X}'$ is an algebraic stack separated and of finite type over $Y$. See Limits of Stacks, Lemma 102.6.2. Clearly it suffices to prove the theorem for $\mathcal{X}'$, hence we may assume $\mathcal{X} \to Y$ is separated and of finite presentation.

Assume $\mathcal{X} \to Y$ is separated and of finite presentation. By Limits of Spaces, Proposition 70.8.1 we can write $Y = \mathop{\mathrm{lim}}\nolimits Y_ i$ as the directed limit of a system of Noetherian algebraic spaces with affine transition morphisms. By Limits of Stacks, Lemma 102.5.1 there is an $i$ and a morphism $\mathcal{X}_ i \to Y_ i$ of finite presentation from an algebraic stack to $Y_ i$ such that $\mathcal{X} = Y \times _{Y_ i} \mathcal{X}_ i$. After increasing $i$ we may assume that $\mathcal{X}_ i \to Y_ i$ is separated, see Limits of Stacks, Lemma 102.4.2. Then it suffices to prove the theorem for $\mathcal{X}_ i \to Y_ i$. This reduces us to the case discussed in the next paragraph.

Assume $Y$ is Noetherian. We may replace $\mathcal{X}$ by its reduction (Properties of Stacks, Definition 100.10.4). This reduces us to the case discussed in the next paragraph.

Assume $Y$ is Noetherian and $\mathcal{X}$ is reduced. Since $\mathcal{X} \to Y$ is separated and $Y$ quasi-separated, we see that $\mathcal{X}$ is quasi-separated as an algebraic stack. Hence the inertia $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is quasi-compact. Thus by Morphisms of Stacks, Proposition 101.29.1 there exists a dense open substack $\mathcal{V} \subset \mathcal{X}$ which is a gerbe. Let $\mathcal{V} \to V$ be the morphism which expresses $\mathcal{V}$ as a gerbe over the algebraic space $V$. See Morphisms of Stacks, Lemma 101.28.2 for a construction of $\mathcal{V} \to V$. This construction in particular shows that the morphism $\mathcal{V} \to Y$ factors as $\mathcal{V} \to V \to Y$. Picture

Since the morphism $\mathcal{V} \to V$ is surjective, flat, and of finite presentation (Morphisms of Stacks, Lemma 101.28.8) and since $\mathcal{V} \to Y$ is locally of finite presentation, it follows that $V \to Y$ is locally of finite presentation (Morphisms of Stacks, Lemma 101.27.12). Note that $\mathcal{V} \to V$ is a universal homeomorphism (Morphisms of Stacks, Lemma 101.28.13). Since $\mathcal{V}$ is quasi-compact (see Morphisms of Stacks, Lemma 101.8.2) we see that $V$ is quasi-compact. Finally, since $\mathcal{V} \to Y$ is separated the same is true for $V \to Y$ by Morphisms of Stacks, Lemma 101.27.17 applied to $\mathcal{V} \to V \to Y$ (whose assumptions are satisfied as we've already seen).

All of the above means that the assumptions of Limits of Spaces, Lemma 70.13.3 apply to the morphism $V \to Y$. Thus we can find a dense open subspace $V' \subset V$ and an immersion $V' \to \mathbf{P}^ n_ Y$ over $Y$. Clearly we may replace $V$ by $V'$ and $\mathcal{V}$ by the inverse image of $V'$ in $\mathcal{V}$ (recall that $|\mathcal{V}| = |V|$ as we've seen above). Thus we may assume we have a diagram

where the arrow $V \to \mathbf{P}^ n_ Y$ is an immersion. Let $\mathcal{X}'$ be the scheme theoretic image of the morphism

and let $Y'$ be the scheme theoretic image of the morphism $V \to \mathbf{P}^ n_ Y$. We obtain a commutative diagram

(See Morphisms of Stacks, Lemma 101.38.4). We claim that $\mathcal{V} = V \times _{Y'} \mathcal{X}'$ and that Lemma 106.10.2 applies to the morphism $\mathcal{X}' \to Y'$ and the open subspace $V \subset Y'$. If the claim is true, then we obtain

with $X \to \overline{X}$ an open immersion, $\overline{g}$ and $h$ proper, and such that $|V|$ is contained in the image of $|g|$. Then the composition $X \to \mathcal{X}' \to \mathcal{X}$ is proper (as a composition of proper morphisms) and its image contains $|\mathcal{V}|$, hence this composition is surjective. As well, $\overline{X} \to Y' \to Y$ is proper as a composition of proper morphisms.

The last step is to prove the claim. Observe that $\mathcal{X}' \to Y'$ is separated and of finite type, that $Y'$ is quasi-compact and quasi-separated, and that $V$ is quasi-compact (we omit checking all the details completely). Next, we observe that $b : \mathcal{X}' \to \mathcal{X}$ is an isomorphism over $\mathcal{V}$ by Morphisms of Stacks, Lemma 101.38.7. In particular $\mathcal{V}$ is identified with an open substack of $\mathcal{X}'$. The morphism $j$ is quasi-compact (source is quasi-compact and target is quasi-separated), so formation of the scheme theoretic image of $j$ commutes with flat base change by Morphisms of Stacks, Lemma 101.38.5. In particular we see that $V \times _{Y'} \mathcal{X}'$ is the scheme theoretic image of $\mathcal{V} \to V \times _{Y'} \mathcal{X}'$. However, by Morphisms of Stacks, Lemma 101.37.5 the image of $|\mathcal{V}| \to |V \times _{Y'} \mathcal{X}'|$ is closed (use that $\mathcal{V} \to V$ is a universal homeomorphism as we've seen above and hence is universally closed). Also the image is dense (combine what we just said with Morphisms of Stacks, Lemma 101.38.6) we conclude $|\mathcal{V}| = |V \times _{Y'} \mathcal{X}'|$. Thus $\mathcal{V} \to V \times _{Y'} \mathcal{X}'$ is an isomorphism and the proof of the claim is complete. $\square$