The Stacks project

Proof. Let $\mathcal{Y}' \to \mathcal{Y}$ be a flat morphism of algebraic stacks. Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Choose a scheme $V'$ and a surjective smooth morphism $V' \to \mathcal{Y}' \times _\mathcal {Y} V$. We may and do assume that $V = \coprod _{i \in I} V_ i$ is a disjoint union of affine schemes and that $V' = \coprod _{i \in I} \coprod _{j \in J_ i} V_{i, j}$ is a disjoint union of affine schemes with each $V_{i, j}$ mapping into $V_ i$. Let

1. $\mathcal{Z} \subset \mathcal{Y}$ be the scheme theoretic image of $f$,

2. $\mathcal{Z}' \subset \mathcal{Y}'$ be the scheme theoretic image of the base change of $f$ by $\mathcal{Y}' \to \mathcal{Y}$,

3. $Z \subset V$ be the scheme theoretic image of the base change of $f$ by $V \to \mathcal{Y}$,

4. $Z' \subset V'$ be the scheme theoretic image of the base change of $f$ by $V' \to \mathcal{Y}$.

If we can show that (a) $Z = V \times _\mathcal {Y} \mathcal{Z}$, (b) $Z' = V' \times _{\mathcal{Y}'} \mathcal{Z}'$, and (c) $Z' = V' \times _ V Z$ then the lemma follows: the inclusion $\mathcal{Z}' \to \mathcal{Z} \times _\mathcal {Y} \mathcal{Y}'$ (Lemma 101.38.4) has to be an isomorphism because after base change by the surjective smooth morphism $V' \to \mathcal{Y}'$ it is.

Proof of (a). Set $R = V \times _\mathcal {Y} V$. By Properties of Stacks, Lemma 100.9.11 the rule $\mathcal{Z} \mapsto \mathcal{Z} \times _\mathcal {Y} V$ defines a $1$-to-$1$ correspondence between closed substacks of $\mathcal{Y}$ and $R$-invariant closed subspaces of $V$. Moreover, $f : \mathcal{X} \to \mathcal{Y}$ factors through $\mathcal{Z}$ if and only if the base change $g : \mathcal{X} \times _\mathcal {Y} V \to V$ factors through $\mathcal{Z} \times _\mathcal {Y} V$. We claim: the scheme theoretic image $Z \subset V$ of $g$ is $R$-invariant. The claim implies (a) by what we just said.

For each $i$ the morphism $\mathcal{X} \times _\mathcal {Y} V_ i \to V_ i$ is quasi-compact and hence $\mathcal{X} \times _\mathcal {Y} V_ i$ is quasi-compact. Thus we can choose an affine scheme $W_ i$ and a surjective smooth morphism $W_ i \to \mathcal{X} \times _\mathcal {Y} V_ i$. Observe that $W = \coprod W_ i$ is a scheme endowed with a smooth and surjective morphism $W \to \mathcal{X} \times _\mathcal {Y} V$ such that the composition $W \to V$ with $g$ is quasi-compact. Let $Z \to V$ be the scheme theoretic image of $W \to V$, see Morphisms, Section 29.6 and Morphisms of Spaces, Section 67.16. It follows from Lemma 101.38.2 that $Z \subset V$ is the scheme theoretic image of $g$. To show that $Z$ is $R$-invariant we claim that both

are the scheme theoretic image of $\mathcal{X} \times _\mathcal {Y} R \to R$. Namely, we first use Morphisms of Spaces, Lemma 67.30.12 to see that $\text{pr}_0^{-1}(Z)$ is the scheme theoretic image of the composition

Since the first arrow here is surjective and smooth we see that $\text{pr}_0^{-1}(Z)$ is the scheme theoretic image of $\mathcal{X} \times _\mathcal {Y} R \to R$. The same argument applies that $\text{pr}_1^{-1}(Z)$. Hence $Z$ is $R$-invariant.

Statement (b) is proved in exactly the same way as one proves (a).

Proof of (c). Let $Z_ i \subset V_ i$ be the scheme theoretic image of $\mathcal{X} \times _\mathcal {Y} V_ i \to V_ i$ and let $Z_{i, j} \subset V_{i, j}$ be the scheme theoretic image of $\mathcal{X} \times _\mathcal {Y} V_{i, j} \to V_{i, j}$. Clearly it suffices to show that the inverse image of $Z_ i$ in $V_{i, j}$ is $Z_{i, j}$. Above we've seen that $Z_ i$ is the scheme theoretic image of $W_ i \to V_ i$ and by the same token $Z_{i, j}$ is the scheme theoretic image of $W_ i \times _{V_ i} V_{i, j} \to V_{i, j}$. Hence the equality follows from the case of schemes (Morphisms, Lemma 29.25.16) and the fact that $V_{i, j} \to V_ i$ is flat. $\square$