Lemma 101.38.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $g : \mathcal{W} \to \mathcal{X}$ be a morphism of algebraic stacks which is surjective, flat, and locally of finite presentation. Then the scheme theoretic image of $f$ exists if and only if the scheme theoretic image of $f \circ g$ exists and if so then these scheme theoretic images are the same.

Proof. Assume $\mathcal{Z} \subset \mathcal{Y}$ is a closed substack and $f \circ g$ factors through $\mathcal{Z}$. To prove the lemma it suffices to show that $f$ factors through $\mathcal{Z}$. Consider a scheme $T$ and a morphism $T \to \mathcal{X}$ given by an object $x$ of the fibre category of $\mathcal{X}$ over $T$. We will show that $f(x)$ is in fact in the fibre category of $\mathcal{Z}$ over $T$. Namely, the projection $T \times _\mathcal {X} \mathcal{W} \to T$ is a surjective, flat, locally finitely presented morphism. Hence there is an fppf covering $\{ T_ i \to T\}$ such that $T_ i \to T$ factors through $T \times _\mathcal {X} \mathcal{W} \to T$ for all $i$. Then $T_ i \to \mathcal{X}$ factors through $\mathcal{W}$ and hence $T_ i \to \mathcal{Y}$ factors through $\mathcal{Z}$. Thus $x|_{T_ i}$ is an object of $\mathcal{Z}$. Since $\mathcal{Z}$ is a strictly full substack, we conclude that $x$ is an object of $\mathcal{Z}$ as desired. $\square$

Comment #8099 by Shizhang on

Line 3 of the proof: should it be $f(x)$ is in fact in the fibre category...''?

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