Lemma 100.38.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a quasi-compact morphism of algebraic stacks. Let $\mathcal{Z} \subset \mathcal{Y}$ be the scheme theoretic image of $f$. Then $|\mathcal{Z}|$ is the closure of the image of $|f|$.

**Proof.**
Let $z \in |\mathcal{Z}|$ be a point. Choose an affine scheme $V$, a point $v \in V$, and a smooth morphism $V \to \mathcal{Y}$ mapping $v$ to $z$. Then $\mathcal{X} \times _\mathcal {Y} V$ is a quasi-compact algebraic stack. Hence we can find an affine scheme $W$ and a surjective smooth morphism $W \to \mathcal{X} \times _\mathcal {Y} V$. By Lemma 100.38.5 the scheme theoretic image of $\mathcal{X} \times _\mathcal {Y} V \to V$ is $Z = \mathcal{Z} \times _\mathcal {Y} V$. Hence the inverse image of $|\mathcal{Z}|$ in $|V|$ is $|Z|$ by Properties of Stacks, Lemma 99.4.3. By Lemma 100.38.2 $Z$ is the scheme theoretic image of $W \to V$. By Morphisms of Spaces, Lemma 66.16.3 we see that the image of $|W| \to |Z|$ is dense. Hence the image of $|\mathcal{X} \times _\mathcal {Y} V| \to |Z|$ is dense. Observe that $v \in Z$. Since $|V| \to |\mathcal{Y}|$ is open, a topology argument tells us that $z$ is in the closure of the image of $|f|$ as desired.
$\square$

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