Lemma 101.38.6. 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 101.38.5 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 100.4.3. By Lemma 101.38.2 $Z$ is the scheme theoretic image of $W \to V$. By Morphisms of Spaces, Lemma 67.16.3 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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)