The Stacks project

Lemma 97.11.11. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. Let $\mathcal{Z}$ be an algebraic stack satisfying the equivalent conditions of Lemma 97.11.3 and let $\mathcal{Z} \to \mathcal{X}$ be a monomorphism such that the image of $|\mathcal{Z}| \to |\mathcal{X}|$ is $x$. Then the residual gerbe $\mathcal{Z}_ x$ of $\mathcal{X}$ at $x$ exists and $\mathcal{Z} \to \mathcal{X}$ factors as $\mathcal{Z} \to \mathcal{Z}_ x \to \mathcal{X}$ where the first arrow is an equivalence.

Proof. Let $\mathcal{Z}_ x \subset \mathcal{X}$ be the full subcategory corresponding to the essential image of the functor $\mathcal{Z} \to \mathcal{X}$. Then $\mathcal{Z} \to \mathcal{Z}_ x$ is an equivalence, hence $\mathcal{Z}_ x$ is an algebraic stack, see Algebraic Stacks, Lemma 91.12.4. Since $\mathcal{Z}_ x$ inherits all the properties of $\mathcal{Z}$ from this equivalence it is clear from the uniqueness in Lemma 97.11.7 that $\mathcal{Z}_ x$ is the residual gerbe of $\mathcal{X}$ at $x$. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 97.11: Residual gerbes

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Your email address will not be published.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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 06MX. Beware of the difference between the letter 'O' and the digit '0'.