The Stacks project

Lemma 100.5.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a monomorphism of algebraic stacks. Then the diagram

\[ \xymatrix{ \mathcal{I}_\mathcal {X} \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \mathcal{I}_\mathcal {Y} \ar[r] & \mathcal{Y} } \]

is a fibre product square.

Proof. This follows immediately from the fact that $f$ is fully faithful (see Properties of Stacks, Lemma 99.8.4) and the definition of the inertia in Categories, Section 4.34. Namely, an object of $\mathcal{I}_\mathcal {X}$ over a scheme $T$ is the same thing as a pair $(x, \alpha )$ consisting of an object $x$ of $\mathcal{X}$ over $T$ and a morphism $\alpha : x \to x$ in the fibre category of $\mathcal{X}$ over $T$. As $f$ is fully faithful we see that $\alpha $ is the same thing as a morphism $\beta : f(x) \to f(x)$ in the fibre category of $\mathcal{Y}$ over $T$. Hence we can think of objects of $\mathcal{I}_\mathcal {X}$ over $T$ as triples $((y, \beta ), x, \gamma )$ where $y$ is an object of $\mathcal{Y}$ over $T$, $\beta : y \to y$ in $\mathcal{Y}_ T$ and $\gamma : y \to f(x)$ is an isomorphism over $T$, i.e., an object of $\mathcal{I}_\mathcal {Y} \times _\mathcal {Y} \mathcal{X}$ over $T$. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 100.5: Inertia stacks

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Your email address will not be published.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 06R5. Beware of the difference between the letter 'O' and the digit '0'.