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 $T$. $\square$

There are also:

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

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).