Lemma 100.5.5. Let $\pi : \mathcal{X} \to \mathcal{Y}$ and $f : \mathcal{Y}' \to \mathcal{Y}$ be morphisms of algebraic stacks. Set $\mathcal{X}' = \mathcal{X} \times _\mathcal {Y} \mathcal{Y}'$. Then both squares in the diagram

\[ \xymatrix{ \mathcal{I}_{\mathcal{X}'/\mathcal{Y}'} \ar[r] \ar[d]_{ \text{Categories, Equation}\ (04Z4) } & \mathcal{X}' \ar[r]_{\pi '} \ar[d] & \mathcal{Y}' \ar[d]^ f \\ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[r] & \mathcal{X} \ar[r]^\pi & \mathcal{Y} } \]

are fibre product squares.

**Proof.**
The inertia stack $\mathcal{I}_{\mathcal{X}'/\mathcal{Y}'}$ is defined as the category of pairs $(x', \alpha ')$ where $x'$ is an object of $\mathcal{X}'$ and $\alpha '$ is an automorphism of $x'$ with $\pi '(\alpha ') = \text{id}$, see Categories, Section 4.34. Suppose that $x'$ lies over the scheme $U$ and maps to the object $x$ of $\mathcal{X}$. By the construction of the $2$-fibre product in Categories, Lemma 4.32.3 we see that $x' = (U, x, y', \beta )$ where $y'$ is an object of $\mathcal{Y}'$ over $U$ and $\beta $ is an isomorphism $\beta : \pi (x) \to f(y')$ in the fibre category of $\mathcal{Y}$ over $U$. By the very construction of the $2$-fibre product the automorphism $\alpha '$ is a pair $(\alpha , \gamma )$ where $\alpha $ is an automorphism of $x$ over $U$ and $\gamma $ is an automorphism of $y'$ over $U$ such that $\alpha $ and $\gamma $ are compatible via $\beta $. The condition $\pi '(\alpha ') = \text{id}$ signifies that $\gamma = \text{id}$ whereupon the condition that $\alpha , \beta , \gamma $ are compatible is exactly the condition $\pi (\alpha ) = \text{id}$, i.e., means exactly that $(x, \alpha )$ is an object of $\mathcal{I}_{\mathcal{X}/\mathcal{Y}}$. In this way we see that the left square is a fibre product square (some details omitted).
$\square$

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