The Stacks project

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). $\square$