Lemma 105.3.7. Let $(f, f') : (\mathcal{X} \subset \mathcal{X}') \to (\mathcal{Y} \subset \mathcal{Y}')$ be a morphism of thickenings of algebraic stacks. Then $\mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X}' \times _{\mathcal{Y}'} \mathcal{X}'$ is a thickening and the canonical diagram

\[ \xymatrix{ \mathcal{X} \ar[r]_-\Delta \ar[d] & \mathcal{X} \times _\mathcal {Y} \mathcal{X} \ar[d] \\ \mathcal{X}' \ar[r]^-{\Delta '} & \mathcal{X}' \times _{\mathcal{Y}'} \mathcal{X}' } \]

is cartesian.

**Proof.**
Since $\mathcal{X} \to \mathcal{Y}'$ factors through the closed substack $\mathcal{Y}$ we see that $\mathcal{X} \times _\mathcal {Y} \mathcal{X} = \mathcal{X} \times _{\mathcal{Y}'} \mathcal{X}$. Hence $\mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X}' \times _{\mathcal{Y}'} \mathcal{X}'$ is isomorphic to the composition

\[ \mathcal{X} \times _{\mathcal{Y}'} \mathcal{X} \to \mathcal{X} \times _{\mathcal{Y}'} \mathcal{X}' \to \mathcal{X}' \times _{\mathcal{Y}'} \mathcal{X}' \]

both of which are thickenings as base changes of thickenings (Lemma 105.3.4). Hence so is the composition (Lemma 105.3.5). Since $\mathcal{X} \to \mathcal{X}'$ is a monomorphism, the final statement of the lemma follows from Properties of Stacks, Lemma 99.8.6 applied to $\mathcal{X} \to \mathcal{X}' \to \mathcal{Y}'$.
$\square$

## Comments (2)

Comment #1916 by Matthew Emerton on

Comment #1986 by Johan on