Lemma 106.3.4. Let \mathcal{Y} \subset \mathcal{Y}' be a thickening of algebraic stacks. Let \mathcal{X}' \to \mathcal{Y}' be a morphism of algebraic stacks and set \mathcal{X} = \mathcal{Y} \times _{\mathcal{Y}'} \mathcal{X}'. Then (\mathcal{X} \subset \mathcal{X}') \to (\mathcal{Y} \subset \mathcal{Y}') is a morphism of thickenings. If \mathcal{Y} \subset \mathcal{Y}' is a first order thickening, then \mathcal{X} \subset \mathcal{X}' is a first order thickening.
Proof. See discussion above, Properties of Stacks, Section 100.3, and More on Morphisms of Spaces, Lemma 76.9.8. \square
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