Lemma 100.33.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. There is a maximal open substack $\mathcal{U} \subset \mathcal{X}$ such that $f|_\mathcal {U} : \mathcal{U} \to \mathcal{Y}$ is smooth. Moreover, formation of this open commutes with
precomposing by smooth morphisms,
base change by morphisms which are flat and locally of finite presentation,
base change by flat morphisms provided $f$ is locally of finite presentation.
Proof. Choose a commutative diagram
where $U$ and $V$ are algebraic spaces, the vertical arrows are smooth, and $a : U \to \mathcal{X}$ surjective. There is a maximal open subspace $U' \subset U$ such that $h_{U'} : U' \to V$ is smooth, see Morphisms of Spaces, Lemma 66.37.9. Let $\mathcal{U} \subset \mathcal{X}$ be the open substack corresponding to the image of $|U'| \to |\mathcal{X}|$ (Properties of Stacks, Lemmas 99.4.7 and 99.9.12). By the equivalence in Lemma 100.16.1 we find that $f|_\mathcal {U} : \mathcal{U} \to \mathcal{Y}$ is smooth and that $\mathcal{U}$ is the largest open substack with this property.
Assertion (1) follows from the fact that being smooth is smooth local on the source (this property was used to even define smooth morphisms of algebraic stacks). Assertions (2) and (3) follow from the case of algebraic spaces, see Morphisms of Spaces, Lemma 66.37.9. $\square$
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