Lemma 101.25.4. Let f : \mathcal{X} \to \mathcal{Y} be a morphism of algebraic stacks. Let \mathcal{Z} \to \mathcal{Y} be a surjective flat morphism of algebraic stacks. If the base change \mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z} is flat, then f is flat.
Proof. Choose an algebraic space W and a surjective smooth morphism W \to \mathcal{Z}. Then W \to \mathcal{Z} is surjective and flat (Morphisms of Spaces, Lemma 67.37.7) hence W \to \mathcal{Y} is surjective and flat (by Properties of Stacks, Lemma 100.5.2 and Lemma 101.25.2). Since the base change of \mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z} by W \to \mathcal{Z} is a flat morphism (Lemma 101.25.3) we may replace \mathcal{Z} by W.
Choose an algebraic space V and a surjective smooth morphism V \to \mathcal{Y}. Choose an algebraic space U and a surjective smooth morphism U \to V \times _\mathcal {Y} \mathcal{X}. We have to show that U \to V is flat. Now we base change everything by W \to \mathcal{Y}: Set U' = W \times _\mathcal {Y} U, V' = W \times _\mathcal {Y} V, \mathcal{X}' = W \times _\mathcal {Y} \mathcal{X}, and \mathcal{Y}' = W \times _\mathcal {Y} \mathcal{Y} = W. Then it is still true that U' \to V' \times _{\mathcal{Y}'} \mathcal{X}' is smooth by base change. Hence by our definition of flat morphisms of algebraic stacks and the assumption that \mathcal{X}' \to \mathcal{Y}' is flat, we see that U' \to V' is flat. Then, since V' \to V is surjective as a base change of W \to \mathcal{Y} we see that U \to V is flat by Morphisms of Spaces, Lemma 67.31.3 (2) and we win. \square
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