Lemma 106.5.1. Consider a commutative diagram
\xymatrix{ (\mathcal{X} \subset \mathcal{X}') \ar[rr]_{(f, f')} \ar[rd] & & (\mathcal{Y} \subset \mathcal{Y}') \ar[ld] \\ & (\mathcal{B} \subset \mathcal{B}') }
of thickenings of algebraic stacks. Assume
\mathcal{Y}' \to \mathcal{B}' is locally of finite type,
\mathcal{X}' \to \mathcal{B}' is flat and locally of finite presentation,
f is flat, and
\mathcal{X} = \mathcal{B} \times _{\mathcal{B}'} \mathcal{X}' and \mathcal{Y} = \mathcal{B} \times _{\mathcal{B}'} \mathcal{Y}'.
Then f' is flat and for all y' \in |\mathcal{Y}'| in the image of |f'| the morphism \mathcal{Y}' \to \mathcal{B}' is flat at y'.
Proof.
Choose an algebraic space U' and a surjective smooth morphism U' \to \mathcal{B}'. Choose an algebraic space V' and a surjective smooth morphism V' \to U' \times _{\mathcal{B}'} \mathcal{Y}'. Choose an algebraic space W' and a surjective smooth morphism W' \to V' \times _{\mathcal{Y}'} \mathcal{X}'. Let U, V, W be the base change of U', V', W' by \mathcal{B} \to \mathcal{B}'. Then flatness of f' is equivalent to flatness of W' \to V' and we are given that W \to V is flat. Hence we may apply the lemma in the case of algebraic spaces to the diagram
\xymatrix{ (W \subset W') \ar[rr] \ar[rd] & & (V \subset V') \ar[ld] \\ & (U \subset U') }
of thickenings of algebraic spaces. See More on Morphisms of Spaces, Lemma 76.18.4. The statement about flatness of \mathcal{Y}'/\mathcal{B}' at points in the image of |f'| follows in the same manner.
\square
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