Lemma 105.5.1. 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 75.18.4. 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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