Lemma 68.3.3. Let $S$ be a scheme. Let $Y \to X$ be a representable morphism of algebraic spaces over $S$. Let $U \to X$ be a morphism from a scheme to $X$. If the fibres of $U \to X$ are universally bounded, then the fibres of $U \times _ X Y \to Y$ are universally bounded.

**Proof.**
This is clear from the definition, and properties of fibre products. (Note that $U \times _ X Y$ is a scheme as we assumed $Y \to X$ representable, so the definition applies.)
$\square$

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