Lemma 66.3.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $V \to U$ be a morphism of schemes over $S$, and let $U \to X$ be a morphism from $U$ to $X$. If the fibres of $V \to U$ and $U \to X$ are universally bounded, then so are the fibres of $V \to X$.

Proof. Let $n$ be an integer which works for $V \to U$, and let $m$ be an integer which works for $U \to X$ in Definition 66.3.1. Let $\mathop{\mathrm{Spec}}(k) \to X$ be a morphism, where $k$ is a field. Consider the morphisms

$\mathop{\mathrm{Spec}}(k) \times _ X V \longrightarrow \mathop{\mathrm{Spec}}(k) \times _ X U \longrightarrow \mathop{\mathrm{Spec}}(k).$

By assumption the scheme $\mathop{\mathrm{Spec}}(k) \times _ X U$ is finite of degree at most $m$ over $k$, and $n$ is an integer which bounds the degree of the fibres of the first morphism. Hence by Morphisms, Lemma 29.55.5 we conclude that $\mathop{\mathrm{Spec}}(k) \times _ X V$ is finite over $k$ of degree at most $nm$. $\square$

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