Lemma 67.3.5. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Consider a commutative diagram

$\xymatrix{ U \ar[rd]_ g \ar[rr]_ f & & V \ar[ld]^ h \\ & X & }$

where $U$ and $V$ are schemes. If $g$ has universally bounded fibres, and $f$ is surjective and flat, then also $h$ has universally bounded fibres.

Proof. Assume $g$ has universally bounded fibres, and $f$ is surjective and flat. Say $n \geq 0$ is an integer which bounds the degrees of the schemes $\mathop{\mathrm{Spec}}(k) \times _ X U$ as in Definition 67.3.1. We claim $n$ also works for $h$. Let $\mathop{\mathrm{Spec}}(k) \to X$ be a morphism from the spectrum of a field to $X$. Consider the morphism of schemes

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

It is flat and surjective. By assumption the scheme on the left is finite of degree $\leq n$ over $\mathop{\mathrm{Spec}}(k)$. It follows from Morphisms, Lemma 29.56.10 that the degree of the scheme on the right is also bounded by $n$ as desired. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).