Lemma 68.3.4. Let $S$ be a scheme. Let $g : Y \to X$ be a representable morphism of algebraic spaces over $S$. Let $f : U \to X$ be a morphism from a scheme towards $X$. Let $f' : U \times _ X Y \to Y$ be the base change of $f$. If

\[ \mathop{\mathrm{Im}}(|f| : |U| \to |X|) \subset \mathop{\mathrm{Im}}(|g| : |Y| \to |X|) \]

and $f'$ has universally bounded fibres, then $f$ has universally bounded fibres.

**Proof.**
Let $n \geq 0$ be an integer bounding the degrees of the fibre products $\mathop{\mathrm{Spec}}(k) \times _ Y (U \times _ X Y)$ as in Definition 68.3.1 for the morphism $f'$. We claim that $n$ works for $f$ also. Namely, suppose that $x : \mathop{\mathrm{Spec}}(k) \to X$ is a morphism from the spectrum of a field. Then either $\mathop{\mathrm{Spec}}(k) \times _ X U$ is empty (and there is nothing to prove), or $x$ is in the image of $|f|$. By Properties of Spaces, Lemma 66.4.3 and the assumption of the lemma we see that this means there exists a field extension $k'/k$ and a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(k') \ar[r] \ar[d] & Y \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r] & X } \]

Hence we see that

\[ \mathop{\mathrm{Spec}}(k') \times _ Y (U \times _ X Y) = \mathop{\mathrm{Spec}}(k') \times _{\mathop{\mathrm{Spec}}(k)} (\mathop{\mathrm{Spec}}(k) \times _ X U) \]

Since the scheme $\mathop{\mathrm{Spec}}(k') \times _ Y (U \times _ X Y)$ is assumed finite of degree $\leq n$ over $k'$ it follows that also $\mathop{\mathrm{Spec}}(k) \times _ X U$ is finite of degree $\leq n$ over $k$ as desired. (Some details omitted.)
$\square$

## Comments (0)