Lemma 68.3.6. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$, and let $U$ be a scheme over $S$. Let $\varphi : U \to X$ be a morphism over $S$. If the fibres of $\varphi $ are universally bounded, then there exists an integer $n$ such that each fibre of $|U| \to |X|$ has at most $n$ elements.
Proof. The integer $n$ of Definition 68.3.1 works. Namely, pick $x \in |X|$. Represent $x$ by a morphism $x : \mathop{\mathrm{Spec}}(k) \to X$. Then we get a commutative diagram
\[ \xymatrix{ \mathop{\mathrm{Spec}}(k) \times _ X U \ar[r] \ar[d] & U \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r]^ x & X } \]
which shows (via Properties of Spaces, Lemma 66.4.3) that the inverse image of $x$ in $|U|$ is the image of the top horizontal arrow. Since $\mathop{\mathrm{Spec}}(k) \times _ X U$ is finite of degree $\leq n$ over $k$ it has at most $n$ points. $\square$
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