Lemma 65.47.8. Let $S$ be a scheme. Let $X$ and $Y$ be algebraic space over $S$ with $|X|$ and $|Y|$ irreducible. Then $X$ and $Y$ are birational if and only if there are nonempty open subspaces $U \subset X$ and $V \subset Y$ which are isomorphic as algebraic spaces over $S$.

**Proof.**
Assume $X$ and $Y$ are birational. Let $f : U \to Y$ and $g : V \to X$ define inverse dominant rational maps from $X$ to $Y$ and from $Y$ to $X$. After shrinking $U$ we may assume $f : U \to Y$ factors through $V$. As $g \circ f$ is the identity as a dominant rational map, we see that the composition $U \to V \to X$ is the identity on a dense open of $U$. Thus after replacing $U$ by a smaller open we may assume that $U \to V \to X$ is the inclusion of $U$ into $X$. By symmetry we find there exists an open subspace $V' \subset V$ such that $g|_{V'} : V' \to X$ factors through $U \subset X$ and such that $V' \to U \to Y$ is the identity. The inverse image of $|V'|$ by $|U| \to |V|$ is an open of $|U|$ and hence equal to $|U'|$ for some open subspace $U' \subset U$, see Properties of Spaces, Lemma 64.4.8. Then $U' \subset U \to V$ factors as $U' \to V'$. Similarly $V' \to U$ factors as $V' \to U'$. The reader finds that $U' \to V'$ and $V' \to U'$ are mutually inverse morphisms of algebraic spaces over $S$ and the proof is complete.
$\square$

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