Lemma 33.42.1. Let $f : X \to Y$ be a morphism of schemes. Let $X^0$ denote the set of generic points of irreducible components of $X$. If
$f$ is separated,
there is an open covering $X = \bigcup U_ i$ such that $f|_{U_ i} : U_ i \to Y$ is an open immersion, and
if $\xi , \xi ' \in X^0$, $\xi \not= \xi '$, then $f(\xi ) \not= f(\xi ')$,
then $f$ is an open immersion.
Proof. Suppose that $y = f(x) = f(x')$. Pick a specialization $y_0 \leadsto y$ where $y_0$ is a generic point of an irreducible component of $Y$. Since $f$ is locally on the source an isomorphism we can pick specializations $x_0 \leadsto x$ and $x'_0 \leadsto x'$ mapping to $y_0 \leadsto y$. Note that $x_0, x'_0 \in X^0$. Hence $x_0 = x'_0$ by assumption (3). As $f$ is separated we conclude that $x = x'$. Thus $f$ is an open immersion. $\square$
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