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

1. $f$ is separated,

2. there is an open covering $X = \bigcup U_ i$ such that $f|_{U_ i} : U_ i \to Y$ is an open immersion, and

3. 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. $\square$

Comment #4207 by HayamaKazuma(羽山籍真) on

（2）Is it be f|_{U_i}: U_i \rightarrow Y ?

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