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
Comments (2)
Comment #4207 by HayamaKazuma(羽山籍真) on
Comment #4389 by Johan on