**Proof.**
The morphism $f$ is proper by Morphisms, Lemma 29.41.7. Thus $f(X)$ is closed and images of closed points are closed. Let $y \in Y$ be the image of a closed point in $X$. Then $f^{-1}(\{ y\} )$ is a closed subset of $X$ not containing any of the generic points of irreducible components of dimension $1$ by condition (3). It follows that $f^{-1}(\{ y\} )$ is finite. Hence $f$ is finite over an open neighbourhood of $y$ by More on Morphisms, Lemma 37.44.2 (if $Y$ is Noetherian, then you can use the easier Cohomology of Schemes, Lemma 30.21.2). Since we've seen above that there are enough of these points $y$, the proof is complete.
$\square$

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