Lemma 29.8.5. Let $f : X \to S$ be a morphism of schemes. Suppose that $X$ has finitely many irreducible components. Then $f$ is dominant (if and) only if for every irreducible component $Z \subset S$ the generic point of $Z$ is in the image of $f$. If so, then $S$ has finitely many irreducible components as well.

Proof. Assume $f$ is dominant. Say $X = Z_1 \cup Z_2 \cup \ldots \cup Z_ n$ is the decomposition of $X$ into irreducible components. Let $\xi _ i \in Z_ i$ be its generic point, so $Z_ i = \overline{\{ \xi _ i\} }$. Note that $f(Z_ i)$ is an irreducible subset of $S$. Hence

$S = \overline{f(X)} = \bigcup \overline{f(Z_ i)} = \bigcup \overline{\{ f(\xi _ i)\} }$

is a finite union of irreducible subsets whose generic points are in the image of $f$. The lemma follows. $\square$

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