What about the reducible fibers of a surjective morphism?

Question

If i have a surjective morphism $\phi :S \rightarrow C$, where $S$ is a smooth complex algebraic projective surface and $C$ a smooth projective curve, what can i say about the number of the fibers of $\phi$? is it true that the set of reducible fibers of $\phi$ is finite?

Answer 1

No, this is not true as written. For example, suppose you have a surjective morphism $S \rightarrow \Gamma$ with $\Gamma$ a curve, and then $\Gamma \rightarrow C$ a nontrivial finite morphism of curves. Then almost all fibres of the composite morphism $\phi: S \rightarrow C$ will be disconnected, in particular reducible.

On the other hand, if $\phi$ has connected fibres, then this is true. See Shafarevich II.6.1, Theorem 1 ("First Bertini Theorem"). That says that if $S$ and $C$ are irreducible and $\phi$ has connected fibres, then there is a dense open set $U \subset C$ such that $\phi^{-1}(p)$ is irreducible for all $p \in U$. Then $C \setminus U$ is a finite set of points, which is exactly what you want.