Let $S$, $S'$ be (smooth projective) algebraic surfaces over an algebraically closed field $k$ and let $f : S \to S'$ be a morphism between them. Suppose that $f(S) = S'$. Is it possible that $f$ contracts infinitely many curves to a point?
(note that this is false, if $f$ is a birational morphism)
Motivation: I am trying to solve the following problem:
Let $S$ be a surface containing uncountably many smooth rational curves. Then $S$ is ruled.
The solution found on the internet (cf. http://homepages.math.uic.edu/~fsuzuk2/beauville_solutions.pdf, B (10), p. 7) considers two cases. In the first case we assume that $\dim \textrm{Alb}(S) \ge 1$. The map $\alpha : S \to \textrm{Alb}(S)$ contracts all the $\mathbb P^1$'s and thus the author concludes that $\dim \alpha(S) = 1$. Why?
No, this is not possible. The basic result you need is the following. (A reference is Shafarevich, Basic Algebraic Geometry 1, I.6.3 Theorem 7 (ii).)
Theorem : Let $f: X \rightarrow Y$ be a surjective morphism of quasiprojective varieties with $\operatorname{dim} X=n$ and $\operatorname{dim} Y=m$. Then there exists a nonempty open subset $U \subset Y$ such that $\operatorname{dim} f^{-1}(y) = n-m$ for all $y \in U$.
In your situation, the Theorem guarantees that there is a nonempty open subset $U \subset S'$ such that for any $u \in U$ the fibre $f^{-1}(u)$ has dimension 0. Let $Z = S' \setminus U$.
Now consider the closed subset $f^{-1}(Z) \subset S$. Note that all the curves contracted by $f$ are contained in this set.
Choose any irreducible component $W$ of this closed set. Since $Z$ is a proper subset of $S'$, the set $W$ is a proper subset of $S$. By irreducibility of $S$, the set $W$ therefore has dimension at most 1, so it is an irreducible curve. Since $f^{-1}(Z)$ has finitely many irreducible components, there are finitely many contracted curves.