Let $f:X\rightarrow Y$ be a morphism of normal projective varieties. Let $S\subseteq X$ be a surface admitting a morphism $g:S\rightarrow C$ to a curve $C$ such that any fiber of $g$ is a curve.


Assume that the general fiber of $g$ is contracted to a point by $f$. Is it true that any fiber of $g$ is contracted by $f$? 



Morphisms contracting a family of curves
Actually, this is true in more generality:


* You don't need $f, X$, or $Y$ to be normal or projective (this is of course, trivial, they are all red herrings). You only need $g$ to be proper (I suppose you meant that $S$ is a projective surface, so that's covered).
* If all the fibers of $g$ are connected and **any** fiber of $g$ is contracted, then all fibers are contracted. This is known as the "There are no bowties in algebraic geometry" theorem (or more conventionally called "Rigidity Lemma").
* From your formulation it seems that you are assuming that $C$ is irreducible. Because of the previous point you only need connected, but that is obviously necessary.


The proof is not too hard, and you should try. I believe this was first proved by Mumford when $S=F\times C$ and $g$ is the projection. A good place to look for the proof of this more general statement is Lemma 1.6 in [Kollár-Mori98](https://books.google.com/books?id=YrysxvPbBLwC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q=rigidity%20lemma&f=false).


