Morphisms contracting a family of curves

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 $f$?

• Yes, that is true. Consider the inverse image under $f$ of a general hyperplane section. The intersection with $S$ is a curve ... – Jason Starr Jun 24 '15 at 22:04

• 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.
• Hi Roy, this sounds good. It's a nice proof assuming that $Y$ is projective. The statement is still true assuming that $g$ is proper and $f$ is arbitrary (plus the connectivity assumptions). (Which is what I had in mind all along so I thought that there cannot be such a simple proof. But there is one with $Y$ projective). Cheers! :) – Sándor Kovács Jun 26 '15 at 3:30