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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 $f$?

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    $\begingroup$ Yes, that is true. Consider the inverse image under $f$ of a general hyperplane section. The intersection with $S$ is a curve ... $\endgroup$
    – Jason Starr
    Commented Jun 24, 2015 at 22:04

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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.

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  • $\begingroup$ Indeed, it seems Jason's hint suffices. $\endgroup$
    – roy smith
    Commented Jun 25, 2015 at 23:29
  • $\begingroup$ Roy: I am not sure how that works if you only know that a special fiber is contracted. In that case the general hyperplane section will miss the point where that special fiber maps and thus does not give any information. No? $\endgroup$
    – Sándor Kovács
    Commented Jun 25, 2015 at 23:42
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    $\begingroup$ Thanks Sandor: let's try this: If a general hyperplane section H misses the point to which a special fiber contracts, then the pullback curve Z of H (in S) misses that fiber, hence its image by g does not surject onto C. Thus its image meets C only in a finite number of points (if C is irreducible). Hence Z meets only a finite number of fibers of g and thus consists of a union of a finite number of them. are we getting there? i.e. then most fibers of g map to points via f, so we are in the situation of the generic fiber of g being collapsed. Does this scan? $\endgroup$
    – roy smith
    Commented Jun 26, 2015 at 3:09
  • $\begingroup$ 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! :) $\endgroup$
    – Sándor Kovács
    Commented Jun 26, 2015 at 3:30

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