Let $f:X\rightarrow Y$ be a flat morphism of normal projective varieties with fibers of positive dimension (in particular all the fibers are connected and of the same dimension).

Let $g:X\rightarrow Z$ be another morphism with connected fibers of positive dimension. Assume that $g$ contracts the general fiber of $f$ to a point of $Z$. Is it true that then $g$ must contract any fiber of $f$ to a point of $Z$?

In other words, is the map

$$Y\rightarrow \mathbb{Z},\: y\mapsto dim(g(f^{-1}(y)))$$



Another proof: The assumptions imply that $X\times_YX$ is irreducible and that the two composite maps $X\times_YX\rightrightarrows X\xrightarrow{g}Z$ coincide over the generic point of $Y$. By density they must be equal (I have to assume $Z$ separated here), so $g$ is constant along the fibers of $f$ (and, by descent, $g$ factors through $Y$).


For each point in $Y$, take a curve $C \subseteq Y$ that passes through that point and the generic point. Let $X'= X \times_{Y} C$. Then the generic fiber of $f': X' \to C$ is still irreducible, and because $f$ is flat no irreducible component is contained in a special fiber, so $X'$ is irreducible.

Consider the image of $X'$ in $C \times Z$. It is closed and irreducible, and is zero-dimensional over the generic point of $C$. Hence it is an irreducible curve with a nonconstant map to $C$. Thus every fiber of the map to $C$ is zero-dimensional.

Hence every fiber of the map from the image of $X$ in $Y \times Z$ to $Y$ is zero-dimensional, as desired.

  • $\begingroup$ What do you mean exactly by no irreducible component is contained in a special fiber? What's the map between the image of $X^{'}$ in $C\times Z$ and $C$? $\endgroup$ – user58018 Apr 12 '15 at 21:01
  • $\begingroup$ 1. I mean that every irreducible component of X' maps surjectively to C - a general property of flat maps. Because the generic fiber is irreducible, there is only one irreducible component. 2. The first projection to $C$. $\endgroup$ – Will Sawin Apr 12 '15 at 22:54

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