# do geometric fibers determine scheme-theoretic image?

Let $X,Y,Z$ be reduced algebraic varieties, and let $Y$ and $Z$ be normal. Let $f:X \to Y$ and $g:X \to Z$ two surjective projective morphisms of algebraic varieties such that the geometric fibers of $f$ and $g$ coincide. Is there an isomorphism $h:Y\to Z$ such that $g=h \circ f$?

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I don't understand "$Y$ and $Z$ coincide set-theoretically". – Laurent Moret-Bailly Aug 29 '11 at 11:01
I mean that there's a one-to-one map between the set of closed points of $Y$ and the set of closed points of $Z$. – IMeasy Aug 29 '11 at 11:05
I don't understand, $f$ and $g$ have different targets, so how can they be the same morphism unless $Y=Z$? Do you mean that $Y$ and $Z$ have the same underlying topological space? In that case the answer is still no: Let $Y$ be a non-reduced scheme, $X=Z=Y_{red}$, $f:Y_{red}\to Y$ the inclustion and $g$ the identity. – J.C. Ottem Aug 29 '11 at 11:35
I still don't understand. Any positive-dimensional complex algebraic variety $Y$ has the same cardinality as, say, $[0,1]$. Fixing a bijection, we obtain by transport of structure a complex variety $Y'$ isomorphic to $Y$, with underlying set $[0,1]$. Now, starting with $f$ and $g$ but without your assumption on $Y$ and $Z$, we can apply the above construction and get, by composition, $f':X\to Y'$ and $g':X\to Z'$ where $Y'$ and $Z'$ have the same underlying set. In other words, this assumption on $Y$ and $Z$ is not a restriction. – Laurent Moret-Bailly Aug 29 '11 at 12:40
I think you want the maps to be surjective. Also, I suggest you rephrase the question to ask whether there exists an isomorphism $h: Y \to Z$ such that $g = h \circ f$. – S. Carnahan Aug 29 '11 at 15:23

In positive characteristic, you get a counterexample by taking $X=Y=Z=$ the affine line (say), $f$ the identity and $g$ the Frobenius map.
Assume now that the ground field is algebraically closed of characteristic zero. Consider the map $(f,g):X\to Y\times Z$. Its image $\Gamma$ is a closed subvariety of $Y\times Z$. The assumption on the fibers exactly means that both projections from $\Gamma$ to $Y$ and $Z$ are bijective (on closed points). Since they are proper they must be (in char. zero) finite birational, hence (by normality) isomorphisms. So, $\Gamma$ is the graph of the isomorphism we are looking for.