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Let $X$ be a smooth projective variety. Is there a criterion (apart from the definition) for the existence of a projective curve $C$ and a proper surjective morphism $\pi:X \to C$?

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Note that you can always blow up $X$ so that such a map $\pi$ exists, even with $C=\mathbb{P}^1$ and very mildly singular fibers (a "Lefschetz pencil"). This is enough for many applications, since for example cohomology of $X$ injects into the cohomology of the blow-up. –  Piotr Achinger Jan 17 '13 at 19:40
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Just for the record, the fact that you can choose $C=\mathbb P^1$ is not too deep since you can just map $C$ to $\mathbb P^1$ and replace $\pi$ with the composition. –  Sándor Kovács Jan 17 '13 at 22:05
    
mathoverflow.net/questions/35429/… –  M P Jan 18 '13 at 11:44

3 Answers 3

Another interesting theorem in this direction is Castelnuovo-de Franchis theorem. It says that if you have two linearly independent holomorphic 1-forms $\omega_1,\omega_2$ with $\omega_1\wedge\omega_2=0$ on $X$, then there exists a morphism $f:X\to C$ with $C$ a smooth curve of genus at least 2 and forms $\omega_i'$ on $C$ such that $\omega_i=f^*\omega_i'$.

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Yes. This is equivalent to the existence of a non-trivial divisor $D$ on $X$ such that there exists two distinct members of the linear system $|D|\ni D_1,D_2$ that are disjoint: $D_1\cap D_2=\emptyset$.

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By a theorem of Gromov and Schoen if the fundamental group of X is a proper amalgamated product or HNN extension then X maps surjectively to a curve .

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Furthermore, $X$ maps surjectively to a hyperbolic orbicurve. –  Misha Jan 17 '13 at 23:25

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