Rational map to a non-uniruled manifold Let X, Y be smooth projective (over IC), let f:X..>Y be a rational map. Assume Y is not uniruled. Is it true that f will be regular over a non-empty open subset of Y? Funny..
 A: I think you have already answered this yourself, for the most part. We have a birational morphism $X' \to X$ with some exceptional locus $E$ inside $X'$, which is some finite union of ruled components $\mathbb{P}^1 \times Z_i$. We also have the (proper) map $f': X' \to Y$, and the restriction of the map to any $\mathbb{P}^1 \times Z_i \to Y$ does not factor through $\mathbb{P}^1 \times Z_i \to Z_i$, so the image inside $Y$ will be uniruled. As such, the image of $E$ inside $Y$ is some closed set $V$ which by hypothesis cannot be all of $Y$. Restricting to the complement $U \subset Y$, the map $f'^{-1}(U) \to U$ has domain away from the exceptional locus inside $X'$, which is what you wanted, yes?
Also, the analysis you gave in your answer almost works, I think, except there is a slight problem in that you can only take $p$ to be very general (in the complement of a countable union of closed proper subsets) rather than merely general. For example, a very general K3 surface has infinitely many rational curves despite not being uniruled, so the locus in $Y$ to which your analysis applies would be precisely the complement of these infinitely many rational curves.
A: By 'regular over U' I meant f is regular everywhere over U. Sometimes this is called 'almost holomorphic'. For example, a standard projection IP_2->IP_1 is not almost holomorphic.
OK, some motivation: After blowing up X we obtain X' and a holomorphic map X'->Y. Positive dimensional fibers of X'->X are rationally chain connected, i.e., rational curves connect any two points. On the other hand, Y is not covered by rational curves by assumption.
Let p be a general point in Y and denote by F_p the fiber of X'->Y over p. The above remarks should imply: if F_p meets a positive dimensional fiber of X'->X, then it already contains this fiber (p general!). This should mean X->Y is a fibration over some non-empty open subset of Y.
It is either false or well known, I don't know. 
