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Sep 9, 2011 at 20:37 vote accept aglearner
Jun 12, 2011 at 23:43 comment added aglearner mrw, thanks, I just wanted to confirm.
Jun 12, 2011 at 19:12 comment added mrw Was there a problem with my answer above? Assume that $\pi:X\to Y$ if proper, surjective and has connected fibers. Then any holomorphic function on each fiber is constant. Therefore $\pi_*\mathcal O_X=\mathcal O_Y$. Viceversa, assume that $\pi_*\mathcal O_X=\mathcal O_Y$ and let $f:X\to Z$ and $g:Z\to Y$ the Stein factorization. Then $f_*\mathcal O_Z=\mathcal O_Z$ but $g_*\mathcal O_Z=\mathcal O_Y$ only if $g$ is an isomorphism, otherwise $g_*\mathcal O_Z$ would have rank greater than $1$.
Jun 11, 2011 at 21:44 comment added aglearner Sylvain, thanks! This answers indeed my first question. What about the second one?
Jun 11, 2011 at 1:43 history answered Sylvain Bonnot CC BY-SA 3.0