Non-flat seesaw

Let $$f\colon X \to Y$$ be a dominant morphism between normal projective algebraic varieties; assume $$f_*\mathcal {O}_X=\mathcal{O}_Y$$; let’s also assume that it is defined over the complex numbers, but I do not think this is relevant for the question. I do NOT assume that $$f$$ is flat.

Let $$L$$ be a line bundle on $$X$$ which is trivial along all fibres of $$f$$. Is $$L$$ the pull-back of a line bundle from $$Y$$?

(I would expect that $$f_*L$$ is a line bundle, and the natural map $$f^*f_*L\to L$$ is an isomorphism. When $$f$$ is flat, this is the seesaw principle.)

I am very interested in the case where $$f$$ is a birational map.

Thanks!

• You should assume that $Y$ is normal and that $f$ is dominant with connected fibers, otherwise there are easy counterexamples. Nov 23 '18 at 20:56
• It seems likely that there should be counterexamples even with $X$ a surface and $f$ the contraction of a curve $C$: it suffices to find a line bundle on $X$ which is trivial on $C$ but not on some infinitesimal neighbourhood. (I have not worked out an explicit example...)
– naf
Nov 24 '18 at 9:20
• @ulrich Sounds convincing. Should the answer be positive if we assume $R^1 f_* \mathcal{O}_X = 0$? In this case if $L$ is trivial on the fiber, it should be trivial on all of its infinitesimal neighborhoods. Nov 24 '18 at 10:09
• @PiotrAchinger: Indeed: working over $\mathbb{C}$ in the analytic topology, this follows by using the exponential sequence (and the Leray spectral sequence) to compute the Picard group.
– naf
Nov 24 '18 at 12:24