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Let $\phi:X \to Y$ be a projective morphism of smooth varieties. Under what condition on $\phi$ does there exist a section to the morphism $\phi$? The example that I have in mind is when $Y$ is an irreducible component of a Hilbert scheme of curves, $X$ is a component of flag Hilbert scheme and $\phi$ is a projection onto one of its components.

This question has been asked before, but the answer is not very clear to me.

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As said before, 'existence of a section' is tantamount to 'existence of a rational point', which is one of the most difficult problems around. –  Damian Rössler Feb 14 at 7:51
    
One can only hope to say something intelligent in special cases. If you add more details about your specific case of interest, perhaps we can help you. –  Daniel Loughran Feb 14 at 11:13
    
@Loughran, Rossler: Would it help if I said everything were over the complex numbers? I would imagine that there would be rational points over $\mathbb{C}$! –  user46578 Feb 14 at 11:21
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@user46578: the issue to to find a rational point over the function field of $Y$. –  Artie Prendergast-Smith Feb 14 at 11:27
    
@Prendergast-smith: I see. Thanks. –  user46578 Feb 14 at 11:34

1 Answer 1

Some comments

(1) On this site there was a question "Why a smooth surjective morphism of schemes admits a section etale-locally?"; so smoothness gives some kind of nice property here.

(2) Even for smooth morphisms, over complex numbers one may not have a Zariski local section (even in the case when the fiber is a projective space).

(3) For an arbitrary morphism there is no hope for a global section - consider the case of a blowup of a point on a projective space.

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