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In response to the affirmative answer to this question using symplectic methods, I am wondering if the equivalent statement holds in positive characteristic?

Explicitly, over an algebraically closed field $k$ of positive characteristic $p > 0$, does every smooth projective morphism $f : X \to \mathbb{P}^1$ admit a section?

If we do not require $k$ to be separably closed, the question becomes uninteresting because of examples such as $\mathbb{P}^1_{k'} \to \mathbb{P}^1_k$ for any finite separable $k'/k$.

My guess is there should exist an example with no section. My idea was to find a surface admitting a nontrivial elliptic fibration over $\mathbb{P}^1$ with smooth fibers. Do there exist such counterexamples?

Ben C
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    I believe the symplectic proof of existence of holomorphic sections of a projective smooth morphism to $\mathbb{CP}^1$ also works if there is a single critical value. Glue the $2$-sphere to an "opposite" $2$-sphere along the boundaries of little disks centered at the critical values, and form the fibration over this new $2$-sphere by glueing the original fibration to its "opposite" fibration over the boundary of the disks. So one could ask about the positive characteristic case with a single critical value, but this fails since the fundamental group of $\mathbb{A}^1$ is nontrivial. – Jason Starr Nov 09 '21 at 11:13

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EDIT: It was wrong.

I think such examples exist, take as $f:Y\rightarrow \mathbb P^1$ any family of supersingular abelian varieties over $\overline{\mathbb F_p}$ which is non-constant but constant up-to isogeny (e.g. take the Moret-Bailly pencil of supersingular abelian surfaces). You cannot have a non-constant morphism $\mathbb P^1\rightarrow Y$, since the group $Y(\overline{\mathbb F_p}(t))$ is torsion (since the family is isotrivial up to isogeny) and any non-constant morphism $\mathbb P^1\rightarrow Y$ would correspond to a non-torsion element in $Y(\overline{\mathbb F_p}(t))$.

Emiliano Ambrosi
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