If I am reading this post correctly, any smooth projective $\mathbb{C}$morphism of schemes $X\rightarrow \mathbb{P}^1$ admits a section. I am afraid of the topological argument presented there. Is there an algebraic proof of this fact? What happens over fields other than $\mathbb{C}$?
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2$\begingroup$ If $k$ is an algebraically closed field of characteristic zero and $f:X\to \mathbb{P}^1$ is a smooth projective morphism then $f$ will have a section. You can prove this by "reducing" to the case that $k=\mathbb{C}$. This is an example of the socalled "Lefschetz principle". To get a feeling of how this goes, assume for simplicity that $k\subset \mathbb{C}$. Now, consider a section $\sigma:\mathbb{P}^1_{\mathbb{C}}\to X_{\mathbb{C}}$ of $f_{\mathbb{C}}$. $\endgroup$ – Ariyan Javanpeykar Jul 12 at 15:58

2$\begingroup$ Now, use "spreading out" (of your section) over a variety over $k$, and then specialize the spreadout section to get a section of $f$. $\endgroup$ – Ariyan Javanpeykar Jul 12 at 15:59

$\begingroup$ It is not so clear to me what one should expect when $k=\overline{\mathbb{F}_p}$. $\endgroup$ – Ariyan Javanpeykar Jul 12 at 16:10

$\begingroup$ Since any smooth morphism admits a section locally in the étale topology, I am wondering if one can not prove this using: i) the simply connectedness of $\mathbb{A}^1$ ii) the fact that any rational map between projective curves can be extended to an actual morphism? $\endgroup$ – Libli Jul 15 at 8:29