Does every smooth, projective morphism to $\mathbb{C}P^1$ admit a section? Possibly this has already been asked, but it came up again in this question of Daniel Litt.  Does every smooth, projective morphism $f:Y\to \mathbb{C}P^1$ admit a section, i.e., a morphism $s:\mathbb{C}P^1\to Y$ such that $f\circ s$ equals $\text{Id}_{\mathbb{C}P^1}$?  
Edit. As Ariyan points out, this article proves that there are at least $3$ singular fibers of $f$ if either $Y$ has nonnegative Kodaira dimension, Theorem 0.1 of Viehweg-Zuo, or if the fibration is non-isotrivial with general fiber either general type or with $\omega_f$ semi-positive, Theorem 0.2 by Möller-Viehweg-Zuo.  This suggests an approach to proving the conjecture, at least assuming the uniruledness conjecture (negative Kodaira dimension implies uniruled): take the MRC quotient and then apply the Minimal Model Program to try to reduce to these theorems.  Unfortunately, both formation of the MRC quotient and the Minimal Model Program are likely to introduce singularities . . . 
Second Edit. As Ben Wieland points out, this is false in the category of compact, complex manifolds.  The examples are interesting (to me) because they also come up in showing "rationally connected" fibrations over a Riemann surface in the analytic category need not admit sections.  Begin with the $\mathbb{C}^\times$ -torsor $T$ over $\mathbb{P}^1$ associated to any nontrivial invertible sheaf, e.g., $\mathcal{O}_{\mathbb{P}^1}(1)$.  Now let $q:T\to Y$ be the fiberwise quotient by multiplication by some element $\lambda\in \mathbb{C}^\times$ of modulus $\neq 1$.  Projection of $T$ to $\mathbb{P}^1$ factors through this quotient, $f:Y\to \mathbb{P}^1$.  Although every fiber of $f$ is the same Hopf surface elliptic curve, there is no section: if there were, its inverse image in $T$ would be a disjoint union of sections of $T$ (since $\mathbb{P}^1$ is simply connected), and $T$ has no sections.     
A Positive Answer by Paul Seidel. 22 September 2017.  I received a communication from Paul Seidel that he knows how to prove this using methods of symplectic topology.
 A: Yes, using some symplectic geometry. Let's say we had $X \subset {\mathbb C}P^n \times {\mathbb C}P^1$, with projection to $\mathbb{C} P^1$ a smooth morphism (meaning, for topologists, a proper holomorphic submersion; all fibres are smooth). With the restriction of the standard Kaehler form, this becomes a symplectic fibre bundle over ${\mathbb C}P^1 = S^2$. It is known that the Gromov-Witten invariant counting sections of any such fibration (with suitable incidence conditions) is always nonzero. This is proved by reversing the orientation of $S^2$ to get another such fibration, and then considering the fibre union of those two as a degeneration of the trivial fibration. References:

*

*The UCDavis math arXiv front paper with ID 9511.5111 (ADDED: this might be Seidel's $\pi_1$ of symplectic automorphism groups and invertibles in quantum homology rings, https://arxiv.org/abs/dg-ga/9511011)


*François Lalonde, Dusa McDuff, Leonid Polterovich, Topological rigidity of Hamiltonian loops and quantum homology, https://arxiv.org/abs/dg-ga/9710017


*Dusa McDuff, Quantum Homology of fibrations over $S^2$, https://arxiv.org/abs/math/9905092
PS: Slight edits, I hope this clarifies (and that I have not misunderstood the question, of course). I added another reference (the third one): see Theorem 1.5 there, and note that the map takes values in the invertible elements of quantum cohomology (hence is nonzero).
