Let $\Sigma\_g$ be the fundamental group of the closed orientable surface of genus $g\ge 2$; let $B\_n$ be the braid group on $n\ge 3$ braids; let $S\_n$ be the symmetric group on $n$ letters; let $p:B\_n\to S\_n$ be the canonical epimorphism.


Does every homomorphism $f:\Sigma\_g\to S\_n$ lift to $B\_n$? That is, is there a homomorphism $\bar f:\Sigma\_g\to B\_n$ such that $f=p\circ\bar f$?


This is known for $n=3$, the proof is by ad hoc elementary computations
(Hector, Meigniez, Matsumoto, "Ends of leaves of Lie foliations", J. Math. Soc. Japan 57 (2005), no. 3, 753--779.) Is it true for every $n$? The question is crucial for the construction of some Lie foliations.



Morphism from a surface group to a symmetric group, lifted to the braid group
In certain very special cases, I think this can be answered. 
In particular, if $f:\Sigma\_g \twoheadrightarrow S\_n$ is onto, $n\geq 4$ and $g \gg 0$, then Theorem 6.20 of [Dunfield-Thurston](http://rdcu.be/tf30) implies that the map $f$ is determined up to the action of the mapping class group by the image $f\_\ast:H\_2(\Sigma\_g) \to H\_2(S\_n) \cong \mathbb{Z}/2$ [(when $n\geq 4$)](https://en.wikipedia.org/wiki/Symmetric_group#Homology). In the case that the image is zero, then $f$ can be chosen to factor through a handlebody of genus $g$, and since the map factors through a free group, this will lift to $B\_n$ since there is no obstruction. 


If $f\_\ast$ is non-trivial, then there will be a lift iff $\mathbb{Z}\oplus \mathbb{Z}/2 = H\_2(B\_n)\twoheadrightarrow H\_2(S\_n)=\mathbb{Z}/2$ is surjective. I'm pretty sure that this
is true, and it should be represented by a torus whose fundamental
group is generated by $\sigma\_1, \sigma\_3$ in the [standard braid group generators.](https://en.wikipedia.org/wiki/Braid_group#Generators_and_relations) One need only check that this torus maps homologically non-trivially into $H\_2(S\_n)$, which I think follows from the [presentation of the double cover of $S\_n$](https://en.wikipedia.org/wiki/Covering_groups_of_the_alternating_and_symmetric_groups). 


If $f$ is not onto, then one could still attempt to apply Theorem 6.20 to its image. Let $f(\Sigma\_g)= H < S\_n$. Then Theorem 6.20 implies that for $g$ large enough, $f$ is classified up to the mapping class group by the image of $f\_\*: H\_2(\Sigma\_g)\to H\_2(H)$, up to the action of $Out(H)$. Let $\tilde{H} = p^{-1}(H)$ be the preimage of $H$ in $B\_n$. If $p\_{|\ast}: H\_2(\tilde{H}) \to H\_2(H)$ is not onto (again, up to the action of $Out(H)$), then one could find a counterexample. I think there's a good chance of such a subgroup existing.


