Morphism from a surface group to a symmetric group, lifted to the braid group 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.
 A: 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 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 $n\geq 4$). 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. 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 $S_n$.   
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.
A: As BS pointed out, the question runs since the 70's. For $n=3, 4$, after Petersen the answer is positive, since $S_n$ is a solvable group (his example 5.8). For $n\ge 5$, the question is open (see Melikhov, problem 1.1); Melikhov even asks if every generic smooth map $\Sigma_1\to\Sigma_2$ between two orientable surfaces lifts to an embedding $\Sigma_1\to\Sigma_2\times R^2$.
Hansen, V.L. "Embedding finite covering spaces into trivial bundles" Math. Ann. 236, 3 (1978), 239-243, doi:10.1007/BF01351369 
Petersen, Peter "Fatness of covers". J. Reine Angew. Math. 403 (1990), 154–165, MR1030413.
Melikhov, S.A."Transverse fundamental group and projected embeddings" Proc. Steklov Inst. Math. (2015) MR3488789
