This is probably trivial but has been bothering me all day.


Suppose $f:\Sigma\_g\to \mathbb{S}^2$ is a $g+1$ fold branched conformal map with $\Sigma\_g$ a connected genus $g$ surface and $f$ having $4$ branch points (all of order $g+1$).


Is it true that $f:\Sigma\_g^\* \to (\mathbb{S}^2)^\*$ (the cover obtained by removing the branch points) is a regular cover (i.e. so the deck group acts transitively on the fibers of $f$)?


I thought this was obvious but then realized I didn't know how to prove it (admittedly this is a bit outside my area of expertise). Part of the problem is that I don't have a good example of a branched map from a connected surface which is not a regular cover away from the branch points.


**Edit**


Based on some comments of Sebastian. It would be enough for my purposes to prove that the conformal structure of $\Sigma\_g$ was uniquely determined by the cross-ratio of the branch points. However, its not clear to me whether this can be shown without knowing that the cover is regular...



Branched Regular Cover over 4-times punctured sphere
The answer is *no*:


The monodromy group of the cover, as a permutation group on the fiber $f^{-1}(b)$ of a non-branch point $b$, is generated by four elements $s\_1,s\_2,s\_3,s\_4$ with $s\_1s\_2s\_3s\_4=1$. By Riemann's existence theorem, the question is equivalent to the following, with $n=g+1$:



> 
> Let $s\_1,s\_2,s\_3,s\_4$ be $n$-cycles in the symmetric group on $n$ letters with $s\_1s\_2s\_3s\_4=1$. Does this imply that the group generated by $s\_1, s\_2,s\_3, s\_4$ has order $n$?
> 
> 
> 


This does not hold for $n=4$: Take for example $s\_1=(1, 2, 3, 4)$, $s\_2=(1, 3, 4, 2)$, $s\_3=s\_2^{-1}$, $s\_4=s\_1^{-1}$. Then $s\_1s\_2s\_3s\_4=1$, but $s\_1s\_2=(2, 4, 3)$.


This works for any $n\ge4$: Simply take $s\_1$ an $n$-cycle and $s\_2$ another $n$-cycle which is not a power of $s\_1$, and $s\_3$, $s\_4$ the inverses of $s\_2$ and $s\_1$ as above.


(Note: I use the right action of permutations, so in $s\_1s\_2$, one *first* applies $s\_1$, and then $s\_2$ afterwards.)


