A morphism of curves is said to be Galois if the corresponding extension of function fields is Galois.


Let $f:X\longrightarrow Y$ be a finite morphism of smooth projective connected curves over $\mathbf{C}$ (or Riemann surfaces if you prefer) with branch locus $B\subset Y$ and ramification locus $R\subset X$. There exists a finite morphism $g:W\longrightarrow X$ of smooth projective curves such that the composition $f\circ g:W\longrightarrow Y$ is Galois. Take the Galois closure for example. 


Can we say something about the branch locus of $g$? For example, when does it lie in $R$? 


In my problem I would like to be able to choose $g$ such that the branch locus of $g$ lies in $R$. I have a feeling this is not always possible unfortunately.


The following might help actually. In my problem, $Y$ is the projective line and $f$ is a Belyi morphism. Thus, what I would like to see is that I can choose $g:W\longrightarrow X$ such that the composition $f\circ g$ is Belyi and Galois. Again, this seems unlikely to be possible.



Branch locus of the Galois closure of a Belyi morphism
This is to address the last part of your question -- about Belyi morphism. The answer is YES: you can always get $g: W\to X$ such that $f\circ g: W\to \mathbb CP^1$ is Belyi.


Indeed, $X\to \mathbb CP^1$ is Belyi if it ramifies only over $0,1,\infty$. Let us now conisder $\hat X=f^{-1}(\mathbb CP^1\setminus (0,1,\infty))$. Clearly, $f\_\*\pi\_1(\hat X)$ is a finite index subgroup of $\pi\_1(\mathbb CP^1\setminus (0,1,\infty))=F\_2$. The map $f:X\to \mathbb CP^1$ is Galois iff $f\_\*\pi\_1(\hat X)$ is normal in $F\_2$. 


Now, take a normal subgroup $N$ in $F\_2$ of finite index, contained in $f\_\*(\pi\_1(\hat X))$
and take the corresponding cover $\mathbb CP^1\setminus (0,1,\infty)$, call it $\hat W$. The completion $W$ of $\hat W$ is what you are looking for. 


**Added.** In order to find a normal subgroup of finite index we use the following standard lemma (I forgot the name):


Lemma. Let $G$ be any group and $G\_1$ be any finite index subgroup. Then there exists $N\subset G\_1$ of finite index, and such that $N$ is normal in $G$. 


Proof. $G$ is acting on the left on the finite set of cosets $G/G\_1$. Let $N$ be the subgroup of $G$ that acts trivially on $G/G\_1$. Clearly $N$ is normal in $G$ and the index of $N$ is at most $\#(G/G\_1)!$


**EDIT** (by SK): fixed LaTeX 


