For equivalence of unbranched coverings of topological spaces, there is a criteria:


Two coverings (unbranched) $p\_1\colon Y\_1\rightarrow X$ and $p\_2\colon Y\_2\rightarrow X$ are equivalent iff for some $q\in X$ and $\bar{q\_1}\in p\_1^{-1}(q)$ and $\bar{q\_2}\in p\_2^{-1}(q)$, the induced subgroups $p\_\*\pi\_1(Y\_1,\bar{q\_1})$ and $p\_\*\pi\_1(Y\_2,\bar{q\_2})$ are conjugate in $\pi\_1(X,q)$. 


Is there any criteria (particularly group theoretic) for equivalence of branched coverings of Riemann surfaces? 



Equivalence of Branched Coverings
The answer is **yes**: the equivalence class of the covering is detected by the monodromy representation of the fundamental group of the base minus the branch locus, up to conjugacy.


More precisely, let $f \colon X \to Y$ be a (possibly branched) covering of degree $d$ of Riemann surfaces. Choosing a point $y\_0 \in Y$ not lying in the branch locus $B$, there is a monodromy representation $$\rho \colon \pi\_1(Y-B, y\_0) \to S\_d \quad (\*)$$
whose image is *transitive* (since $X$ is assumed to be connected). Moreover, if we choose a different base point it is easy to check that the map $\rho$ varies only up to conjugacy in $S\_d$.


Then there is the following well-known



> 
> **Theorem.**
>  There exists a one-to-one correspondence between branched coverings $f \colon X \to Y$ of degree $d$ whose branch points lie in $B$ and group homomorphisms of type $(\*)$ with transitive image (up to conjugacy in $S\_d$).
> 
> 
> 


For further details, see Miranda's book *Algebraic curves and Riemann surfaces*, especially Chapter 4.


