Consider a compact Riemann surface $X$. If $p_1:Y\longrightarrow X $ and $p_2:Z\longrightarrow X$ are two topological coverings of $X$, then it is univocally defined a structure of Riemann surface on both $Y$ and $Z$ such that $p_1$ and $p_2$ are holomorphic. Now if these two coverings are equivalent, then there is a homeomorphism $g: Y\longrightarrow Z$ such that $p_2\circ g=p_1$, but is $g$ a biholomorphism? In other terms two equivalent coverings of a compact Riemann surfaces are isomorphic in the category of Riemann surfaces?
I think that the answer is yes because in classical complex analysis if $g$ and $f$ are two holomorphic functions with $g$ non costant and moreover if $F$ is a continuous function such that $g\circ F=f$, then $F$ is holomorphic. I can't formalize this argument in the case of Riemann surfaces.