Riemann surfaces and coverings Suppose i have two riemann surfaces $S_{g_1}$ $S_{g_2}$ of genus $g_1$ and $g_2$, i want to find a criteria when first surface covers the second. My guess is that $2-2g_1 = k(2-2g_2)$(euler characteristics).My question is in topological setting, so im interested in classical coverings. So i have already proven, using triangulations , the necessity of it,but im stuck with sufficiency. Any ideas?
 A: One can prove sufficiency using some covering space theory combined with the classification of surfaces.
To start, one needs to know that $\pi_1(S_{g_2})$ contains subgroups of every finite index $k \ge 2$. This is not hard to see using surjectivity of the Hurewicz homomorphism $\pi_1(S_{g_2}) \to H_1(S_{g_2}) \approx \mathbb Z^{2{g_2}}$, just compose that with your favorite surjective homomorphism $\mathbb Z^{2{g_2}} \mapsto C_k$ onto the order $k$ finite cyclic group $C_k$, and then take the kernel of the composition.
Next, one uses covering space theory to produce a degree $k$ covering map $p : \widetilde S \to S_g$ with connected covering space $\widetilde S$.
Next, one proves that $\chi(\widetilde S) = k \chi(S_g) = k(2-2{g_2})$. I suspect from your post that you already know how to do this. One can lift a triangulation of $S_{g_2}$ to get a triangulation of $\widetilde S$, and then one can verify that the number of simplices of each dimension in $\widetilde S$ is equal to $k$ times the number of cells of that dimension in $S_{g_2}$.
Finally, one uses the classification of surfaces: since $S_{g_1}$ and $\widetilde S$ are both orientable and have the same Euler characteristic, there exists a homeomorphism $S_{g_1} \mapsto \widetilde S$. Composing that homeomorphism with the covering map $\widetilde S \to S_{g_2}$ one obtains the desired covering map $S_{g_1} \mapsto S_{g_2}$.
