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?

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 $S_{g_1} \mapsto S_{g_2}$.
