I've found an interesting exercice that I don't know how to approach. It goes like this.
We have a topological space which is Hausdorff, compact, connected and locally homeomorphic to $\mathbb{R}^{2}$. It is covered by hexagons and heptagons. There are two hexagons and one heptagon on each vertex. Assuming it is orientable and that there are fewer than 30 hexagons, which surface are we talking about?
So, I have thought about doing a complete triangulation (giving an orientation to each triangle) and then trying to reconstruct the surface as I would with the torus. But I've looked for it and it seems the tilling is that of a hyperbolic soccerball, and then I don't know how to keep going. Then I thought about using the Euler characteristic and I tried counting the faces, vertices and edges. No success either.
So, I'd welcome a little help!
You were indeed correct to take a look at the Euler Characteristic. Recall that for a surface, $\chi(X) = V-E+F$. Now let n be the number of hexagons and m be the number of heptagons. The number of faces is then $F= n+m$. Now each hexagon has 6 edges and each heptagon has 7 edges, and furthermore each edge is shared by two of the figures. Therefore, the number of edges is $E = \frac{6n +7m}{2}$. Similarly, each hexagon has 6 vertices and each heptagon has 7 vertices, and each vertex is shared by three of the figures.Therefore, $V = \frac{6n +7m}{3}.$
Now an orientable surface of genus g has Euler characteristic $2-2g$. In our case, $2-2g = \frac{6n +7m}{3} - \frac{6n +7m}{2} + n+m$. Which reduces to $12-12g = -m.$ Now for every heptagon there are 2 hexagons. Therefore, m can be at most 15. Therefore, $m=12$ and $g=2$.