Suppose $X$ is a closed compact surface (i.e. a compact two-dimensional manifold without boundary).
- If $X$ has genus $g$, what is the minimum number of (homeomorphic images of) open balls required to cover $X$?
- If $X$ can be covered by $n$ (homeomorphic images of) open balls, is there an upper bound for the genus of $X$? If so, what's the upper bound?
One can show that if $X$ can be covered by $2$ open balls, then $X$ has to be a sphere (it's a fun exercise so I won't spoil it with further details). Furthermore, it's not too hard to show that a torus can be covered by $3$ open balls. I'm not sure what the situation is for more balls/higher genus though, so I'm content to leave this one with the internet!
I believe that you only need $3$ open balls to cover any closed oriented compact surface $X$ of $g\geq 1$.
Consider the standard fundamental polygons of $X$, it is a CW complex by identifying the boundary of a $2$-cell with the wedge sum of $2g$ circles. Suppose the intersection point of the $2g$ circles is $p$. Let $B_1$ be the $2$-cell of $X$, $B_2$ be a open neighborhood of $p$ which is homeomorphic to an open $2$-ball, then $X-(B_2\cup B_2)$ is a disjoint union of $2g$ paths, each path is homeomorphic to the closed interval. Finally, choose $B_3$ to cover these $2g$ paths, we are done.