algebraic-topologygeneral-topology
Let $\Sigma_n$ be a connected sum of $n$ tori for $n \in \mathbb{N}$. So basically, $\Sigma_0$ is the sphere, $\Sigma_1$ is the torus and then it's basically connected tori and the whole thing has $n$ holes. (I hope you get the picture.) I want to know for which values of $n$ does the sphere $S^2$ cover $\Sigma_n$.
I think the $\Sigma_n$ can be written as $S^2\times S^2\times \cdots \times S^2$ with $n$ many $S^2$'s, but does it mean that $S^2$ can cover the $\Sigma_n$ for all values of $n$?
Intuitively: deform the "top" and "bottom" of the cube into circles. Remove two opposite vertical edges. Then the wireframe will clearly be a connected sum of three tori. But adding in those two missing edges is like adding one of the missing edges is clearly like adding a "diameter" to one of the torus components, and this is just adding another torus.
Slightly more formally, start with the "tubed" base square, and add edges one by one. If you count how many times you needed to add a handle, as opposed to just stretching out an "appendage", you will come up with four, which makes five in total. In fact, you could start with just a single vertex. Then the number of times you'd have added a handle would be exactly five.
It's easy to see that you added a handle at those times when you added an edge without also adding a vertex. That should tell you what the general formula for the number of handles is. You can verify that it works in case of tetrahedron (by hand, as in case of the cube). For dodecahedron, you'd probably better stick to applying it.
One image solution: here $g_1-1=2$, $g_2-1=6$, $n=3$. If $g_1$ is greater, just add holes in the middle. If $n$ is greater, just go on with the snake. As pointed out by John Hughes, you clearly need $g_1$ (and hence $g_2$) to be strictly positive.
Best Answer
If $X$ is an $n$-fold cover of $Y$ then (under mild hypotheses which are satisfied here, say, for finite CW complexes) we have $\chi(X) = n \chi(Y)$ where $\chi$ is the Euler characteristic. The Euler characteristic of $\Sigma_g$ is $2 - 2g$, so it follows that if $\Sigma_g$ covers $\Sigma_h$ (any such covering map is necessarily finite, because $\Sigma_g$ is compact) then $1 - h$ must divide $1 - g$.
Setting $g = 0$ gives that $1 - h$ must divide $1$ (and the quotient must be positive, since it must be the degree of the cover) which gives $h = 0$. In other words, the only closed orientable surface that $S^2$ covers is itself.
An alternate argument is thinking about universal covers. $S^2 = \Sigma_0$ is simply connected, $T^2 \cong \Sigma_1$ has universal cover $\mathbb{R}^2$, and for $g \ge 2$, $\Sigma_g$ has universal cover the upper half plane $\mathbb{H} \cong \mathbb{R}^2$ (e.g. by the uniformization theorem although one can also write down a covering map explicitly). In particular the universal cover of $\Sigma_g$ is noncompact for $g \ge 1$ (this is is equivalent to the fundamental group being infinite) and so cannot be $S^2$.