covering of a connected genus? 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 $n$?
 A: 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 $S^2$.
A: Firstly, $\Sigma_n$ cannot be written as a product $S^2 \times \cdots \times S^2$.
Secondly, using the fact that the Euler characteristic of a covering is the product of the Euler characteristic of the base space and the degree of the cover, we see that $S^2$ can only cover $\Sigma_0$ (i.e., itself) since $\chi(\Sigma_n) = 2 - 2n$.
