$\mathbb{S}^2$ and $\mathbb{RP}^2$ are the only compact surfaces with finite number of covers.

Question

Prove that the only compact surfaces with a finite number of covers are $\mathbb{S}^2$ and $\mathbb{RP}^2$

I have that the characteristic of this surfaces is $2,1$ therefore since a covering should have characteristic of the cover $\chi(\overline{X}) = k \chi(X)$ where $k$ is the number of sheets (which is finite since $\overline{X}$ is compact) the only posibility is for $\chi(\overline{X})$ to be a positive number but there are only a finite number of surfaces with positive characteristic.

However, what about the rest?

Answer 1

Hint: Just look the fundamental group of a surface $S$, and its universal cover $S'$. The covers of $S$ are $S'/H$ where $H$ is a subgroup of $\pi_1(S)$.

For the torus for example, $\pi_1(S)=\mathbb{Z}\oplus\mathbb{Z}$, you have infinitely many quotients $(\mathbb{Z}\oplus\mathbb{Z})/H$ for examples take $H_p=\mathbb{Z}\oplus p\mathbb{Z}$.

For oriented surfaces of genus >2, the abelianization of the fundamental group is ${\mathbb{Z}}^{2g}$ you can use a similar method.

Answer 2

General guidance rather than an explicit answer:

For all integers $n > 0$, the torus admits an $n$-fold cover (by a torus). The Klein-bottle admits a 2-fold cover (by a torus), hence a $2n$-fold cover.

The two-holed torus admits a 2-fold cover by a 3-holed torus (it may take a few moments to see why, and I'm going to leave this part as an exercise for you).

Once you see that (and indeed, show that a $k$-holed torus is covered by a $(2k-1)$-holed torus), you get the infinity of covers by just stacking up these things.

That leaves the connected sum $P^2 \# T^2 \# \ldots \# T^2$ for you to think about. Hint: find the cover corresponding to the generator of $\pi_1(P^2)$ to start with.

Note: your approach, with Euler characteristics, is definitely the wrong way to approach the general problme. Your argument correctly shows that these have only a finite number of covers, yes ... but it's useless for showing that there are many covers, because mere divisibility of Euler characteristics doesn't imply the existence of a cover. (Does the Klein bottle cover the torus, for instance?)