# Is there a connected $n$-sheeted covering space of a closed orientable surface for any positive integer $n$?

Let $$S$$ be a closed orientable surface. For any positive integer $$n$$, is there a connected $$n$$-sheeted covering space of $$S$$? This is certainly not true if $$S=S^2$$ because $$S^2$$ is simply-connected. The result is true for $$S=T^2$$, since $$\pi_1(T^2)=\Bbb Z^2$$ has an index $$n$$ subgroup for each $$n$$. However I cannot handle the other cases. I know that this is equivalent to finding an index $$n$$ subgroup of $$\pi_1(S)$$ but the $$\pi_1$$ for the other surfaces are quite complicated, I think. Any hints?

• See Example 1.41 of Hatcher's Algebraic Topology. Commented Jun 17, 2021 at 22:24

This lay out works for any surfaces of genus $$\geq 1$$ and any $$n$$.
Since the first homology is the abelianization of the fundamental group, and $$H_1(S)=\mathbb{Z}^{2g}$$, where $$g$$ is the genus, we see that we have a surjective composition $$\pi_1(S) \rightarrow H_1(S) \rightarrow \mathbb{Z}$$, where this last map is just projection onto the first coordinate.
Since taking the preimage of a subgroup $$H \leq G$$ under a surjective map preserves index, we deduce that $$\pi_1(S)$$ contains an index n subgroup given by the preimage of $$n\mathbb{Z}$$ under the above composite. As you mention, covering space theory implies we have the covering you ask for.