# Covering of torus by orientable surface of genus g

I want to know whether there is a way of proving that $S_g$ (the orientable surface of genus $g$) covers $S_1$ (the torus) if and only if $g=1$ without invoking the Euler characteristic. I know that any covering $S_g \rightarrow S_1$ induces an injection on fundamental groups, but does this also necessarily yield an injection on the abelianizations for example? In that case, the assertion would follow.

I also found the older post Surface of genus $g$ does not retract to circle (Hatcher exercise) in which the top answer asserts that a certain map does induce an injection on abelianized fundamental groups, but I do not understand why.

## 1 Answer

If $S_g \to S_1$ is a covering, then the induced map on fundamental groups $\pi_1(S_g) \to \pi_1(S_1)$ is injective. Therefore $\pi_1(S_g)$ is isomorphic to a subgroup of $\pi_1(S_1)$. As $\pi_1(S_1)$ is abelian, every subgroup of $\pi_1(S_1)$ is abelian. For $g > 1$, $\pi_1(S_g)$ is non-abelian and therefore there is no covering $S_g \to S_1$.

• Thanks! I was being silly, thinking that the fundamental group of the torus was the free product. – Thomas Bakx Apr 10 '18 at 14:44