$S^1/G$ homeomorphic to $S^1$ when $G$ is a finite group acting freely on $S^1$?

Question

Let $G$ be a finite group acting freely on $S^1$. I saw from here(proof of proposition 1) that $S^1/G$ homeomorphic to $S^1$. Which theorem is used here? Is there a direct proof of this result?

Answer 1

One way to prove this is via the classification of $1$-manifolds. As JoeJohnson said in a deleted answer, any finite group acting freely on a manifold $M$ gives a cover $M\rightarrow M/G$ where $M/G$ is a topological manifold (without boundary) (and $M/G$ is smooth if $M$ is smooth and the action is by diffeomorphisms).

When $M = S^1$, $M/G$ must be compact since it's the continuous image of $S^1$ under the covering map. Then, from the classification of $1$-manifolds, it follows that $M/G$ must be homeomorphic to $S^1$ since it's the only compact $1$-manifold without boundary.