Let $R^2$ be the plane, and let G act on it with orientation preserving homeomorphisms, and assume that
- every orbit of G is a discrete subset in $R^2$
- G acts freely: $(\forall g \in G, g \neq e)$, $(\forall x \in R^2)$ $xg \neq x$.
Is it true that $R^2/G$ is a manifold with the factor topology, and G determines a covering to it?
In EMS: Geometry II$^1$, it is stated in a slightly more general way:
If $\Gamma$ is a discrete group of orientation-preserving homeomorphisms of a surface $X$, then the mapping it: $\pi: X \rightarrow X/\Gamma$ is a ramified covering (Kerekjarto [1923]$^2$)
So the statement may be true. But the source is a German textbook. Can anyone prove it, and/or give English sources, or provide a counter example?
1: Gamkrelidze, R. V. (ed.); Vinberg, E. B. (ed.), Geometry II: spaces of constant curvature. Transl. from the Russian by V. Minachin, Encyclopaedia of Mathematical Sciences. 29. Berlin: Springer-Verlag. 254 p. (1993). ZBL0786.00008.
2: von Kerékjártó, B., Vorlesungen über Topologie. I.: Flächentopologie. Mit 80 Textfiguren., Berlin: J. Springer, (Die Grundlehren der mathematischen Wissenschaften. Bd. 8.) VII u. 270 S. gr. $8^\circ$ (1923). ZBL49.0396.07.).
The point is that we may quickly prove that the action is properly discontinuous, that is for every compact $K \subset \mathbb R^2$, $I(K)=\{g \in G \mid gK\cap K \neq \emptyset \}$ is finite. To see this choose an alleged counterexample $K$ ( i.e., $K$ is a compact subset of $\mathbb R^2$ and $I(K)$ is infinite )and note that the set $$\bigcup_{g \in I(K)}gK$$ has diameter at most $3$ times the diameter of $K$ and contains $I(K)k$ for every $k \in K$. Then for any $k \in K$ we have that $I(K)k$ is (by freeness) an infinite bounded subset of $\mathbb R^2$ and thus must have a limit point. This contradicts the discreteness of orbit hypothesis.