Are concrete Riemann surfaces Riemann domains over $\mathbb C$? While reading about Riemann surfaces I stumbled upon the following two definitions. A Riemann surface of a complete analytic function is an example of both definitions while Abstract Riemann Surface is a generalisation of both definitions. So it seems like every (non singular, concrete) Riemann surface can be considered as Riemann domain over $\mathbb C$ or maybe vice versa. But I am unable figure it out. I am also curious about the historical origins of these definitions.
Riemann surface:
A subset $S \subseteq \mathbb C^2$ is called a (non-singular, concrete) Riemann surface if for each point $s \in S$, there is a neighbourhood $U$ of $s$ and a holomorphic function $F$ on $U$ with non-zero gradient at $s$ such that $S \cap U$ is the zero set of $F$.
Riemann domain over $\mathbb C$:
A topological space (Hausdorff and connected) $G$ is called a Riemann domain over $\mathbb C$ if there exists a local homeomorphism map $\pi:G \to \mathbb C$.
 A: There are theorems of Gunning and Narasimhan (Gunning, R. C., and Raghavan Narasimhan. "Immersion of open Riemann surfaces." Mathematische Annalen 174.2 (1967): 103-108.) and a theorem of Narasimhan (Narasimhan, R. "Imbedding of open Riemann surfaces", Göttingen Nachrichten, No. 7 (1960), pp. 159-165; also see American Journal of Mathematics Vol. 82, No. 4 (Oct., 1960), pp. 917-934) that almost(!) demonstrate an equivalence.
The first paper proves (quoting from the paper):
Theorem: Any (connected) open Riemann surface $X$ admits a holomorphic 
immersion into the complex plane; that is, there is a holomorphic mapping $F:X\to\mathbb{C}$ which is a local homeomorphism.
In other words, given any surface $S$ as in the original post above there is a holomorphic mapping $F:S\to\mathbb{C}$ exhibiting it as a $G$ as in the original post.
In the second paper it is proved that (quoting from the third paper!) "any open Riemann surface has a non-singular imbedding in $\mathbb{C}^3$".
In general, going from $\mathbb{C}^3$ to $\mathbb{C}^2$ will introduce some singularities. So going back all the way to $\mathbb{C}^2$ may not be possible.
A possible reason for such a obstruction is given below, but it is not a counter-example until proven!
For example, a smooth compact Riemann surface of any genus has an embedding in $\mathbb{CP}^3$. However, a general smooth compact Riemann surface of genus 6 does not have an embedding in $\mathbb{CP}^2$.
One can produce curves of degree 6 in $\mathbb{CP}^2$ with 4 nodes whose desingularisations are compact Riemann surfaces of genus 6. The 4 nodes will (in general) not lie on a line, so one cannot get an open Riemann surface of the type above by removing a $\mathbb{CP}^1$. Note, however, that this is not a proof that it cannot be done since there are more holomorphic functions on an open Riemann surfaces than those obtained by restrictions of meromorphic algebraic functions on the compact one.
