I refer to Chapter II.3 of Rick Miranda - Algebraic curves and Riemann surfaces, which I understand says in Proposition II.3.9 that injective holomorphic maps $F: X \to Y$ between Riemann surfaces $X$ and $Y$, both of which are connected but not necessarily compact, are isomorphisms onto their images $F(X)$. (Connected is part of the definition of Riemann surfaces in this book. Not sure about other books.)
Question 1: Must (the underlying set of) a Riemann surface necessarily not be a singleton? I kind of assume this for the next questions 2 and 3.
Question 2: How is $F(X)$ necessarily a Riemann surface?
This is what I tried:
$X$ is connected and so since $F$ is continuous (since holomorphic implies continuous), $F(X)$ is connected (since connectedness is preserved under continuity).
$F$ is not constant: If $F$ were constant, then $F(X)$ is a singleton. Since $F$ is injective, we have that $X$ is a singleton if $F(X)$ is a singleton. I conclude $F$ is not constant if Riemann surfaces cannot be singletons (see Question 1).
$X$ is of course open in itself, and so since $F$ is open (by Proposition II.3.8, which applies, assuming (2)), we have that $F(X)$ is open.
$F(X)$ is Hausdorff and 2nd countable since these properties are inherited from $Y$.
Finally, for the atlas, Miranda has a recipe for open connected subsets of Riemann surfaces to be Riemann surfaces.
Question 3: If we now suppose $X$ is compact, then is $F$ surjective?
This is what I tried:
Proposition II.3.11 says that a non-constant holomorphic map with a compact domain is surjective has its image/range as compact.
If Riemann surfaces cannot be singletons (see Question 1), then I think $F$ injective implies $F$ non-constant and therefore, Proposition II.3.11 is applicable.
Question 4: To possibly strengthen Question 1, must (the underlying sets of) Riemann surfaces be either uncountable or empty since Riemann surfaces are locally holomorphic/ homeomorphic/ diffeomorphic to open subsets of the complex plane $\mathbb C$?