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:

  1. $X$ is connected and so since $F$ is continuous (since holomorphic implies continuous), $F(X)$ is connected (since connectedness is preserved under continuity).

  2. $F$ 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).

  3. $X$ 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.

  4. $F(X)$ is Hausdorff and 2nd countable since these properties are inherited from $Y$.

  5. 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:

  1. Proposition II.3.11 says that a non-constant holomorphic map with a compact domain is surjective has its image/range as compact.

  2. 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 $\mathbb C$?

  • 1
    $\begingroup$ If $X=\{x\}$ then $\{x\}$ is the only chart around $x$, but there's no homeomorphism $\{x\}\to D$ where $D$ is the open unit disk in $\mathbb C$. $\endgroup$ Sep 5, 2020 at 8:08
  • $\begingroup$ @RandyMarsh Right, that's what I thought. Thanks. Follow-up 1. post as answer for question 1? i'll upvote even if you don't answer the others. Follow-up 2. So this extends to non-empty countable i.e. Question 4 is answered affirmatively? Follow-up 3. Is that riemann surfaces are not singletons indeed relevant for answering questions 2 and 3 affirmatively? $\endgroup$
    – BCLC
    Sep 5, 2020 at 8:11
  • $\begingroup$ Also, $X=\{x\}$ is compact, however the "smallest" compact Riemann surface is the Riemann sphere. $\endgroup$ Sep 5, 2020 at 8:11
  • $\begingroup$ @RandyMarsh ummmm....ok thanks...i guess smallest is in the sense that riemann sphere is holomorphically embedded or something in every compact riemann surface? $\endgroup$
    – BCLC
    Sep 5, 2020 at 8:46
  • $\begingroup$ Smallest in the sense of (topological) genus. It is the unique compact Riemann surface with genus $0$. $\endgroup$ Sep 5, 2020 at 9:43

1 Answer 1

  1. Yes. If $X=\{x\}$ then $\{x\}$ is the only chart around $x$, but there's no homeomorphism $\{x\}\to D$ where $D$ is the open unit disk in $\mathbb C$.

  2. In general $F(X)$ will not be a Riemann surface. However, if $F$ is injective, then $F(X)$ will be: Let $w\in F(X)$, let $w=F(z)$ and $(U,\phi\colon U\to V)$ a chart around $z\in X$. Then $F(U)$ is an open neighbourhood of $w$ in $F(X)$ and the map $\phi\circ F^{-1}\colon F(U)\to V$ is a homeomorphism, hence a chart for $F(X)$. This, of course, fails when $F$ is not injective. The compatibility of charts is immediate from $(\phi_1\circ F^{-1})\circ(\phi_2\circ F^{-1})^{-1}=\phi_1\circ\phi_2^{-1}$ (when defined).

  3. Yes, this follows by Proposition 3.11.

  4. They have exactly the cardinality of $\mathbb C$. They must be uncountable and thus can't be empty. This is a general fact about connected Hausdorff manifolds, see e.g. https://mathoverflow.net/questions/67962/cardinality-of-connected-manifolds

  • $\begingroup$ Thanks Randy Marsh! $\endgroup$
    – BCLC
    Sep 5, 2020 at 9:49

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