A holomorphic function $\phi:X\to Y$, where $X$ and $Y$ are Riemann surfaces, is described in the following way:

For $a\in X$, let $a\in U_1$, where $U_1$ is an open set. Let $C_1:U_1\to V_1$ be a chart. Similarly, let $f(a)\in U_2$, and let $C_2:U_2\to V_2$ be a chart. Then $\phi$ is holomorphic iff $(C_2\circ \phi\circ C_1^{-1}): V_1\to V_2$ is holomorphic.

Another definition of holomorphic functions that is given is the following:

For any holomorphic function $f$ on $Y$, if $f\circ \phi$ is also holomorphic on $X$, then $\phi$ is holomorphic.

How are the two definitions equivalent?


marked as duplicate by Shailesh, Leucippus, user91500, Claude Leibovici, user223391 Sep 18 '16 at 18:56

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  • $\begingroup$ Can you do at least one direction? If you need a reference, it is done in Bredon's "Topology and Geometry" for smooth mappings, but it's the same for holomorphic. $\endgroup$ – xyzzyz Sep 15 '16 at 20:52
  • $\begingroup$ This statement is false as written. You need that Y is an open surface . Then it is Stein (not easy) and thus will embed holomorphically into $C^n$. $\endgroup$ – Moishe Kohan Sep 16 '16 at 0:03
  • $\begingroup$ @Gianluca no, it is not a duplicate. $\endgroup$ – Moishe Kohan Sep 16 '16 at 0:13
  • $\begingroup$ @xyzzyz what you are suggesting will not work, because it would prove an incorrect claim. $\endgroup$ – Moishe Kohan Sep 16 '16 at 0:15

Let me assume that $Y$ is connected in order to simplify the discussion. First of all, the answer to your question is negative if $Y$ is compact, simply because then $Y$ has only constant holomorphic functions (which I understand as functions $Y\to {\mathbb C}$). Now, assume that $Y$ is noncompact. By a theorem of Behnke and Stein:

H. Behnke, K. Stein, Entwicklung analytischer Funktionen auf Riemannschen Flachen, Math. Ann. Vol. 120 (1949) p. 430–461.

the surface $Y$ is Stein, i.e. admits a holomorphic embedding in ${\mathbb C}^n$ for some $n$. In particular, given $q\in Y$ there exists a holomorphic function $g: Y\to {\mathbb C}$ such that $g'(q)\ne 0$. Restricting $g$ to a small neighborhood $V$ of $q$ we obtain that $g: V\to {\mathbb C}$ is biholomorphic to its image. Therefore, there exists a local chart $C_2=g|V$ at $q$ such that the composition $C_2\circ f \circ C_1^{-1}$ is holomorphic implying that $f$ is holomorphic in the sense of the definition 1.

Remark. Note that this question is different (but suffers from a deficiency similar to that of) the question

Show that a function from a Riemann Surface $g:Y\to\mathbb{C}$ is holomorphic iff its composition with a proper holomorphic map is holomorphic.

The difference between the two questions is that in the linked question it was asked to prove holomorphicity of $h: Y\to {\mathbb C}$ given holomorphicity of the composition $$ X\stackrel{f}{\to} Y\stackrel{h}{\to} {\mathbb C}$$ where $f$ was the given proper function ($Y$ was assumed connected). In that question the missing assumption was that $f$ is nonconstant: (non)compactness of $Y$ was irrelevant.


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