Forster gives a proof of the local behavior of holomorphic mappings as the following. Let $f: X\to Y$ be a nonconstant holomorphic mapping where $X, Y$ are Riemann surfaces. Suppose $a\in X$ and $b=f(a)$, then there exist charts $(U, \varphi)$ on $X$ and $(U',\varphi')$ on $Y$ such that

(1) $a\in U, \varphi(a)=0$, $b\in U',\varphi'(b)=0$;

(2) $f(U)\in U'$;

(3) The map $F=\varphi'\circ f\circ\varphi^{-1}:V\to V'$ is given by $f(z)=z^k$ for all $z\in V$.

First, he notes that there exist charts $\varphi_1: U_1\to V_1$ and $\varphi': U'\to V'$ that suffice the first two properties if we replace $(U,\varphi)$ by $(U_1, \varphi_1)$. Then he claims that $f_1=\varphi'\circ f\circ\varphi^{-1}$ is nonconstant by identity theorem. Since $f_1(0)=0$, there exists some $k$ such that $f(z)=z^kg(z)$ for $z\in V\ni 0$, where $g(0)\neq 0$ is holomorphic. Then there is some holomorphic $h$ such that $h^k=g$...

I could understand the part after this construction. There are three steps that are not that clear to me. First, the existence of $\varphi_1:U_1\to V_1$. Second, in what way the identity theorem is applied to our case so that we can conclude $f_1$ is nonconstant (my guess is that if $f_1$ is constant then $\varphi^{-1}\circ f_1\circ\varphi$ will be constant and since this composition coincides with $f$ on the open set $U$, $f$ will then be constant by identity theorem, a contradiction)? Third, where does the function $h$ come from?

I tried to read some other sources, but the proof here seems to be pretty standard. Thanks in advance. Any help will be appreciated.


1 Answer 1


Existence of $\varphi_1:U_1 \rightarrow V_1$: Forster's wording is a little confusing. Since $Y$ is a Riemann surface, there is a chart $(\psi, U')$ around $b$. Since $f$ is continuous, there is an open set $W$ with $a \in W$ and $f(W) \subset U'$. There is also a chart $(\varphi, U)$ around $a$. Set $U_1 = W \cap U$ and $\varphi_1 = \varphi|U_1$. Then replacing $(U, \varphi)$ with $(U_1, \varphi_1)$ satisifes (i) and (ii).

The identity theorem: Pretty much what you said. Since $f$ is non-constant, by the identity theorem it is non-constant on the open set $U_1$. Since $\varphi_1$ is surjective from $U_1$ to $V_1$, $f \circ \varphi_1^{-1}$ is non-constant on $V_1$. And since $f \circ \varphi_1^{-1}(V_1) \subset U'$ and $\psi$ is 1-1 on $U'$, $f_1 = \psi \circ f \circ \varphi_1^{-1}$ is non-constant on $V_1$.

The function $h$: since $g(0) \neq 0$, $g$ is nonzero in a disk containing $0$, and therefore has a well-defined logarithm there, so set $h(z) = e^{\log(g(z))/k}$.


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