# Forster's proof of the local behavior of holomorphic mappings

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.

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 $$h(z) = e^{\log(g(z))/k}$$.