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Let $\mathbb{D}$ be the unit disk. I am trying to prove the following statement

Let $U\subsetneq \mathbb{C}$ be a simply connected domain. Then there exists an injective holomorphic map $F:U\rightarrow \mathbb{D}$.

The first part of the proof goes something like this:

(1) WLOG we can assume that $0\notin U$.

(2) Let $z_0\in U$ be fixed. Define $f$ on $U$ as $$f(z)=\int_{z_0}^{z}\frac{1}{w} dw$$ Since $U$ is simply connected and $0\notin U$, the map $f$ is well defined.

I have to prove that $f$ is injective. How to do this?

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  • $\begingroup$ You want to prove Riemann mapping theorem! $\endgroup$ – Nosrati Dec 3 '17 at 6:53
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The idea is that $f$ is an antiderivative of $\frac 1z$ and therefore "essentially" a logarithm (which is injective as the inverse function of the exponential function). The following argument however is quite elementary, avoiding logarithms and their ambiguities (branches) completely.

From the definition of $f$ it follows that $f'(z) = \frac 1z$ on $U$. If we define $g$ on $U$ as $g(z) = ze^{-f(z)}$ then $$ g'(z) = e^{-f(z)} - z f'(z) e^{-f(z)} = 0 \\ \Longrightarrow g(z) = C $$ for some constant $C$, and therefore $e^{f(z)} = \frac 1C z$.

The injectivity of $f$ follows immediately.

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