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In the Ahlfors' Complex Analysis chapter about the Riemann Mapping Theorem section 6.1.3, page 233, he states and proves this theorem:

Theorem 3. Suppose that the boundary of a simply connected region $\Omega$ contains a line segment $\gamma$ as a one-sided free boundary arc. Then the [Riemann mapping] function $f(z)$ which maps $\Omega$ onto the unit disk can be extended to a function which is analytic and one-to-one on $\Omega \cup \gamma$. The image of $\gamma$ is an arc $\gamma'$ on the unit circle.

In this particular context by "free boundary arc" he means an open interval $a < x < b$ of the real line.
I understand the proof of the existence of the extension: use the version of Schwarz reflection which only requires continuity of $\text{Re} \log f(z)$ at the boundary.

What I don't understand is his argument that $f$ is injective on $\gamma$. Here's how it goes:

We note further that $f'(z) \ne 0$ on $\gamma$. Indeed, $f'(x_0) = 0$ wold imply that $f(x_0)$ were a multiple value, in which case the two subarcs of $\gamma$ that meet at $x_0$ would be mapped on arcs that form an angle $\pi/n$ with $n \ge 2$; this is clearly impossible. If, for instance, the upper half disks are in $\Omega$, then $$\partial \log |f| / \partial y = - \ \partial \arg f / \partial x < 0$$ on $\gamma$, and $\arg f$ moves constantly in the same direction. This proves that the mapping is one-to-one on $\gamma$.

My attempt to clarify: In order to show that $f'(x_0)$ is not zero, I figured that if it were, we could factor $f(z)-f(x_0) = [(z-x_0)g(z)]^n$ where $g(x_0) \ne 0$. I think if you pick the right arc in the domain $\Omega \cup \gamma$, the image of that arc contains a point outside the closed unit disk.
Even if $f'(x) \ne 0$ on $\gamma$, I still don't see why the image of $\gamma$ can't "wrap all the way around the circle" and come back to the same point, and so be locally injective without being globally one-to-one.

Any help is greatly appreciated.

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Concerning the non-vanishing of $f'$ on $\gamma$, Ahlfors' argument is incorrect (even the best make mistakes sometimes), $f^{-1}$ would map two arcs of the unit circle to two arcs at an angle of $\pi/n$, not $f$ maps two subarcs of $\gamma$ to two arcs intersecting at an angle of $\pi/n$, cf. also this question.

Once the non-vanishing of $f'$ on $\gamma$ is established, the injectivity follows thus:

Suppose there were $x_1 \neq x_2$ in $\gamma$ with $f(x_1) = f(x_2) = w$. Then $f$ maps small disks $D_{\rho_k}(x_k)$ conformally to neighbourhoods of $w$. If $\rho_k$ is small enough, we have

$$f\bigl(D_{\rho_k}(x_k) \cap \Omega\bigr) = \mathbb{D}\cap f\bigl(D_{\rho_k}(x_k)\bigr)$$

for $k\in \{1,2\}$. But then every point in the nonempty open set

$$\mathbb{D}\cap f\bigl(D_{\rho_1}(x_1)\bigr) \cap f\bigl(D_{\rho_2}(x_2)\bigr)$$

would have at least two preimages in $\Omega$, thus $f$ would not be injective on $\Omega$.

Since $f$ is by assumption injective on $\Omega$ (it is, after all, a biholomorphic map between $\Omega$ and $\mathbb{D}$), the assumption that $f(x_1) = f(x_2)$ for some $x_1 \neq x_2$ in $\gamma$ must have been false.

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  • $\begingroup$ Thank you - again. I blame the error on his typist. ;) $\endgroup$ – bryanj Jun 15 '14 at 1:48
  • $\begingroup$ So it boils down to: locally one-to-one, plus open mapping. Got it. $\endgroup$ – bryanj Jun 15 '14 at 1:52

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