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Be $U\subseteq \mathbb{C}$ simply connected region with $U\neq\mathbb{C}$ and $a,b\in U$, $a\neq b$. Is there an biholomorphism $f:U\longrightarrow U$ with $f(a)=b$ and $f(b)=a$?

I know that, by the Riemann mapping theorem, there are unique isomorphisms $h:U\longrightarrow D(0,1)$ with $h(a)=0$, and $g:U\longrightarrow D(0,1)$ with $g(b)=0$. And considering $Id:D(0,1)\longrightarrow D(0,1)$ automorphism, I lead to two cases:

If $g(a)=h(b)$, then I can define $f=g^{-1}\circ Id \circ h $, becouse $f$ is an isomorphism by composition of isomorphisms, $f(a)=g^{-1}\circ Id \circ h(a)=g^{-1}( h(a))=g^{-1}(0)=b$ and $f(b)=g^{-1}\circ Id \circ h(b)=g^{-1}(h(b))=g^{-1}(g(a))=a$.

If $g(a)\neq h(b)$ is where I have problems. I think that I must build an automorphism $\phi$ of the disc that take $g(a)$ into $h(b)$ so $f=g^{-1}\circ \phi\circ h$.

I also know that the set of automorphisms of the disc is $\{e^{i\theta}\phi_{a} (z)=e^{i\theta}\dfrac{z-a}{1-\bar{a}z},\theta\in\mathbb{R}, a\in D(0,1)\}$.

Is there an automorphism of the disc like that?

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  • $\begingroup$ [Psst! It's -morphism, not -morfism! ;) ] $\endgroup$ – Shaun Nov 23 '13 at 12:11
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Let $h: U\longrightarrow D(0,1 ) $ be isomorphism such that $h(a) =0.$ Then You can take $\varphi : U\longrightarrow U $, $$\varphi (z) =h^{-1} \left(\frac{h(b)-h(z)}{1-\overline{h(b)} h(z)} \right) .$$

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  • $\begingroup$ Yes, that works! thank you! $\endgroup$ – Irene Nov 24 '13 at 14:45

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