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Why is every conformal bijection between disks actually a linear fractional transformation?

I thought I could justify this claim with the following idea.

Suppose $f$ is a conformal bijection from a disk $A$ to a disk $B$. Let $z_0\in A$ be arbitrary. Now there is a LFT $g$ from the unit disk to $A$ mapping 0 to $z_0$. Also, there is a LFT $h$ from the unit disk to $B$ mapping $0$ to $f(z_0)$. So altogether, $F=h^{-1}\circ f\circ g$ is a bijection on the unit disk fixing $0$, so by Schwarz' lemma, $|F(z)|\leq |z|$. Since $F^{-1}$ shares the same property, we have $\vert F^{-1}(F(z))\vert=|z|\leq |F(z)|$, so $|F(z)|=|z|$, so by Schwarz' lemma, $F(z)=cz$ for some $c$. So $F$ is a LFT, and thus $f$ is as well.

Is this valid? I was surprised to conclude $|F(z)|=|z|$ for all $z$, I wasn't expecting to find $F$ to be an isometry. Thanks all.

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  • $\begingroup$ If you're proposing $F$ is a rotation, it shouldn't surprise you that $|F(x)|=|z|$. I believe the proof in books is similar to what you've written. $\endgroup$ – ShawnD Mar 15 '12 at 3:17
  • $\begingroup$ The conformity of a linear fractional transformation is dependent on that it preserves circles. $\endgroup$ – mathon Jul 14 '12 at 8:17
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Yes this is valid. ${}{}{}{}{}{}{}{}{}{}{}$

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  • $\begingroup$ I don't know if it that obvious, why does the conclusion that $F$ is a LFT imply also that $f $ is also a LFT? $\endgroup$ – J. Kyei Dec 1 '17 at 4:30
  • $\begingroup$ Compositions of LFTs are LFTs. $\endgroup$ – leo Dec 1 '17 at 5:16
  • $\begingroup$ Oh Okay, In that case can I say $f=h\circ F\circ g^{-1}$ $\endgroup$ – J. Kyei Dec 1 '17 at 6:45

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