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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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    The conformity of a linear fractional transformation is dependent on that it preserves circles.2012-07-14

1 Answers 1

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Yes this is valid. ${}{}{}{}{}{}{}{}{}{}{}$

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    Oh Okay, In that case can I say $f=h\circ F\circ g^{-1}$2017-12-01