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(This question might turn out to be too elementary for this site,
if so I'm sorry, but I can't find the answer anywhere.)


Does there exist a function $\; f : \{z\in \mathbb{C} : |z| < 1\} \to \mathbb{C} \;$

such that $f\hspace{.01 in}$ is a quasiconformal bijection and $f^{-1}$ is quasiconformal?

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8  
The plane and a disk are not quasiconformally equivalent, see e.g. page 11 in math.qc.edu/~zakeri/papers/ahl-bers.pdf. Incidentally, the inverse of a quasiconformal map is automatically quasiconformal. –  Igor Belegradek Sep 20 '11 at 2:08

1 Answer 1

If you mean whether there is a quasiconformal mapping from the unit disk D onto the whole complex plane C, then the answer is NO. As you have noticed, the inverse mapping of a quasiconformal mapping is always quasiconformal. One can easily prove that f(R^n)=R^n when f is an entire quasiconformal mapping. {Here one can use the fact that quasiconformal mappings are locally quasisymmetric and so it must be global quasisymmetric (since the domain is R^n). For quasisymmetric mapping, it is easy to see that it must send unbounded set to unbounded set.}

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