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We just consider puntured unit disk $\triangle^{*}$ in $\mathbb{C}$. $f$ is a bounded quasiconformal map on $\triangle^{*}$. Why $f$ can extend to the origin,becoming quasiconformal map on the whole disk?

rk:It is easy to see we only have to show that we can define $f(0)$ such that $f$ is continous.Then quasiconformal quasiconformality of new $f$ follows easily!
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We just consider puntured unit disk $\triangle^{*}$ in $C$. \mathbb{C}$. $f$ quasiconformal map on $\triangle^{*}$. \triangle^{*}$. Why $f$ can extend to the origin,becoming quasiconformal map on the whole disk?

rk:It is easy to see we only have to show that we can define $f(0)$ such that $f$ is continous.Then quasiconformal of new $f$ follows easily!
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We just consider puntured unit disk $\triangle^{}$ \triangle^{*}$ in $mathbb{C}$. C$. $f$ quasiconformal map on $triangle^{}$.Why \triangle^{*}$. Why $f$ can extend to the origin,becoming quasiconformal map on the whole disk?

rk:It is easy to see we only have to show that we can define $f(0)$ such that $f$ is continous.Then quasiconformal of new $f$ follows easily!
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