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!

3 added 11 characters in body

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!

2 added 4 characters in body; deleted 5 characters in body; deleted 3 characters in body

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!

1