every continuous map $S^1 \rightarrow S^1$ can be extended to continuous map $B^2 \rightarrow B^2$ Let $S^1$ denote the unit circle, and $B^2$ denote the closed unit disk. I came across this question and got stuck:
Q:) Every continuous map $f :S^1 \rightarrow S^1$ can be extended to continuous map $B^2 \rightarrow B^2$. If $f$ is a homeomorphism, then we can choose the extension to be a homeomorphism also.
I tried the obvious approach, namely for $x\in B^2-{0},$ define the extension by $x\rightarrow ||x||.f(x/||x||)$, but this does not seem to work,or I cannot figure out why this might work. What am I missing?
Thanks in advance. 

The map 
$$
x\mapsto\begin{cases}||x||f(x/||x||), &x\ne0\\0,&x=0\end{cases}
$$
is obviously continuous on $B\setminus0$, where it is a composition of the continuous mappings $x\mapsto x/||x||$, $f$, and multiplication by $||x||$. So we need to prove only the continuity at 0, which is obvious as well, because $||(||x||f(x/||x||)||=||x||$.
