every continuous map $S^1 \rightarrow S^1$ can be extended to continuous map $B^2 \rightarrow B^2$

Question

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.

Answer 1

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 $||(||x||f(x/||x||)||=||x||$.

Answer 2

Riccardo's comment is still useful you just need to be clever in how you apply it. Every continuous map $f\colon S^1\to S^1$ can be extended to a map $\tilde{f}\colon S^1\to B^2$ by composing with inclusion $i\colon S^1\to B^2$ of the boundary, so $$\tilde{f}=f\circ i$$ and then because $B^2$ is contractible, $\tilde{f}$ must be nullhomotopic, therefore the map can be extended to a map $B^2\to B^2$ because of the theorem which says that any map from the circle to a space is nullhomotopic if and only if that map can be extended to the disc.

For the second part, just note that the identity on the circle can be extended to a homemorphism of the disc (also the identity) and so, using the fact that a homeomorphism of the circle is homotopic to the identity on the circle (or reflection), you can easily form you map in pieces (do the outer annulus first via the relevant homotopy and then 'fill in the hole' with a disc which corresponds to the identity.)