How to extend continous function from $S^1\to S^1$ to $D^2\to D^2$ continously

Question

Let $f:S^1\to S^1$ is the continous map then I have to prove that there exist continous extension $\bar f$ of f such that $\bar f:D^2\to D^2 $ is continous map where $D^2$ is closed disc in $\mathbb R^2$

I had following idea .

we can map origin to origin.then suppose there is ray emaniting form origin to some point a, its image is ray emaniting form origin to f(a) and same will follows.

But I could not able to write explicit form of function.

Is my idea is correct? Can anyone please help me how to write explicitly map.

Any Help will be appreciated

Answer 1

It might be easier to use polar coordinates. We think of $f$ as a continuous periodic function $f(\theta)$ defined on $\mathbb{R}$ (or $[0,2\pi]$). Then define a function on $D^2$ by $$\overline{f}(r,\theta)=rf(\theta)$$ where $0<r\leq 1$ and $0\leq\theta \leq 2\pi$. Also define it at the origin by $\overline{f}(0)=0$.

Answer 2

$D^2$ is the cone of $S^1$, that is, it is homeomorphic to the quotient of $S^1\times [0,1]$ where you collapse $S^1\times \{0\}$ to a single point, and in this model the inclusion $S^1\to D^2$ is just the inclusion $S^1\to S^1\times \{1\}$.

This gives a straightforward way of extending these maps : write an element of $D^2$ as $r z$ with $z\in S^1, r\in [0,1]$ and define $\overline{f}(rz) := rf(z)$.

If you don't know about the above homeomorphism, you can try to prove it, or you can check the continuity by hand, it's not hard.