Extending diffeomorphism to disk I am trying to prove the following
If $f:S^1 \to S^1$ is a diffeomorphism it can be extended to a diffeomorphism $F: D^2 \to D^2$.
But I can't seem to prove it. I proved it for homeomorphisms using polar coordinates: $F(r,\theta) = (r, f(\theta))$ and then $F$ is clearly a homeomorphism. I'm not sure how to prove that it is also a diffeomorphism. Please help me prove this. 
The derivatives I computed are these:
$$ {\partial F \over \partial r}(r,\theta) = (\begin{array}{c}1 \\ 0\end{array})$$ 
and
$$ {\partial F \over \partial \theta}(r,\theta) = (\begin{array}{c}0 \\ f'(\theta)\end{array})$$ 
But how to use the derivatives to prove the function is a diffeomorphism? Or would having the derivatives be sufficient? Problem is, the derivatives would only exist away from $(0,\theta)$. 

Maybe someone could provide me a reference to a book where this is a
  theorem given with proof. A reference to a book would be very helpful
  to me in general to help me learn this topic.

 A: Naively, given $f:\mathbb{S}^1\to\mathbb{S}^1$, we'd like to define $F:\mathbb{D}^2\to\mathbb{D}^2$ in polar coordinates as
$$ F(r,\theta) = (r,f(\theta)).$$
In fact, $F$ is continuous, but only smooth if $f$ is a rigid motion of $\mathbb{S}^1$. 
So some care must be taken: we need to "swirl" the function so that it approaches $0$ in a smooth, rather than kinky, manner.

We'll use the fact that a smooth degree-$\pm 1$ self-map of $\mathbb{S}^1$ is smoothly isotopic to $\pm I$.[1] For simplicity, assume $f$ is orientation-preserving. Let $H(t,\theta)$ be as defined below. Now put
$$F(r,\theta) = (r, H(r,\theta)) = (r, r^2f(\theta) + (1-r^2)\theta).$$
This is a homeomorphism of $\mathbb{D}^2$, and everywhere smooth except possibly at $r=0$. There are several ways to check differentiability, but perhaps the most straightforward is to just verify that for any $\theta$, 
$$\lim_{\epsilon\to 0}\frac{1}{\epsilon}\bigg( F(\epsilon,\theta) - F(0,0) - (\epsilon,\theta) \bigg) = 0.$$
Note that $F$ is differentiable at zero because we used $t^2$ in the isotopy instead of $t$.[2]

[1] Any map $f:\mathbb{S}^1\to\mathbb{S}^1$ lifts to a map $\widetilde{f}:\mathbb{R}\to\mathbb{R}$ which satisfies $\widetilde{f}(x+1) = f(x) + \deg f$. If $f$ is a degree $1$ homeomorphism, then it lifts to a strictly increasing map $\widetilde{f}$ which satisfies $\widetilde{f}(x+1) = \widetilde{f}(x) + 1$. Now define $\widetilde{H}$ as 
$$\widetilde{H}(t,x) = t^2\widetilde{f}(x) + (1-t^2)x.$$
Observe that $\widetilde{H}(t,x+1) = \widetilde{H}(t,x) + 1$, so it descends to a map $H:I\times \mathbb{S}^1\to\mathbb{S}^1$. For each $t$, $\widetilde{H}_x(t,x) = (1 - t^2) + t^2\widetilde{f}'(x) > 0$ so $H$ is a diffeomorphism of $\mathbb{S}^1$ for each fixed $t$. Thus $H$ is a smooth isotopy between $f$ and the identity map.
[2] I haven't thought about verifying higher-order derivatives, but if it's an issue, instead of $r^2$ use a bump function which is $0$ on a neighborhood of $0$, $1$ on a neighborhood of $1$, and $C^\infty$. If $\phi$ is such a function (it can be constructed using bump functions), then define $$\widetilde{H}(t,x) = \phi(t)\widetilde{f}(x) + (1-\phi(t))x$$
and the proof proceeds as before, except that $F$ is manifestly $C^\infty$ everywhere.
A: The radial extension does not work, unless $f$ is a rigid motion of the boundary. But the harmonic extension works, by the Radó–Kneser–Choquet theorem. Duren's book Harmonic mappings in the plane has a careful exposition of this theorem. 
Harmonic extension works only in this setting, extending from $S^1$ to $D^2$. Extending an $S^2$ diffeomorphism to a $D^3$ diffeomorphism is hard. Extending an $S^6$ diffeomorphism to a $D^7$ diffeomorphism is impossible in general.
