Suppose for $n\geq 1$, you have a smooth map $f\colon S^{n-1}\to S^{n-1}$. Viewing $S^{n-1}=\partial D^n$, is it possible to extend $f$ to a smooth map $\hat{f}\colon D^n\to D^n$, $D^n$ being the closed $n$-ball?
I noticed it extends it to the punctured disk by defining $\hat{f}(x)=|x|f(x/|x|)$. Can we do better and get an extension on all of $D^n$? Thanks.
can $S^{n-1} \to S^{n-1}$ be extended to $D^n \to D^n$ ? The answer is yes in the topological category - easily. In the smooth category you want to extend it linearly, so that you get a function $D^n \to D^n$ which is everywhere smooth but in the origin. Next thing you do is taking a smooth bump function defined on the complement of a neighborhood of 0 and takes 1 on the boundary. Taking the product of this real valued bump function and the original extension (as a function to $D^n \subset R^n$) you will get a smooth function everywhere (as product) and this resulting function will also be smooth at the origin, since it is constant on a neighborhood of the origin. done.