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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.

Yong Pan
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  • If the dimension is $n=2$, can you multiply by $x^d$ or $\overline{x}^d$ to get a degree 0 map ? Now, for a smooth degree 0 map $f$, I believe you can get a smooth homotopy (as an approximation of any homotopy) so that you can extend $f$ to the closed disk. – Roland Oct 10 '14 at 07:51
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    Why not $g(|x|)f(x/|x|)$ where $g$ is a smooth function with all derivatives vanishing at $0$, such as $e^{1-1/|x|}$? –  Oct 10 '14 at 08:06

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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.

Daniel Valenzuela
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  • No, it's true in the smooth category, too. Using a bump function, you can in fact make the function $0$ on a neighborhood of the origin. – Ted Shifrin Oct 10 '14 at 12:27
  • That actually sounds right. Thank you I will edit my post. Sorry for that. – Daniel Valenzuela Oct 10 '14 at 12:37
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    Just as a remark if anybody is interested why I wrote this wrong statement so quick without thinking enough: I thought about the problem: can an isomorphism on $S^{n-1} \to S^{n-1}$ be extended to an isomorphism $D^n \to D^n$. Then the answer is yes in $Top$ and no in $Diff$ since otherwise the Poincaré Conjecture would hold in the smooth category or in other words, this would imply that there don't exist any exotic spheres! sorry for inconvenience. – Daniel Valenzuela Oct 10 '14 at 12:49