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It is well known that topological manifolds of dimension $\leq 3$ admit a unique smooth structure. In particular, homeomorphic open subsets of $\mathbb{R}^n$ are diffeomorphic for $n \leq 3$.

So my question is: given a homeomorphism $h:U \rightarrow V$ with $U, V \subset \mathbb{R}^n$, $n \leq 3$ open, is there a natural way to "smooth" it out to get a diffeomorphism?

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    $\begingroup$ What does "naturally" mean? I think the answer to your question depends on this; one can prove the existence of a smoothing operation but I am skeptical it can be made terribly explicit. $\endgroup$
    – mme
    Commented Aug 11, 2021 at 20:12
  • $\begingroup$ Yeah, I didn't have a precise meaning of "natural" in mind, but I was hoping for something explicit/constructive. $\endgroup$
    – PrimeNumbers
    Commented Aug 13, 2021 at 17:14

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The question asks, in particular, if there is a "natural" map from $Homeo(\mathbb{R})$ to $Diff(\mathbb{R})$. "Natural" should probably imply "continuous" and identity on smooth maps. Then the standard topological fact about retracts implies that such a map does not exist.

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  • $\begingroup$ I agree on "continuous." But, for example, convolution with a Gaussian might qualify as "natural" but isn't the identity on smooth maps. Of course it doesn't work in dimension > 1, so the point may be moot. $\endgroup$
    – Martin M. W.
    Commented Aug 12, 2021 at 16:00
  • $\begingroup$ "Natural" is not defined. But if a function is already smooth, the "smoothing" of it should not change the function. $\endgroup$
    – markvs
    Commented Aug 12, 2021 at 16:23
  • $\begingroup$ Continuity and identity on smooth maps definitely sound like something I'd want to see. I'm not particularly familiar with the topologies usually put on the $Homeo$ and $Diffeo$ groups, but I will look into it. Thanks! $\endgroup$
    – PrimeNumbers
    Commented Aug 13, 2021 at 17:20

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