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Suppose $M_1$ and $M_2 $ each $C^{\infty}$ manifolds and admits an atlas with once chart. Prove $M_1$ and $M_2$ are homeomorphic iff they are $C^\infty$ diffeomorphic.

One direction is obvious. But given that those manifolds are homeomorphic I'm not sure how to prove that they are $C^\infty$ diffeomorphic. Given that they are indeed homeomorphic there exists a homeomorphism from each chart of the compatible atlas on the compatible manifold. Im not sure how can I conclude that this hommeomorphism is actually $\infty $ times differentialbe with $\infty$ times differentiable inverse.

Any help would be appreciated. Thanks in advance.

DirichletIsaPartyPooper
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  • Since the charts give diffeomorphisms from $M_1$ and $M_2$ to open subsets of $\mathbb R^n$, this is equivalent to the claim that any two homeomorphic, open subsets of $\mathbb R^n$ are diffeomorphic. – Danu Apr 22 '22 at 14:39
  • @Danu Why do we know that the charts give diffeomorphisms from $M_1$ and $M_2$ to open subsets of $\mathbb{R}^n$? Is'nt the diffeomorphisms talks about the transition charts? (since there is no meaning for differentiability over abstract manifolds), also I am trying to prove this claim and then to conclude the equivalent claim about open subsets in $\mathbb{R}^n$. – DirichletIsaPartyPooper Apr 22 '22 at 14:43
  • Charts are diffeomorphisms basically by definition since it is they themselves that define the smooth structure. The claim about open subsets in $\mathbb R^n$ is false (see here: https://mathoverflow.net/questions/114528/are-homeomorphic-open-subsets-of-mathbbrn-also-diffeomorphic). – Danu Apr 22 '22 at 14:47
  • @Danu My claim is equivalent to the claim that 2 open subsets of $\mathbb{R}^n$ are homeomorphic iff they are $C^\infty$ diffeomorphic (and not just diffeomorphic). – DirichletIsaPartyPooper Apr 22 '22 at 14:52
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    I don't think this is true. See, for example, small exotic $\mathbb{R}^4$s. – Jason DeVito - on hiatus Apr 23 '22 at 19:31

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