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The question:

Find $f$ such that $\{(\mathbb{R},f)\}$ is a one chart atlas for a topological manifold on the set $\mathcal{M}=\mathbb{R}$, but not a $C^r$-manifold for $r>0$.

Attempt:

$\mathbb{R}$ is topological space that is Hausdorff, second-countable and locally Euclidean.

Question: A topological manifold does not really depend on the atlas, right? Just the initial set $\mathcal{M}$ that you given to work with.

Let $$f:\mathbb{R}\to\mathbb{R} \quad f(x) = \begin{cases} x & x<0 \\ 2x & x\geq 0 \end{cases} $$

$f$ is injective, therefore it generates a chart.

$f$ is bijective, continuous and $f^{-1}$ is continuous so it is homeomorphic.

$f$ is not differentiable so $f$ is not diffeomorphic.

Thus we have a $\mathcal{C}$-manifold.

Is this correct?

EDIT:

Let $(\mathbb{R},g)$ be another chart covering $\mathbb{R}$ so that $\{(\mathbb{R},f),(\mathbb{R},g)\}$ is also an atlas of $\mathbb{R}$. We want $f$ such that for some atlas map $g$ $$g \circ f^{-1} : f(\mathbb{R})\to g(\mathbb{R})$$ is homeomorphic and for but not diffeomorphic.

Let $g$ be the identity map and let

Let $$f:\mathbb{R}\to\mathbb{R} \quad f(x) = \begin{cases} x & x<0 \\ \frac{x}{2} & x\geq 0 \end{cases} $$ such that $$(g\circ f^{-1})(x) = \begin{cases} x & x<0 \\ 2x & x\geq 0 \end{cases} $$

$f$ is injective and $g\circ f$ is homeomorphic but not diffeomorphic because the derivative at $x=0$ does not exist.

EDIT II:

Let $(\mathbb{R},g)$ be another chart covering $\mathbb{R}$ so that $\{(\mathbb{R},f),(U,g)\}$ is also an atlas of $\mathbb{R}$. We want $f$ such that for any atlas map $g$ $$g \circ f^{-1} : f(\mathbb{R})\to g(\mathbb{R})$$ is not diffeomorphic.

So we need to show that for any $g:U\subset\mathbb{R}\to\mathbb{R}$, with $U$ open and $g$ injective: $g \circ f^{-1}$ is not bijective or not differentiable.

berrygreen
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  • You would prove that it is not $C^1$ manifold if you show that you can't extend that atlas to a $C^1$ atlas, i.e. an atlas in which all transition functions are $C^1$. In a one-chart atlas there are no transition functions. Therefore it is trivially $C^1$ so far. But what happens if I add a chart formed by a flat function $g$? Then $g\circ f^{-1}$ can be differentiable just fine. –  Apr 28 '18 at 11:45
  • Notice that your 'Edit' only shows that the atlas formed by the identity and the one formed by $f$ are not in the same $C^1$ equivalence class. This doesn't mean that the $C^1$ equivalence class of the atlas formed by $f$ doesn't contain other functions. –  Apr 28 '18 at 12:18
  • Let $g(x)=e^{-1/x}$ for $x>0$, $g(x)=-e^{-1/x}$ for $x<0$ and $g(0)=0$. Then $g\circ f^{-1}(x)=e^{-2/x}$ for $x>0$, $g\circ f^{-1}(0)=0$, and $g\circ f^{-1}(x)=-e^{-1/x}$ for $x<0$. You can check that $g\circ f^{-1}$ is a diffeomorphism. –  Apr 28 '18 at 12:18
  • The thing is that the problem is tricking you. The claim is actually false. If they give a 1-chart atlas, you can get a maximal $C^k$ atlas that contains it for all $k$. Here is one link with some references. –  Apr 28 '18 at 12:29
  • @deyore Thanks for all the help. I think the point of the problem was to understand what it is asking rather than try to find a solution. – berrygreen Apr 28 '18 at 12:37

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