TITLE: The generated $C^{\infty}-$structure $\mathcal{A}_{\phi_{3}}\subset \mathcal{A_{id_{\mathbb R}}}$
QUESTION [1 upvotes]: Show that the  $C^{\infty}-$structure $\mathcal{A}_{\phi_{3}}$ generated from the  homeomorphism $\phi_{3}(x)=x^3$ is strictly smaller than the $C^{\infty}-$structure generated from $id_{\mathbb R}$, namely $\mathcal{A_{id_{\mathbb R}}}$. 
Show that $\mathcal{A}_{\phi_{3}}=\{\kappa \circ \phi_{3}: \kappa \in \mathcal{A_{id_{\mathbb R}}}\} \subset \mathcal{A_{id_{\mathbb R}}}$ and that this subset is strict. 
Definitions: Note that the $C^{\infty}-$structure $\mathcal{A}_{\phi_{3}}$ is simply all maps $\kappa$ so that $\kappa \circ \phi_{3}^{-1}\in C^{\infty}$
My ideas: For the first equality: 
Note for $\kappa \in \mathcal{A}_{\phi_{3}}$, it follows that $\kappa = (\kappa \circ \phi_{3}^{-1}) \circ \phi_{3}$
and further note that $(\kappa \circ \phi_{3}^{-1})\in C^{\infty}$ and hence $(\kappa \circ \phi_{3}^{-1})\circ id_{\mathbb R}\in C^{\infty}$, thus $(\kappa \circ \phi_{3}^{-1})\in \mathcal{A}_{id_{\mathbb R}}$ and hence $\kappa \in \{\kappa \circ \phi_{3}: \kappa \in \mathcal{A_{id_{\mathbb R}}}\}$, thus $\mathcal{A}_{\phi_{3}}\subseteq\{\kappa \circ \phi_{3}: \kappa \in \mathcal{A_{id_{\mathbb R}}}\}$
Note also that for any $\kappa \circ \phi_{3}$ where $\kappa \in \mathcal{A}_{id_{\mathbb R}}$, we have that $(k \circ \phi_{3})\circ \phi_{3}^{-1}=k \circ (\phi_{3}\circ \phi_{3}^{-1})=k\circ id_{\mathbb R}\in C^{\infty}$ and hence $k \circ \phi_{3}\in \mathcal{A}_{\phi_{3}}$, hence we have proven the equality $\mathcal{A}_{\phi_{3}}=\{\kappa \circ \phi_{3}: \kappa \in \mathcal{A_{id_{\mathbb R}}}\}$ and now I am struggling to prove that the inclusion $\{\kappa \circ \phi_{3}: \kappa \in \mathcal{A_{id_{\mathbb R}}}\}\subset \mathcal{A_{id_{\mathbb R}}}$ is not only true but also that it is strict. 
Any ideas?
Furthermore, use the above to show that the homeomorphism of $1-$dimensional $C^{\infty}-$manifolds $id_{\mathbb R}: (\mathbb R,\mathcal{A}_{id_{\mathbb R}})\to (\mathbb R,\mathcal{A}_{\phi_{3}})$ is $C^{\infty}$, however, its inverse is not even $C^{1}$.

REPLY [2 votes]: To show that the inclusion $\{\kappa \circ \phi_{3}: \kappa \in \mathcal{A_{id_{\mathbb R}}}\}\subset \mathcal{A_{id_{\mathbb R}}}$ is strict. Take the map $i(x)=x \in \mathcal{A_{id_{\mathbb R}}}$. You already showed that $\mathcal{A}_{\phi_{3}}=\{\kappa \circ \phi_{3}: \kappa \in \mathcal{A_{id_{\mathbb R}}}\}$ so for $i(x)$ to be an element of $\mathcal{A}_{\phi_{3}}$ we need to find $\kappa \in \mathcal{A_{id_{\mathbb R}}}$ such that $\kappa\circ\phi_{3}(x)=i(x)$. We know that $\phi_3$ is a homemorphisim and its inverse is given by $f(x)=x^{\frac{1}{3}}$. Thus if $i(x)$  is an element of $\mathcal{A}_{\phi_{3}}$ then $f \in \mathcal{A_{id_{\mathbb R}}}$ witch is false. 
There is a point that I am confused. In the definitions that I know $id_{\mathbb R}: (\mathbb R,\mathcal{A}_{id_{\mathbb R}})\to (\mathbb R,\mathcal{A}_{\phi_{3}})$ is not differentiable, as $x^{\frac{1}{3}}= \phi_3^{-1} \circ id \circ id(x)$ is not. We do have that $id_{\mathbb R}:(\mathbb R,\mathcal{A}_{\phi_{3}})  \to (\mathbb R,\mathcal{A}_{id_{\mathbb R}})$ is $C^{\infty}$ as $x^3=id \circ id \circ \phi_{3}$ is.
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