Several statements about $\mathbb{R}$ with chart defined by $f(x)=x^3$ I think I managed to show this statements but I am not sure about it. Since this is common problem in differentiable manifolds I was wondering if anybody has (or may write) a solution. 
Let $X$ be a copy of the real line $R$ and let $f:X \to R$ be $f(x)=x^3$. Taking $f$ as a chart, this defines a smooth structure on X. Prove or disprove the following statements. 


*

*X is diffeomorphic to R


My solution: By equivalence of definition of manifold using charts and functional structures we can say that $X$ has functional structure $F(U)= \{l\circ f | l \in C^\infty(f(U))\}$, so by $(R,C^\infty) \to (X,F_x)$ where $x \to x^{1/3}$ we have diffeomorphism becasue for each $g$ (which is equal to $l\circ f$ for some $l$)  in $F_x$ there is $g\circ f^{-1} \in C^\infty$ (which is equal to $l\circ f \circ f^{-1} = l \in C^\infty$ ) Similarly each function in $C^\infty (U)$ we identify with $F(U)$ just from construction ($F(U)= \{l\circ f | l \in C^\infty(f(U))\}$)


*The identity map $Id: X \to R$ is a diffeomorphism


Solution: No, since function $k$ sending $x$ to $x$ is in $(R,C^\infty)$ but $k\circ Id^{-1}=k$ is not in $F(R)$ because $k\circ x^{1/3} = x^{1/3} \not\in (R,C^\infty)$ 


*$f$ together with identity map comprise an atlas


Solution: $f^{-1} \not\in C^\infty$ thus we don't have proper change of coordinates.


*The last question I will post later


I hope you understand my proofs and have yours! Thank you!
You can find all of these in Bredon's Topology and Geomerty p.76
 A: My comments are below (all of your solutions are correct, so what is below is not much more than paraphrasing what you have said in my own words):


*

*The function $\phi: X\to \mathbb{R}$ defined by the rule $\phi(x) = x^{1/3}$ is smooth (as a map between smooth manifolds).

*The "identity map" $X\to \mathbb{R}$ is given by $\psi(x) = x^3$ in local coordinates (i.e., $\psi: \mathbb{R}\to \mathbb{R}$); although $\psi$ is smooth, $\psi^{-1}$ is not smooth (at $0$) so the identity map is not a diffeomorphism $X\to \mathbb{R}$.

*Yes, as you say, the transition map (corresponding to $f$ and the identity) is given by $\psi$ (see 2. above) in local coordinates, which is not a diffeomorphism for the reason explained (also in 2.).


Hope this helps! 
(As a supplementary remark, it is worth pointing out that a 1-manifold admits a unique smooth structure up to diffeomorphism; so the classification of topological 1-manifolds is equivalent to the classification of smooth 1-manifolds. In fact, this statement is true for $n$-manifolds where $n\leq 3$, and fails for $n=4$ (where deep phenomena occur which in a sense are still poorly understood). For example, $\mathbb{R}^4$ admits uncountably many smooth structures (the only $n$ for which $\mathbb{R}^n$ admits an exotic (i.e., nonstandard) smooth structure is $n=4$), and it is the open (smooth) Poincaré conjecture as to whether or not $\mathbb{S}^4$ admits any exotic smooth structures.) 
