Counterexample for a differentiable structure I'm trying to understand the definition of a differentiable structure.
Is it correct that $x\rightarrow x$ and $x \rightarrow x^3$ doesn't form a diffeomorphism, since $x\rightarrow x^{1/3}$ isn't differentiable in $0$?
 A: You're talking about a common example when introducing the notion of differentiable structure in $\mathbb{R}$. We will see that even though the underlying topological space is the same, what is usually seen as a differentiable mapping with one structure need not be differentiable when considering two different structures.
To be explicit, I am assuming you are considering two different differentiable structures in $\mathbb{R}$. One of them would be given by the atlas
$$A_1 = \{(\mathbb{R}, f)\}$$
where $f(x) = x$ and the other by another atlas
$$A_2 = \{(\mathbb{R}, g)\}$$
where $g(x) = x^3$.
Now, the identity $i: \mathbb{R}\rightarrow \mathbb{R}$ $i(x)=x$ certainly is a homeomorphism of topological spaces. However, as a map between differentiable manifolds, it depends on the structures whether or not it is a diffeomorphism. In particular,
$$
i:(\mathbb{R},A_1) \rightarrow (\mathbb{R},A_2)
$$
as seen in coordinates
$$ g^{-1} \circ i \circ f (s) = s^{1/3}$$
is not a differentiable mapping (not differentiable at the origin). Note that however the inverse
$$
i:(\mathbb{R},A_2) \rightarrow (\mathbb{R},A_1)
$$
is differentiable, as it can be seen in coordinates $f^{-1}\circ i \circ g (\xi) = \xi^3$.
