TITLE: Are diffeomorphic smooth manifolds truly equivalent?
QUESTION [37 upvotes]: It seems to be an often repeated, "folklore-ish" statement, that diffeomorphism is an equivalence relation on smooth manifolds, and two smooth manifolds that are diffeomorphic are indistinguishable in terms of their smooth atlases.
There is a strange counter example in Lee's Introduction to Smooth Manifolds though, let us define two smooth manifolds modelled on the real line.
Let $\mathcal A$ be a smooth maximal atlas on $\mathbb{R}$ that is generated by the global chart $\varphi:\mathbb{R}\rightarrow\mathbb{R}$, $\varphi(x)=x$, and let $\bar{\mathcal A}$ be the maximal smooth atlas on $\mathbb{R}$ generated by the global chart $\bar{\varphi}(x)=x^3$.
The transition function $\varphi\circ\bar{\varphi}^{-1}$ is not smooth, so these two smooth structures are incompatible.
However the map $F:(\mathbb{R},\mathcal A)\rightarrow(\mathbb{R},\bar{\mathcal{A}})$ given by $F(x)=x^{1/3}$ is a diffeo, because $$ (\bar{\varphi}\circ F\circ \varphi^{-1})(x)=x, $$ and this map is smooth.
So the smooth manifolds $(\mathbb{R},\mathcal A)$ and $(\mathbb{R},\bar{\mathcal{A}})$ are diffeomorphic. Yet the two manifolds have incompatible, thus, different smooth structures.
Question: I guess I don't have a clear question, I am just somewhat confused. Because this is a counterexample, it seems to prove that the statement "two diffeomorphic manifolds cannot be told apart by their smooth structures" is wrong.
However how wrong it is? Can we consider the two manifolds given in this example equivalent? Is there any practical difference between the two? Is differential geometry the same on them?

REPLY [16 votes]: Keep in mind, the equivalence relation on differential manifolds is defined by the existence of a diffeomorphism between the two manifolds. This definition does not require that your favorite function be a diffeomorphism, nor anyone else's favorite function. The definition just requires that some function be a diffeomorphism.
For example, $(\mathbb{R},\mathcal{A})$ and $(\mathbb{R},\tilde{\mathcal{A}})$ are diffeomorphic despite that the identity map on $\mathbb{R}$ is not a diffeomorphism. The definition of being diffeomorphic does not require that the identity map on $\mathbb{R}$ be a diffeomorphism (which is what it means for the atlases $\mathcal{A}$ and $\tilde{\mathcal{A}}$ to be compatible). What the definition does require is that some function be a diffeomorphism, and the function $F(x)=x^{1/3}$ does the trick.