I'm new to differential geometry and reading Lee's book Manifold and Differential Geometry.
In the first chapter, he mentioned the following two maps on $\mathbb{R}^n$:
(1) $id: (x_1,x_2\cdots x_n) \rightarrow (x_1,x_2\cdots x_n)$
(2) $\varphi: (x_1,x_2\cdots x_n) \rightarrow (x_1^3,x_2\cdots x_n)$
Then, $\mathcal{A}_1$= { $(\mathbb{R}^n,id)$ } and $\mathcal{A}_2$= { $(\mathbb{R}^n,\varphi)$ } are two differential structure on $\mathbb{R}^n$, and $\mathcal{M}_1=(\mathbb{R}^n, \mathcal{A}_1)$, $\mathcal{M}_2=(\mathbb{R}^n,\mathcal{A}_2)$ are two manifolds.
It easy to verify that $\mathcal{M}_1$ and $\mathcal{M}_2$ have the same induced topology, the standard topology.
$\mathcal{A}_1$ and $\mathcal{A}_2$ are not compatible, for $id\circ \varphi^{-1}:(x_1,x_2\cdots x_n) \rightarrow (x_1^{\frac{1}{3}},x_2\cdots x_n)$ is not differentiable at origin. Therefore, $\mathcal{M}_1$ and $\mathcal{M}_2$ have different differential structure.
My question is: are they diffeomorphic?
According to Lee, the author, they are diffeomorphic through $\varphi$ (page 27).
But I don't think $\varphi$ is a diffeomorphism between them because $\varphi^{-1}$ is not differentiable at origin.
So are they not diffeomorphic?
But according the result of Donaldson and Freedman, each $\mathbb{R}^n$ except $n=4$ (with standard topology) only have one diffeomorphism class, so for any $\mathbb{R}^n$ except $\mathbb{R}^4$, $\mathcal{M}_1$ and $\mathcal{M}_2$ are diffeomorphic.
But why?