In Tu's "An Introduction to Manifolds", Prop 6.10 says:
If $(U,\phi)$ is a chart on the manifold $M$ of dimension $n$, then the coordinate map $\phi:U\to\phi(U)\subset\mathbb{R}^n$ is a diffeomorphism.
However, I come up with a seeming counterexample from exercise 6.1.
Consider $(\mathbb{R},\psi)$ where $\psi = x^{\frac{1}{3}}$ is a homeomorphism but not a diffeomorphism. I don't see why this didn't hold water. Definitely something is wrong… Any help will be appreciated! Thanks in advance!
Best Answer
You have two different copies of $\Bbb{R}$ and they’re playing different roles, but you’re not distinguishing them. The first, lets call it $M$ to emphasize it is the ‘abstract manifold’ has a global chart $\psi:M\to\Bbb{R}$ (this $\Bbb{R}$ on the target is our ‘usual’ $\Bbb{R}$). How do we check the smoothness of the map $\psi$? By definition it means we fix an atlas for $M$ and check that for all charts in this atlas, $\psi$ composed with the inverse chart maps give something smooth in the usual multivariable calculus sense.
In our case, $M$ is by definition equipped with the maximal atlas containing the chart $(M,\psi)$. Thus, to check that $\psi$ is smooth, we have to merely check that $\psi\circ\psi^{-1}:\psi[M]\to\Bbb{R}$ is a smooth map. But this is obviously just the identity map of $\Bbb{R}$, $\text{id}_{\Bbb{R}}:\Bbb{R}\to\Bbb{R}$, so it is smooth in the usual calculus sense.
To repeat: you’re not taking into account that on the domain of $\psi$, that copy of $\Bbb{R}$ is given a smooth structure defined using $\psi$ itself.