I have the same question as this post here. To summarise, the poster introduces an apparent counter-example. Consider the manifold $\mathbb{R}$ and a spanning atlas $(\mathbb{R}, \psi)$ where $\psi = x^{1/3}$. Clearly $\psi$ is homeomorphic, but it is not diffeomorphic. I believe the easy counter example is to see that $\partial_x\psi=(1/3)x^{-2/3}$ which is clearly not defined for $x = 0\in\mathbb{R}$. Hence, the counter-example.
I do not understand the accepted answer on the post. I am also following Loring's book on manifolds so I know that to show that a map $f: M\rightarrow\mathbb{R}$ is smooth on a manifold $M$ then we need to ensure that for every chart $(U, \phi)$ on $M$ then $f\circ\phi^{-1}$ is also $C^{\infty}$.
It seems that the answer to the post is saying that $\psi$ is diffeomorphic since the composition $\psi\circ\psi^{-1} = \mathbb{1}$ is clearly smooth. This agrees with the definition of smoothness just mentioned, but it also seems incompatible with the smoothness I showed in the first paragraph. I don't understand how to resolve this discrepancy; can I get another explanation on why $\psi$ is indeed a diffeomorphism?
As usual, the simplest examples in differential topology are also the most confusing; largely because we use precise and general definitions which are good for proving general statements, but take care to unpack into what should be easy examples.
The crux of the matter is is that we can equip $\mathbb R$ with two smooth structures. The first is the standard one given by $(\mathbb R, I)$, where $I$ is the identity map. The second is the $(\mathbb R, \phi)$, where $\phi$ is the homeomorphism $x\mapsto x^3$ (this defines a smooth structure because $\phi^{-1}\circ \phi=I$, which is smooth, so we just through in any other charts which are smoothly compatible with $\phi$ to obtain a maximal smooth atlas). These smooth structures are not the same (meaning the maximal smooth atlases they induce are different) because $I\circ \phi^{-1}(x)=x^{1/3}$ is not smooth at the origin, it has a cusp. It follows that $(\mathbb R,I)$ and $(\mathbb R, \phi)$ are "different" smooth manifolds, in the sense that their smooth structures are different.
However, these two smooth manifolds are diffeomorphic. Indeed, consider the map: \begin{align} F:(\mathbb R, I)&\longrightarrow (\mathbb R,\phi)\\ x&\longmapsto x^{3} \end{align} We claim that this map is smooth, which means we need to show that $\phi^{-1}\circ F\circ I$ is smooth. This is clear though, as $\phi^{-1}\circ F\circ I(x)=x$ which is a smooth function. It has an inverse given by $F^{-1}(x)=x^{1/3}$, which is also smooth since $I\circ F^{-1}\circ \phi(x)=x$. It follows that $F$ is a diffeomorphism.
So the moral of the story is that even though these two smooth manifolds are indeed distinct, they are the same up to a diffeomorphism, which in differential topology is as good as you're going to get. In fact, every smooth structure on $\mathbb R$ is diffeomorphic to one another, so up to diffeomorphism, there is only one way to make $\mathbb R$ a smooth manifold. This holds in every dimension except $4$, where $\mathbb R^4$ admits uncountably many non diffeomorphic smooth structures.