I'm currently reading from Loring Tu's Introduction to Manifolds, and I've come across the following problem:
My confusion lies primarily with the last comment mentioned in Tu's hint: the identity map $\mathbb{R'} \to \mathbb{R}$ is not smooth (after looking in Tu's list of errata for the book, he meant to say $\mathbb{R'} \to \mathbb{R}$ instead of $\mathbb{R} \to \mathbb{R}$).
This perplexes me a little bit for the following reasons:
Let $i: \mathbb{R'} \to \mathbb{R}$ be the map given by $i(x) = x$.
Tu gives a very clear set of criteria to determine when such a map (an $\mathbb{R}$-valued function defined on a manifold) is smooth:
So, at least in my mind, we have the atlas $ \{ (\mathbb{R}, \psi) \} $ on $\mathbb{R'}$ where $\psi(x) = x^{1/3}$. It's not necessarily the smooth structure on $\mathbb{R'}$, but it's an atlas contained in the smooth structure. It is clear that the inverse of $\psi$ is the map $\psi^{-1}: \mathbb{R} \to \mathbb{R'}$ given by $\psi^{-1}(x) = x^3$. By proposition $6.3(\text{ii})$, $i(x) = x$ will be $C^{\infty}$ if the map $i \circ \psi^{-1}: \mathbb{R} \to \mathbb{R}$ is $C^\infty$.
But $i \circ \psi^{-1}(x)$ is precisely $i \circ \psi^{-1}(x) = x^3$ which is clearly smooth. What am I missing here? I've been trying to find which assumptions I'm making are incorrect, but I can't see them. Any help with this would be greatly appreciated! (Note, I'm not asking for help solving problem 6.1 as I've already found another diffeomorphism between $\mathbb{R}$ and $\mathbb{R'}$. I would just like clarity on identity's lack of smoothness).
Let us consider the map $\iota : \mathbb R \to \mathbb R' ,\iota(x) = x$. Is it a diffeomorphism? This is the case if and only if both $\iota$ and $\iota' = \iota^{-1} : \mathbb R' \to \mathbb R$ are smooth. Clearly as maps between the underlying topological space $\mathbb R$ both $\iota, \iota'$ are the identity.
Proposition 6.3 can only be applied to $\iota'$ (with $M = \mathbb R'$). You can take the atlas $\mathcal A' = \{ \psi\}$ on $\mathbb R'$ which easily shows that $\iota'$ is smooth.
However, $\psi$ is certainly not smooth ($\psi'(0)$ does not exist). As a consequence $\iota$ is not smooth. This can be shown by using Proposition 6.8 (ii). Take the atlas $\mathcal A'$ on $\mathbb R'$ and the atlas $\mathcal A = \{id\}$ on $\mathbb R$. The map $$\psi \circ \iota \circ id^{-1} = \psi$$ is not smooth, thus $\iota$ is not smooth.