Smooth atlas and $C^\infty$ May be this is a stupid question but in this answer: https://math.stackexchange.com/q/4338398, it is said that $\varphi(x,0)=x^\frac{1}{3}$ is a smooth atlas on $N=\{(x,0)\in \mathbb R^2 | x \in \mathbb R \}$ but, shouldn't $\varphi(x,0)$ be $C^\infty$ to be a smooth atlas?
ps. I couldn't "Add a comment" in that question, that's the reason I'm posting it here (I don't know if this is an appropriate procedure, though)
 A: The smoothness of $\varphi$ as a map from the manifold $N$ to $\mathbb{R}$ depends on what smooth structure you put on $N$.
The $\varphi$ is a homeomorphism from $N$ to $\mathbb{R}$ so you directly define the smooth structure on $N$ to be consisting of the chart $(N,\varphi)$ and hence you get smoothness. That is you are simply declaring that this $\varphi$ is a diffeomorphism and thus giving $N$ the topology and smooth structure induced by this map.
A: Here's another way to put it:
Suppose $N$ is a set (no other assumptions) for which there is a bijection $\phi_0: N \rightarrow \mathbb{R}$. You can now define an atlas $\mathcal{A}$ consisting of all bijections $\phi: N \rightarrow \mathbb{R}$ such that $\phi\circ\phi_0^{-1}: \mathbb{R}\rightarrow\mathbb{R}$ is smooth. In this example, let $N$ be the real line and $\phi_0: N \rightarrow \mathbb{R}$ be $\phi_0(x) = x^{1/3}$. This turns $N$ into a smooth $1$-dimensional manifold.
You now have two different manifold structures on $N$, one using this atlas and one using the standard atlas. The identity map from $N$ as a manifold using the first atlas to $N$ as a manifold using the second atlas is not a diffeomorphism. The two manifold structures are in fact diffeomorphic. It's not too hard, using the atlases, to find a diffeomorphism.
