can $(x^3 , y)$ be a chart if $(x , y)$ is a chart on $M$?

Question

Let $M$ be a $2$-dimensional manifold and $(U , \varphi)$ be a chart on $M$ with $\varphi = (x , y)$. Can $(U , \psi)$, with $\psi = (x^3 , y)$, be a chart on $M$? I thought no because if it were a chart, the map $\psi = g \circ \varphi : U \to {\mathbb{R}}^2$ should be a diffeomorphism, with $g : {\mathbb{R}}^2 \to {\mathbb{R}}^2$ given by $g(a , b) = (a^3 , b)$. But it implies that $g$ is also a diffeomorphism, but the first partial derivative of $g^{- 1} : (a , b) \mapsto (a^{\frac13} , b)$ does not exist.

My argument should be wrong, as on can be read on $\textbf{Lee's Smooth Manifold}$ page 65, that $(\tilde{x},\tilde{y})$ in $\mathbb{R}^2$ related by $$ \tilde{x} = x, \qquad \tilde{y} = y+x^3 $$ is a chart if $(x , y)$ denotes the standard coordinates on ${\mathbb{R}}^2$. Where is my mistake?

Answer 1

These things do not contradict each other. In your example, the transition maps for $\phi$ and $\psi$ are $g$ and $g^{-1}$, and $g^{-1}$ is not smooth, as you observed.

But there is no such problem for the other example. The forward transition map is $g: (x,y)\to (x, y+x^3)$ and the inverse is $g^{-1}:(\tilde{x}, \tilde{y})\to (\tilde{x}, \tilde{y}-\tilde{x}^3)$. Both are perfectly nice smooth maps.