I am asked to show that the map $x \to \frac{x}{\sqrt{1 - |x|^2}}$ from an open unit ball to the $R^{2}$ plane is a diffeomorphism.
According to my lecture notes:
"A function $f: M \to N$ is smooth at a point $p \in M$ if there are charts $(U, \alpha)$ for $M$ and $(V, \beta)$ for $N$ with $p \in U$ and $f(p) \in V$ such that $\beta \circ f \circ \alpha^{-1} : \alpha(U) \to \beta(V)$ is smooth at $\alpha(p)$. A function is smooth if it is smooth at all points $p \in M$. A diffeomorphism is a smooth f with a smooth inverse."
What I do not understand is why this exercise does not provide an atlas to select charts from. If a manifold $M$ is an open unit ball and a manifold $N$ is the $R^2$ plane, then what is the set of charts $({U_{x}, \alpha_{x}})$ covering $M$ and what is the set of charts $({V_{x}, \beta_{x}})$ covering $N$?
Because an atlas of an open subset $U$ of $\mathbb{R}^2$ is $U$, since an atlas of a $n$-dimensional manifold is defined by an open covering $(U_i)_{i\in I}$ with homeomorphism $f_i:U_i\rightarrow f(U_i)\subset\mathbb{R}^2$ such that $f_i\circ{f_j}^{-1}$ is a diffeomorhism. If $U$ is an open subset of $\mathbb{R}^2$, you can take the atlas of $U$ which contains only one element, $U$.