Diffeomorphism from disk to plane I want to show that the disk $D = \{(x,y) \in \mathbb R^2 : x^2+y^2 < 1\}$, the open square $K = (-1, 1)^2$ and the whole plane $\mathbb R^2$ are all diffeomorphic to each other. Therefore I want to consider two functions, namely
$$f \colon D \to \mathbb R^2, \quad f(x,y) = \left(\frac{x^2+y^2}{1-x^2-y^2} \cdot x, \frac{x^2+y^2}{1-x^2-y^2} \cdot y \right)$$
and
$$g \colon K \to \mathbb R^2 = \left(\tan \frac{\pi x}{2}, \tan \frac{\pi y}{2}\right).$$
After calculating partial derivatives I know that both are differentiable (just like their inverses), but how to prove that they are even $C^\infty$ (e.g. infinitely often differentiable)?
 A: The first map $f$ above has a problem: $Df(0,0)$ is the $0$ transformation. So it can't be a diffeomorphism.
Let's try $f(re^{it}) = \tan (\pi r/2)e^{it}.$ Then the inverse mapping is $f^{-1}(re^{it}) = [(2\arctan r)/\pi]e^{it}.$ Because $r=(x^2+y^2)^{1/2}$ is $C^\infty$ away from $0,$  as is $e^{it},$ these functions are $C^\infty$ away from $0.$ Near $0,$ they take the form $$re^{it} \to (a_1 + a_2r^2 + a_4r^4 +\cdots)\cdot re^{it},$$ where $a_1 \ne 0.$ That's the same as
$$(x,y) \to (a_1 + a_2(x^2+y^2) + a_4(x^2+y^2)^2 +\cdots)\cdot(x,y)$$
which is clearly $C^\infty$ near $0.$
Now relatively painless here is verifying $Df$ is nonsingular at each point in its domain. Near $0,$ we have $f(x,y) = a_1\cdot(x,y) +O(x^2+y^2).$ So $Df(0,0) = a_1I.$ Away from $0,$ you'll get
$$\partial f/\partial r = (\pi /2)\sec^2 (\pi r/2), \,\,\, \partial f/\partial t = \tan (\pi r/2)(ie^{it}).$$ Never mind the actual values, these are linearly independent vectors, in fact mutually perpendicular. This shows $Df$ is never singlular as desired.
