Define a differentiable bijection How do I define;


*

*a bijection which is also differentiable from $\{ (x,y) | x^2 + y^2 < 1 \}$ to $\mathbb R^2 $.


and


*

*a bijection which is also differentiable from a bounded non-convex subset of $\mathbb R^2$ to $\mathbb R^2 $.


Would I use continuity on the first part?
Would $ f(x,y)=(2x+3y,x+2y) $ be a viable answer for the second one? 
Any help or answers greatly appreciated. 
 A: Try:
$f(x,y)=\left(x\cdot\tan\left(\frac{\pi}{2}\sqrt{x^2+y^2}\right),y\cdot\tan\left(\frac{\pi}{2}\sqrt{x^2+y^2}\right)\right)$
For the second part, try a bijective map of the 2/3 of the disk (non-convex and bounded) into the full disk with a function $g$:

by preserving the magnitude, and multiplying the angle by 3/2.
Then $g\circ f$ will be a bijections from the 2/3 of the disk into $\mathbb R^2$
In polar coordinates:
Answer to the first question:
$f(r\cos\theta,r\sin\theta)=\left(\tan\left(\frac{\pi r}{2}\right)\cos\theta,\tan\left(\frac{\pi r}{2}\right)\sin\theta\right)$ with $0\le r<1$ and $0\le\theta<2\pi$
Mapping from sector to unit ball:
$g(r\cos\theta,r\sin\theta)=\left(r\cos\left(\frac{3\theta}{2}\right), r\cos\left(\frac{3\theta}{2}\right)\right)$ with $0\le r<1$ and $0\le\theta<\frac{4\pi}{3}$
Answer to the second question $h=f\circ g$:
$h(r\cos\theta,r\sin\theta)=\left(\tan\left(\frac{\pi r}{2}\right)\cos\left(\frac{3\theta}{2}\right), \tan\left(\frac{\pi r}{2}\right)\cos\left(\frac{3\theta}{2}\right)\right)$  with $0\le r<1$ and $0\le\theta<\frac{4\pi}{3}$
