Let $a,b\in\mathbb{N}$, can you find (and prove that it is what it is) an explicit diffeomorphism between $$M_{a,b}:=\{(x,y)\in\mathbb R^2:x^{2a}+y^{2b}=1\}$$ and the usual circle $S:=\{(x,y)\in\mathbb R^2:x^2+y^2=1\}$?
Of course, these manifolds are equipped with the unique smooth structure that derives from seeing them as regular levels sets of functions from $\mathbb R^2$ to $\mathbb R$.
This is a question inspired by Problem 5-1 of Lee's "Introduction to Smooth Manifolds".
Best Answer
Denote by $C_{a,b}$ the curve $x^{2a} + y^{2b} = 1$. Let's show that the central projection map from $C_{a,b}$ to another curve $C_{c,d}$ is smooth. Indeed, $(x,y) \in R^{2} \backslash \{(0,0)\}$ maps to $r(x,y)\cdot (x,y)$, where $r = r(x,y)$ is such that $$r^{2c} x^{2c} + r^{2 d} y^{2d} = 1$$ Now, for given $u$, $v$ positive, not both $0$, there exists a unique $r>0$ such that $$r^{2c} u + r^{2d} v = 1$$ and $r$ depends smoothly on $(u,v)$ (implicit function theorem).