Restricted function

Question

Let $A=\{(x,y): x,y\in(-1,1)\}$. Is there a function $f:A\mapsto A$ such that

1. $f(x,0)=(x,x^2)$
2. $f$ differentiable and bijective on $A$.

I have tried a lot of constructions but the problem is in the condition of $f$ is bijective. Any advice? Thanks

Answer 1

Make the domain $[-1,1] \times [-1,1]$. define f(x,-1) = (x,-1), and similarly f(x,1) = (x,1). Now for any point below the curve $y=0$, $(x,y) = \alpha (x,-1) + (1-\alpha) (x,0)$ where $\alpha$ is a number between 0 and 1. Now, define $$ f(x,y) = \alpha f(x,-1) + (1 - \alpha) f(x,0) = \alpha(x,-1) + (x,x^2)$$ and for the $(x,y)$ above $y = 0$ define analogously $$f(x,y) = \alpha f(x,1) + (1 - \alpha) f(x,0)$$ Now if you restrict the $f$ to old $A$, the function is bijective !!

This is a bijection. I have not checked the derivative part (may not be true). I gotta run for a class. Will come back later.