Constructing a diffeomorphism which maps $\Gamma$ to a parabola

99 Views Asked by At

$f : \mathbb{R} \to \mathbb{R}$ be a differentiable function and $$\Gamma = \left \{ (x,y) \in \mathbb{R}^2 | y=f(x) \right \} \subset \mathbb{R}^2$$ its function graph.

a.) Construct a diffeomorphism $\varphi : \mathbb{R}^2 \to \mathbb{R}^2$, which maps $\Gamma$ to the parabola with the equation $y = x^2$.

b.) Show that there is no diffeomorphism $\varphi: \mathbb{R}^2 \to \mathbb{R^2}$ which maps $\Gamma$ on the set with the equation $y = \mid x \mid$

I know from the teacher that the solution for a.) is $\varphi(x,y) \to (x, y+x^2)$. Unfortunately, I don't see how he got to this solution. Can someone explain it to me? I can not imagine how a diffeomorphism can be mapped on a parabola.

And for b.) how do I show that there is no diffeomorphism? Do I have to show that it is not differentiable everywhere?

Thank you very much for your time!