Consider the paraboloid $x^2+y^2=z$. I need to show that this paraboloid is diffeomorphic to a plane. Let $P$ be this paraboloid and $S$ be the plane that I will use.
So, by definition, I need to show that a function $\phi:P\rightarrow S$ satisfies the following. Let $\alpha:U_{1}\subset\mathbb{R}^{2}\rightarrow P$ a local parametrization of $P$, with $p\in\alpha (U_1);\beta:U_{2}\rightarrow S$ a local parametrization of $S$ with $\phi(\alpha(U_{1}))\subset\beta(U_{2})$. We have $\beta^{-1}\, \circ\,\phi\,\circ\,\alpha:U_{1}\rightarrow U_{2}$ differentiable at $q=\alpha^{-1}(p),$ and $\phi$ has a differentiable inverse $\phi^{-1}:S\rightarrow P.$
I take the parametrization of $P$ given by
$$\alpha(u,v)=(v\cos(u),v\sin(u), v^2)\quad u\in[0,2\pi),v\in\mathbb{R}.$$
For $S$, I think that I can take an arbitrary plane to be diffeomorphic to $P$, so, consider the plane $xy$, that is, the set of points with form $(x,y,0), x,y\in\mathbb{R}$. Take the parametrization of $S$ given by
$$\beta(u,v)=(u,v,0),\quad u,v\in\mathbb{R}$$
So, I can define a $\phi:S\rightarrow P$ by $\phi(u,v,0)=(v\cos(u),v\sin(u),v^2)$.
Is this $\phi$ a diffeomorphism?
Best Answer
I think it is easier to take the parametrization $(x,y,x^2+y^2)$, and the chart itself on $P$ will be the diffeomorphism you are looking for.