Let $f:S^2\to \mathbb{R}$ be a positive differentiable function on the unit sphere.Show that $S(f)=\{f(p)p \in \mathbb{R}^3:p \in S^2\}$ is a regular surface and that $\phi:S^2 \to S(f)$ given by $\phi(p)=f(p)p$ is a diffeomorphism.
It's routine to prove that if $\mathbb{x}:U \in \mathbb{R}^2 \to S^2$ is a parametrization, then $f(\mathbb{x}(u,v))\mathbb{x}(u,v):U \to S(f)$ is a parametrization of $S(f)$.If we denote $g=f(\mathbb{x}(u,v))\mathbb{x}(u,v)$, clearly g is differentiable and one-to-one,but things become untrivial when it comes to prove that $dg_{p}$ is one-to-one.
So I am stuck in how to prove that $dg_{p}$ is one-to-one.Could someone give me some advice or proof on it? Any response would be appreciated.
Without loss of generality, we set the parametrization $\mathbb{x}(u,v)=(sinv cosu,sinvsinu,cosv)$, where $U=\{(u,v) \in \mathbb{R}^2:0<u<2\pi,0<v<\pi \}$. If we denote $f(u,v)=f(\mathbb{x(u,v)})$,then $dg_{p}$ would be $$ \left[ \begin{array}{cc|c} f_{u}\sin v\cos u-f(u,v)\sin v\sin u & f_{v}\sin v\cos u+f(u,v)\cos v\cos u\\ f_{u}\sin v\sin u+f(u,v)\sin v\cos u & f_{v}\sin v\sin u+f(u,v)\cos v\sin u\\ f_{u}\cos v & f_{v}\cos v-f(u,v)\sin v\\ \end{array} \right] $$
The derivatives of $g$ are $$g_u=f_ux+fx_u$$ and $$g_v=f_vx+fx_v.$$ It is easy to see that these two vectors are linearly independent for all $(u,v)$, since for all $(u,v)$ the three vectors $x$, $x_u$ and $x_v$ are linearly independent and $f$ does not vanish.