Let $\boldsymbol{\sigma}(u,v)=\big(f(v)\cos u, f(v) \sin u, g(v)\big)$, where $U=[(u,v):0<u<2\pi,~v\in I]$ ($I$ open interval of $\mathbb{R}$) be the parametric surface of revolution by rotating the one-to-one curve $\boldsymbol{\gamma}(v)=(f(v),0,g(v)),~v\in I$ around the $z-$axis (we let $f(v)>0$ for all $v$). Prove that $\boldsymbol{\sigma}^{-1}:\mathrm{Im}\boldsymbol{\sigma}\to U$ is continuous.
Attempt. Let $\boldsymbol{\sigma}^{-1}(x,y,z)=(u,v)$, so we should prove that $u=u(x,y,z)$ and $v=v(x,y,z)$ are continuous on $\mathrm{Im}\boldsymbol{\sigma}$. I have proven that $u(x,y,z)=2\arctan\left(\frac{y}{x+\sqrt{x^2+y^2}}\right)$ while $u\neq \pi $ (i.e. except $(x,0,z)$ for $x<0$) and $u(x,y,z)=2\mathrm{arccot}\left(\frac{y}{-x+\sqrt{x^2+y^2}}\right)$, while $u$ is in a neighborhood of $\pi$, therefore continuous.
On the other hand, for $(x,y,z)\in \mathrm{Im}\boldsymbol{\sigma}$ we have $f(v)=\sqrt{x^2+y^2}$ and $g(v)=z$, so $\boldsymbol{\gamma}(v)=(\sqrt{x^2+y^2},0,z)$ and this is where I got stuck (in order to prove that $v=v(x,y,z)$ is continuous). Of course $\boldsymbol{\gamma}^{-1}$ exists, but should we guarantee that it is continuous?
Thank you for the help.
Let's suppose $g'(v)\neq 0$ for all $v$ in $I$.
Now, let $h(u,v,r)=(rf(v)\cos(u), rf(v)\sin(u), g(v))$ for $r>0$, $0<u<2\pi$ and $v\in I$. One can check this function is injective and that its Jacobian is $rf^2(v)g'(v)\neq 0$ (as $f>0$ and $g'\neq 0$). Thus, $h$ is a diffeomorphism (by the Inverse Function Theorem).
Take any open sets $U\subseteq (0,2\pi), V\subseteq I$.
Then, $$\sigma(U\times V)=h([U\times V\times (0,\infty)]\cap[(0,2\pi)\times I\times \{1\}])=\\h([U\times V\times (0,\infty)])\cap h([(0,2\pi)\times I\times \{1\}])=W\cap \sigma((0,2\pi)\times I),$$
where $W$ is open due to $h$ being a diffeomorphism.
We deduce $\sigma$ is open and thus $\sigma^{-1}$ is continuous.