Let $U$ be open and $M_{f}:=\{(x,y,z) \in \mathbb R^{3}: x^{2}+y^{2}=f(z)^{2}\}$ and define:
$\phi: U \times ]-\pi,\pi[ \to M_{f}\setminus V$ where $\phi(z, \varphi):=(f(z)\cos{(\varphi)}, f(z)\sin{(\varphi)}, z)$
and $f \in C^{1}(U)$ is strictly positive on $U$ and $V=M_{f}\setminus\phi(U \times ]-\pi,\pi[)$
Show that $\phi$ is a homeomorphism.
My idea:
$\phi$ is by definition bijective and continuously differentiable as $f \in C^{1}(U)$ thus all that is left to be shown is that $\phi^{-1}$ is indeed continuous. And this is where I am stuck. The preimage and image of $\phi$ do not have the same dimension so looking at the invertability of $D \phi$ does not make sense.
And if I take any arbitrary points in $M_{f}\setminus V$, say $(a,b,c)$ then I know that there exists $\varphi\in]-\pi,\pi[$ and $z \in U$so that $(a,b,c)=(f(z)\cos{(\varphi)},f(z)\sin{(\varphi)},z)$ but now how do I get to the preimage?
My attempt:
Since we have an arbitrary $(a,b,c) \in M_{f}\setminus V$, we know that there exists $\varphi\in]-\pi,\pi[$ and $z \in U$so that $(a,b,c)=(f(z)\cos{(\varphi)},f(z)\sin{(\varphi)},z)$.
Define $\phi^{-1}(a,b,c)=(\cos^{-1}({\frac{a}{f(c)})},c)$, note that $f(c) > 0$ from above. and therefore $\phi^{-1}(a,b,c)$ is partially differentiable and thereby continuous. Is this correct?
Any help is greatly appreciated
Note that your use of arccosine can only tell you about points in the $ab$-plane in quadrants I and II. If you switch to arcsine, it can only tell you about points in quadrants I and IV. You might benefit from thinking about why atan2() exists. (The definition given there using the tangent half-angle formula might be of immediate interest.)
Also, while the image of $\phi$ is embedded in $\mathbb{R}^3$, the image is just a 2-manifold, so the preimage and image of $\phi$ are both 2-dimensional.