Moebius Strip Parameterization Moebius strip $S$ is classically parameterized as follows:
$$
\phi(u,v)=\left( \left(2-v \sin \left(\frac{u}{2}\right)\right) \sin (u), \left(2-v \sin \left(\frac{u}{2}\right)\right) \cos (u), v\cos \left(\frac{u}{2}\right)\right)
$$
for $0 < u < 2 \pi$ and $-1<v<1$. I denote by $U$ the domain in $\mathbb{R}^2$ described so far. It is obvious that $\phi_u \wedge \phi_v \neq 0$ for any $(u,v) \in U$. How can I prove that Moebius strip is a regular surface with such a parameterization? I need to prove that $\phi$ is a homemomorphism onto its image and $\phi(U)$ is open in $S$. To this regard, it is simple after some computations to see that $\phi$ is injective. Clearly, $\phi$ is also continuous. Moreover, to prove that $\phi(U)$ is open in $S$, we observe that $\phi(U)$ is the Moebius strip minus the segment $v \to (0,2,v)$. At this point, I need to prove that $\phi^{-1}: \phi(U) \to U$ is continuous. Nevertheless, I cannot express in an easy way the relation of the coordinates $(u,v)$ in terms of $(x,y,z)$. In the comments, I tried to solve this point, but you will see that a cumbersome computation arises. How can I solve this exercise?

Your $\phi$ does not produce the segment $v\mapsto (0,2,v)\in S$. Therefore you need a second (similarly looking) chart $\psi$, valid for $-\pi<u<\pi$, that has this segment at $u=0$. The two charts $\phi$ and $\psi$ together then form an atlas of $S$.
That your $\phi$ is a homeomorphism of the open rectangular domain $R$ in the $(u,v)$-plane onto its image is easily checked by inspection: The formulas show that $\phi$ is continuous and injective, and one can also express $\phi^{-1}:\>\phi(R)\to R$ in terms of elementary functions whose continuity is understood.
