Parametric Equation of The Moebius Strip

Question

One of the classical parameterizations of the Moebius strip is the following (see DoCarmo) $$ \psi(u,v)=\left( \left(2-v \sin \left(\frac{u}{2}\right)\right) \sin (u), \left(2-v \sin \left(\frac{u}{2}\right)\right) \cos (u), v\cos \left(\frac{u}{2}\right)\right) $$ with $0 < u < 2 \pi$ and $-1<v<1$. The map $\psi$ is clearly continuous and it is injective. To prove that $\psi$ is a coordinate patch on the Moebius strip, I must show that $\psi$ is a homeomorphism onto its image. To this regard, I want to show that $\psi^{-1}$ is continuous. Nevertheless, I can't find such a map. After denoting by $x,y,z$ the three coordinates of $\psi(u,v)$, I have to express $u$ and $v$ as continuous functions of $x,y,z$. How can I do it?

Answer 1

First, note that $ 2-v\sin\frac{u}2\geq 2-\sin\frac{u}2\geq 1 $ Hence $\arg(y+ix)=\arg(\cos u+i\sin u)=u$up to the usual multiples of $2\pi$ away from a branch cut ray 0 to $\infty$.

Having determined $u$, either

• $\cos\frac{u}2=0$, in which case in a neighbourhood we will have $\sin\frac{u}2\neq 0$ so we use $\sqrt{x^2+y^2}=2-v\sin\frac{u}{2}$ to determine $v=(2-\sqrt{x^2+y^2})\csc\frac{u}2$; or

• $\cos\frac{u}2\neq 0$, in which case we use $z=v\cos\frac{u}2$ to determine $v=z\sec\frac{u}{2}$.

Now check they are actually smooth.