Möbius band parameterizaton: Showing injective So, I'm trying to show that the parameteization function from $\mathbb R^2$ to $\mathbb R^3$ given in the wikipedia page http://en.wikipedia.org/wiki/Mobius_band#Geometry_and_topology is injective on the interior of the domain.   (this is part of a larger exercise I'm doing in a general topology class showing it is homeomorphic to an identification space of the unit square).  Trying to do it by brute force gets me 3 equations with 4 unknowns.  Is there a multivariable analysis way to do this?  I know the implicit function theorem/inverse function theorem can give locally 1-1,  but I need 1-1 on the entire open unit square.
 A: The question is about proving that the mapping $f:(u,v)\mapsto (x,y,z)$ where
$$
\begin{aligned}
x(u,v)&= \left(1+\frac{v}{2} \cos \frac{u}{2}\right)\cos u,\\
y(u,v)&= \left(1+\frac{v}{2} \cos\frac{u}{2}\right)\sin u,\\
z(u,v)&= \frac{v}{2}\sin \frac{u}{2} 
\end{aligned}
$$
is injective on the rectangle $(u,v)\in(0,2\pi)\times[-1,1]$.
Assume that $f(u_1,v_1)=(x,y,z)=f(u_2,v_2)$.
Here always
$$
\sqrt{x^2+y^2}=\left(1+\frac{v}{2} \cos \frac{u}{2}\right)\in[\frac12,\frac32].
$$
So the assumption implies that
$$
\cos u_1=\frac{x}{\sqrt{x^2+y^2}}=\cos u_2
$$
and similarly that
$$
\sin u_1=\frac{y}{\sqrt{x^2+y^2}}=\sin u_2.
$$
But the point $(\cos u,\sin u)$ on the unit circle determines the angle $u$
up to an integer multiple of $2\pi$. Therefore we can conclude that $u_1=u_2=u$.
Furthermore, if $\cos(u/2)\neq0$ then from
$$
\left(1+\frac{v_1}{2} \cos \frac{u}{2}\right)=\sqrt{x^2+y^2}=\left(1+\frac{v_2}{2} \cos \frac{u}{2}\right)
$$
we get $v_1=v_2$ by cancelling the $1$s and dividing by $\cos(u/2)$. On the other hand, if $\cos(u/2)=0$ then $\sin(u/2)\neq0$, and we reach the same conclusion from
$$
\frac{v_1}{2}\sin \frac{u}{2}=z=\frac{v_2}{2}\sin \frac{u}{2}.
$$
The claimed injectivity follows from this.
