On a representation of the Moebius strip Let $f:\, (x,y,z)\in C\mapsto ((x^2-y^2)(2+xz),2xy(2+xz),yz)\in\mathbb{R}^3$. I have to prove that the image of this function is homeomorphic to the Moebius strip. I would like to use the universal property of quotient spacea, but I need to show that $f(x,y,z)=f(x',y',z',)$ if and only if $(x,y,z)=\pm(x',y',z')$. Is there any simple way to prove this? (I got lost in too many calculations).
EDIT
The domain of the map is the cylinder $C=\{(x,y,z):x^2+y^2=1,|z|\leq1\}$.
Definition of Moebius strip: the quotient space $C/\sim$ where $P\sim Q\iff P=\pm Q$.
 A: Write $C = S^1 \times [-1,1]$ with $S^1 = \{ w \in \mathbb C \mid \lvert w \rvert = 1\}$ . Then
$$f(w,z) = ((2+\text{Re}(w)z) w^2,\text{Im}(w)z) \in \mathbb C \times \mathbb R .$$
Therefore it is obvious that $(w,z) = - (w',z')$ implies $f(w,z) = f(w',z')$.
We have $\lvert \text{Re}(w) \lvert \le 1, \lvert z \rvert \le 1$, thus $\lvert \text{Re}(w)z \rvert \le 1$ and therefore $2+\text{Re}(w)z \ge 1 > 0$.
If $f(w,z) = f(w',z')$, then


*

*$(2+\text{Re}(w)z) w^2 = (2+\text{Re}(w')z') (w')^2$

*$\text{Im}(w)z = \text{Im}(w')z'$.


Taking the absolute value on both sides of the first equation, we get $2+\text{Re}(w)z = 2+\text{Re}(w')z'$, i.e. $\text{Re}(w)z = \text{Re}(w')z'$. Add this and $i$ times the second equation and get


*$wz = w'z'$.


Taking the absolute value on both sides of this equation gives $\lvert z \rvert =  \lvert z' \rvert$. Thus $z  = \pm z'$. If $z \ne 0$, we immediately see from 3. that  $(w,z) = \pm (w',z')$. If $z = 0$, the first equation shows that $w^2 = (w')^2$, i.e. $w =  \pm w'$. Thus again $(w,z) = \pm (w',z')$.
