How to find $4$-tuple parametric representation of Möbius strip? We know that a Möbius strip can be constructed as follow:
$$M\cong \frac{\Bbb S^1\times \Bbb S^1}{\Bbb Z_2},$$
where $\Bbb Z_2$ act by permutation. So I think that it is possible  represent each point of $M$ by $4$-tuple $(x_1,y_1,x_2,y_2)$ (see this post for $3$-tuple case). 

How to find this parametric representation of Möbius strip?

 A: We have two kinds of Möbius strips.


*

*Let $[0,1]\times[0,1]=\{(s,t)\mid 0\leq s,t\leq 1\}$. Then the quotient space $([0,1]\times[0,1])/\sim$ is a Möbius strip, where $(s,0)\sim(1-s,1)$.

*Let $S^1\times S^1=\{(e^{2\pi is},e^{2\pi it}) \mid 0\leq s,t\leq 1\}$ be a torus. Then the quotient space $(S^1\times S^1)/\sim$ is a Möbius strip, where $(e^{2\pi is},e^{2\pi it})\sim(e^{2\pi it},e^{2\pi is})$.
[Reference] The accepted answer of "Orbit space of torus homeomorphic to mobius strip"
We want to find a parametrization $f\colon [0,1]\times[0,1]\to(S^1\times S^1)/\sim$ with two variables $(s,t)\in[0,1]\times[0,1]$.


At first define $f_1\colon [0,1-t]\times[0,1]\to S^1\times S^1$ and $f_2\colon [1-t,1]\times[0,1]\to S^1\times S^1$ by
  $$
\begin{cases}
f_1(s,t)=\left( e^{\pi i(1-t+s)}, e^{\pi i(1-t-s)} \right), & \text{$0\leq s\leq 1-t$} \\
f_2(s,t)=\left( e^{\pi i(3-t-s)}, e^{\pi i(1-t+s)} \right), & \text{$1-t\leq s\leq1$}
\end{cases} \tag{*}
$$

Then $f_1(1-t,t)=(e^{2\pi i(1-t)},1)\sim(1,e^{2\pi i(1-t)})=f_2(1-t,t)$. Attaching $f_1$ and $f_2$ along the line $s=1-t$ we have a continuous surjection
$$
f\colon [0,1]\times[0,1] \to (S^1\times S^1)/\sim
$$
Moreover, $f$ induces a homeomorphism between the two Möbius strips 
$$
\tilde f\colon ([0,1]\times[0,1])/\sim \, \to (S^1\times S^1)/\sim
$$
since for all $0\leq s\leq1$,
$$
f(s,0) = f_1(s,0) = (e^{\pi i(1+s)},e^{\pi i(1-s)}) = f_2(1-s,1) = f(1-s,1)
$$

We can rewrite the parametrization (*) with 4-tuples as follows:
  $$
\begin{align*}
f(s,t)=f_1(s,t) = \bigl( &\cos\bigl[\pi(1-t+s)\bigr], \sin\bigl[\pi(1-t+s)\bigr], \\
&\cos\bigl[\pi(1-t-s)\bigr], \sin\bigl[\pi(1-t-s)\bigr] \bigr)
\end{align*}
$$
  for $0\leq s\leq 1-t$, and
  $$
\begin{align*}
f(s,t)=f_2(s,t) = \bigl( &\cos\bigl[\pi(1-t-s)\bigr], \sin\bigl[\pi(1-t-s)\bigr], \\
 &\cos\bigl[\pi(1-t+s)\bigr], \sin\bigl[\pi(1-t+s)\bigr] \bigr)
\end{align*}
$$
  for $1-t\leq s\leq 1$.

Note that $f_1(s,t)$ is the same as $f_2(s,t)$ in the quotient space $(S^1\times S^1)/\sim$.
