Obtaining the Möbius strip as a quotient of $S^1\times[-1,1]$ I am trying to obtain the Möbius strip as a quotient of $S^1\times[-1,1]$, where $S^1$ is, of course, the circle.
My definition of Möbius strip is the quotient of the square $[0,1]\times[0,1]$ by the equivalence relation $(x,0)\sim(1-x,1)$.
I am searching for an identification $f$ between $S^1\times[-1,1]$ and $[0,1]\times[0,1]$ and then I want to prove that $\sim_{p'\circ f}=\sim_p$ with $p(t)=[t]$ by the antipodal relation and $p'(t)=[t]$ by the Möbius strip's relation.
My problem is I can't find $f$.
Could I do it the other way, that is, find an identification between $[0,1]\times[0,1]$ and $S^1\times[-1,1]$?
 A: We can construct the quotient directly by identifying 
$$(z,-1)\sim \left(\frac{1}{z},1\right)$$
We can see that this works correctly by noting that if we map $[0,1]\times [0,1]\to \mathbb{S}^1\times [-1,1]$ via the function 
$$(x,y)\mapsto (e^{2\pi i (x-\frac{1}{2})},2(y-\frac{1}{2}))$$
then the identification $(x,0)\sim (1-x,1)$ is translated to 
$$(e^{2\pi i (x-\frac{1}{2})},-1)\sim (e^{2\pi i (\frac{1}{2}-x)},1)$$
and this coincides with the definition $(z,-1)\sim \left(\frac{1}{z},1\right)$.
EDIT: Note that there is no way to obtain a Mobius strip from $\mathbb{S}^1\times [-1,1]$ by identifying opposite edges. The space constructed here is homeomorphic to the Klein bottle.
We obtain the Mobius strip by identifying $(z,x)\sim (-z,-x)$. To see this, map $[-1,1]\times [-1,1]\to \mathbb{S}^1\times [-1,1]$ by sending $(x,y)\mapsto (e^{\pi i x},y)$. When we identify $(z,x)\sim (-z,-x)$, we are identifying $(e^{\pi i x},y)\sim (e^{\pi i (x+1)},-y)$. Thus if $x<0$ we have $(x,y)\sim (x+1,-y)$. Map the quotient space to the Mobius strip, considered as the quotient of $[0,1]\times [0,1]$ by identifying $(0,y)\sim (1,1-y)$, by sending $[(x,y)]\mapsto [(x+1,\frac{1}{2}(1+y))]$ for $x\leq 0$. Verify that this is a homeomorphism between the quotient space and the Mobius strip.
