Consider the Möbius strip defined by the following equivalence relation on the subspace $[0,1]\times]-a,a[$ of $\mathbb{R}^2$:
$$(x,y)\sim (x',y')\implies (x,y)=(x',y')\vee|x'-x|=1\:\text{and}\:y'=-y$$
So the Möbius strip would be $M=[0,1]\times]-a,a[\setminus\sim$ be the quotient space with the following quotient map:
$$p:[0,1]\times]-a,a[\to M$$
Let$V=p(([0,1]\setminus\{\frac{1}{2}\})\times]-a,a[)$ and $\psi:V\to\psi(V)=]\frac{1}{2},\frac{3}{2}[\times]-a,a[$ be given as follows:
$$\psi([(x,y)])=\begin{cases}(x+1,-y)\:\:\text{if}\:\: x<\frac{1}{2}\\(x,y)\:\:\:\:\:\:\:\:\:\:\:\:\text{if}\:\:\:x>\frac{1}{2}\end{cases}$$
Prove that $\psi$ is well defined and is a homeomorphism.
I am having trouble proving $\psi$ to be continuous and open.
I noticed that if $p\sim p'$ then $\psi(p)=\psi(p')$ so the function is well defined and bijective.
If I build the following translation:
$$f:\psi(V)\to [0,1]\times]-a,a[\\(x,y)\to(x-\frac{1}{2},y)$$
Then I can build the following function:
$$p\circ f(\psi(V)):]\frac{1}{2},\frac{3}{2}[\times]-a,a[\to M$$
I think this last construction could prove continuity if I could assure $\psi$ to be open. However I am not seeing how to define open sets in M.
Question:
How should I solve this problem?
Thanks in advance!
I’m going to try first to give a slightly different approach that may make it a bit clearer exactly what is going on here.
Let $A=(-a,a)$ and $M_0=[0,1]\times A$. The equivalence relation $\sim$ identifies $\{0\}\times A$, the left edge of $M_0$, with $\{1\}\times A$, the right edge of $M_0$, after turning it upside down. It’s that turning upside down, by identifying $\langle 0,y\rangle$ with $\langle 1,-y\rangle$, that corresponds to the half twist in a physical Möbius strip.
Let $U_0=\left[0,\frac12\right)\times A$, $U_1=\left(\frac12,1\right]\times A$, and $U=U_0\cup U_1$; $U$ is what is left of $M_0$ after we cut out the segment $\left\{\frac12\right\}\times A$. Now flip $U_0$ around the $x$-axis (or, if you prefer, reflect it in the $x$-axis) by the map
$$r:U_0\to U_0:\langle x,y\rangle\mapsto\langle x,-y\rangle\;,$$
and then translate the result one unit to the right by the map
$$t:U_0\to\Bbb R^2:\langle x,y\rangle\mapsto\langle x+a,y\rangle\;.$$
Let $U'=U_1\cup(t\circ r)[U_0]$, and define
$$f:U\to U':u\mapsto\begin{cases} (t\circ r)(u),&\text{if }u\in U_0\\ u,&\text{if }u\in U_1\;. \end{cases}$$
Note that any point $\langle 0,y\rangle$ on the left edge of $U_0$ is sent by $t\circ r$ to $\langle 1,-y\rangle$, precisely the point with which it is identified by $\sim$, so we can think of this point as ‘being’ $[\langle 0,y\rangle]=[\langle 1,-y\rangle]$. With this identification in mind it’s not too hard to see that $\psi\circ(p\upharpoonright U)=f$; this may make it a little easier to see what $\psi$ does.
In very informal physical terms, imagine that you’ve actually taken the strip $M_0$, glued it to the table along the strip $[0.49,0.51]\times A$, bent the righthand end up into a half cylinder, so that the edge $\{1\}\times A$ lies directly above the line $x=\frac12$, and then bent the lefthand end up similarly, given it a half twist, and identified the two ends to get a physical Möbius strip. Then $\psi$ amounts to performing the following physical steps:
This is basically my $U'$ and the original $\psi[V]$.
The only points at which there is any real difficulty showing continuity of $\psi$ (or $t\circ r$) are the points where $\sim$ identifies two distinct points of $M_0$. If $\epsilon>0$ is sufficiently small to avoid running into the upper and lower edges of $U_0=\psi[V]$, $(t\circ r)^{-1}[B(\langle 1,y\rangle,\epsilon)]$ is the union of $B(\langle 0,-y\rangle,\epsilon)\cap U_0$ and $B(\langle 1,y\rangle,\epsilon)\cap U_1$, a set that is open in $U$.