Past exam of the options II course of the University of Oxford / 2012 / E / E4 / b

Every iterate $g^n(x,y)=(x+n,(-1{)}^{n}y)$ is an isometry of the strip with its Euclidean subspace metric. Let $a,b$ lie in distinct orbits. Since the first-coordinate difference in $a-g^{n}b$ grows without bound as $|n|\to \infty$, the positive distances $d(a,g^{n}b)$ have a positive minimum $\delta$. Only finitely many indices can give a distance below any fixed bound, and none gives zero.

Choose $0<\epsilon <\delta /3$. Saturate the relative open balls by settingThese are open saturated neighbourhoods of the two orbits. Any two centres $g^{n}a,g^{m}b$ are at least $\delta$ apart, since their distance equals $d(a,g^{m-n}b)$. Their radius-$\epsilon$ balls therefore cannot intersect. Hence $U\cap V=\varnothing$, and part (a)(iv), an instance of positive orbit separation implies a Hausdorff quotient, proves

Let $D=[0,1]\times [-1,1]$ and $q=p\circ i:D\to S/{\mathcal{R}}^{\prime }$. This is continuous. It is onto: given $(x,y)\in S$, choose an integer shift bringing $x$ into $[0,1]$, with the corresponding possible sign change in $y$; the resulting representative lies in $D$.

Two representatives in $D$ are in the same orbit only if their first coordinates differ by an integer. An interior point can therefore be equivalent only to itself. Distinct boundary representatives have first coordinates zero and one, and the iterate $g$ identifies exactly $(0,y)$ with $(1,-y)$. Thus the fibres of $q$ are precisely the equivalence classes defining the Möbius band $M$.

The rectangle $D$ is compact, and the target is Hausdorff by part (i). Applying part (a)(v) to $q$ therefore yieldsThe compactness argument supplies continuity of the inverse without requiring a discontinuous choice of one representative per orbit.