Explicitly give a collection of deck transformations on \(\{(x,y)\mid -1\leq x\leq 1,\ -\infty<y<\infty\}\) such that the quotient is a Möbius band.
Solution
Let \[ S=[-1,1]\times\RR. \]
1. Define\[ \tau:S\longrightarrow S, \qquad \tau(x,y)=(-x,y+1). \]Then the required collection is the infinite cyclic group\[ G=\langle\tau\rangle =\{\tau^n:n\in\ZZ\}, \qquad \tau^n(x,y)=((-1)^n x,y+n). \]
The displayed formula for \(\tau^n\) follows by induction for \(n\ge0\) and from \[ \tau^{-1}(x,y)=(-x,y-1) \] for negative \(n\). Each \(\tau^n\) is a homeomorphism of \(S\), and \[ \tau^m\circ\tau^n=\tau^{m+n}, \] so these maps form an action of \(\ZZ\) on the strip.
2. The action of \(G\) on \(S\) is free and properly discontinuous.
If \[ \tau^n(x,y)=(x,y), \] then comparison of second coordinates gives \(y+n=y\), hence \(n=0\). Thus no nonidentity element fixes a point.
For proper discontinuity, fix \((x_0,y_0)\in S\) and choose a relative-open neighborhood whose \(y\)-coordinates lie in \[ (y_0-1/4,y_0+1/4). \] Its image under \(\tau^n\) has \(y\)-coordinates in \[ (y_0+n-1/4,y_0+n+1/4), \] which is disjoint from the original interval for every \(n\ne0\). Hence this neighborhood is disjoint from all of its nontrivial translates.
3. The rectangle\[ F=[-1,1]\times[0,1] \]is a fundamental domain, and its only quotient identification between the two horizontal boundary levels is\[ (x,0)\sim(-x,1). \]
Given \((x,y)\in S\), choose \(n\in\ZZ\) so that \(y-n\in[0,1]\). Then \(\tau^{-n}(x,y)\) lies in \(F\), so every orbit meets \(F\). An orbit meeting the interior \([-1,1]\times(0,1)\) meets it only once because every nontrivial \(\tau^n\) changes the second coordinate by a nonzero integer.
The two remaining representatives occur at \(y=0\) and \(y=1\). Since \[ \tau(x,0)=(-x,1), \] the bottom point \((x,0)\) is identified with the top point \((-x,1)\). No interior point of either vertical side \(x=\pm1\) is identified with another point of \(F\); those sides remain boundary.
4. Therefore\[ \boxed{S/G\cong\text{the Möbius band}.} \]
3. The endpoints of the two vertical sides are paired by the horizontal-edge identification, so those sides concatenate to the single boundary circle of the Möbius band.
By 2, the quotient map \[ q:S\longrightarrow S/G \] is a covering map, and every element of \(G\) satisfies \(q\circ\tau^n=q\). Thus each of the maps \[ \tau^n(x,y)=((-1)^n x,y+n), \qquad n\in\ZZ, \] is a deck transformation, and their common quotient is the Möbius band.