Let \(X=[0,1]\times[0,1]\) denote the rectangle in \(\mathbb R^2\). Let \(\sim\) be the equivalence relation generated by \((0,p)\sim(1,1-p)\) where \(0\le p\le1\). The quotient space \(X/\sim\) is called the Möbius band. Show that \(S^1\) is a retract of the Möbius band.
Solution
Let \[ q:X=[0,1]\times[0,1]\longrightarrow M=X/\sim \] be the quotient map, where \(M\) is the Möbius band, and set \[ C=q\bigl([0,1]\times\{1/2\}\bigr)\subseteq M. \]
1. The subspace \(C\) is homeomorphic to \(S^1\).
On the horizontal midline, the only nontrivial identification is \[ (0,1/2)\sim(1,1/2), \] because the boundary rule gives \[ (0,p)\sim(1,1-p) \] and \(1-p=1/2\) precisely when \(p=1/2\). Thus \(C\) is the quotient of the interval \([0,1]\) obtained by identifying its two endpoints.
Define \[ \phi:[0,1]\longrightarrow S^1, \qquad \phi(s)=e^{2\pi i s}. \] This map is continuous and identifies exactly \(0\) and \(1\). Hence it factors through a continuous bijection \[ \bar\phi:C\longrightarrow S^1. \] The space \(C\) is compact as the continuous image under \(q\) of the compact interval \([0,1]\times\{1/2\}\), and \(S^1\) is Hausdorff. Therefore the continuous bijection \(\bar\phi\) is a homeomorphism.
2. The map\[ R:X\longrightarrow C, \qquad R(s,t)=q(s,1/2), \]is constant on every equivalence class of \(\sim\).
Away from the two vertical boundary edges, equivalence classes are singletons, so there is nothing to check. For a generating boundary identification \[ (0,p)\sim(1,1-p), \] we have \[ R(0,p)=q(0,1/2) \] and \[ R(1,1-p)=q(1,1/2). \] But \[ (0,1/2)\sim(1,1/2), \] so these two values are equal in \(C\). Thus \(R\) is constant on equivalence classes.
3. The map \(R\) descends to a continuous map\[ r:M\longrightarrow C \]satisfying\[ r\circ q=R. \]
The map \[ (s,t)\longmapsto(s,1/2) \] from \(X\) to \(X\) is continuous, so its composite with \(q\) is continuous and has image in \(C\). Hence \(R:X\to C\) is continuous. By 2 it is constant on the fibers of the quotient map \(q\). The universal property of the quotient topology therefore gives a unique continuous map \[ r:M\to C \] with \(r\circ q=R\).
4. The map \(r\) restricts to the identity on \(C\).
Every point of \(C\) has the form \[ c=q(s,1/2) \] for some \(s\in[0,1]\). Then \[ r(c) =r(q(s,1/2)) =R(s,1/2) =q(s,1/2) =c. \] Thus \[ r|_C=\operatorname{id}_C. \]
5. Therefore \(S^1\) is a retract of the Möbius band.
By 1, \(C\cong S^1\). By <1>3–<1>4, the inclusion \[ C\hookrightarrow M \] has the left inverse \(r:M\to C\). Hence \(C\) is a retract of \(M\), so the Möbius band contains a retract homeomorphic to \(S^1\).