- Main Idea: Deformation retract \(M\) onto its center circle; two spaces that deformation retract onto a common space are themselves homotopy equivalent.
Claim: \(S^1 \cross I \homotopic S^1 \cross \theset{*}\) This is because \(I\) is contractible, so \(I \homotopic \theset{*}\). (Maybe needs further proof)
Claim: \(M \homotopic S^1 \cross \theset{*}\).
If both of these claims hold, then we will have \(M \homotopic S^1 \cross I\) as two spaces that deformation retract onto a common space. Identifying \(M = I \cross I / \sim\) where \((x, 0) \sim (1-x, 1)\), fix \(x=1/2\).
Then consider the subspace \(U = \theset{(1/2, y) \mid y \in [0,1]} \subset M\). Claim: \(U \cong \theset{*} \cross S^1\) for some point \(*\).
\(U\) can be written \(\theset{1/2} \cross (I/\sim)\), and since \((1/2, 0) \sim (1/2,1)\), we have \(I/ \sim = I /\bd I \cong S^1\), so \(U \cong \theset{1/2}\cross S^1\) as desired (taking \(* = \frac{1}{2}\)).
However, we can define a homotopy from \(M\) onto \(U\), in the form of a deformation retract.
Let \(F: M \cross I \into M\) be defined by \(F((x,y), t) = F_t(x,y) = ((1-t)x + \frac{1}{2}t, y)\). Then \(F((x,y), 0) = (x,y) = \id_M\), and \(F((x,y), 1) = (\frac{1}{2}, y) \subseteq U\). Moreover, if \((x,y) \in U\), then \((x,y) = (\frac{1}{2}, y)\) and \(F((x,y), t) = ((1-t)\frac{1}{2} + \frac{1}{2}t, y) = (\frac{1}{2} - t\frac{1}{2} + \frac{1}{2}t, y) = (\frac{1}{2}, y) = (x,y)\), so \(F = \id_U\). This makes \(F\) a deformation retract from \(M\) onto \(U\), and so \(M \homotopic U\).
But then, summarizing our results, we have \(S^1 \cross I \homotopic S^1 \cross \theset{*} \cong S^1 \cross \theset{\frac{1}{2}} = U \homotopic M\), and so \(S^1 \cross I \homotopic M\) as desired.
Solution
1. Model the Möbius band as\[M=[0,1]\times[-1,1]/(0,t)\sim(1,-t).\]
This is the standard quotient model.
2. The homotopy\[H_s([x,t])=[x,(1-s)t]\]is a strong deformation retraction of \(M\) onto its center circle \(C=\{[x,0]\}\).
It is compatible with the quotient relation because \((0,t)\sim(1,-t)\) implies \((0,(1-s)t)\sim(1,-(1-s)t)\). At \(s=1\) every point lies on \(C\), and points of \(C\) are fixed throughout.
3. The cylinder \(S^1\times I\) also deformation retracts onto \(S^1\times\{0\}\).
Use \((z,t,s)\mapsto(z,(1-s)t)\).
4. Since both spaces deformation retract onto a circle,\[\boxed{M\simeq S^1\times I.}\]
Each is homotopy equivalent to \(S^1\).