I can show that the unit square deformation retracts to a line and that line under a particular equivalence relation is homeomorphic to $S^1$.
Let $X = [0,1] \times [0,1]$ and define an equivalence relation $\sim$ on $X$ by $(0,s) \sim (1,1-s)$. Then the Moebius band is defined to be $M := X/\sim$ and we have the canonical quotient map $q: X \to M$ defined by $q(x, y) = [(x, y)]_{\sim}$.
For each $t \in [0,1]$, define the map $r_t: X \to [0,1] \times [0, 1-t]$ by $r_t(x, y) = \left(x, \, y(1-t)\right)$. Then it is clear that $r_t$ is a retraction and we get a deformation retraction of $X$ onto $A := [0,1] \times \{0\}$, $F: X \times [0,1] \to X$ defined by $F(x, t) = r_t(x)$.
I know in general that quotient maps don't preserve deformation retractions (quotient map preserves deformation retraction) but I want to say it does in this case. I attempt to define a deformation retraction from $M$ onto $q(A)$, $G: M \times [0,1] \to M$ by $G(x, t) = (q \circ r_t) (x)$. To show it satisfies the properties of a deformation retraction, I need to show
- For every $[(x,y)] \in M$, $G([x,y], 0) = [x, y]$.
- For every $[(x,y)] \in M$, $G([x,y], 1) \in q(A)$.
- For every $[(a, b)] \in q(A)$, $G([a, b], 1) = [a, b]$.
This issue is that why should $G$ be well-defined?
Best Answer
Your proposed retraction does not descend to the quotient, hence $G$, as it's written is not well-defined. The problem is that it collapses the square to the "bottom" where $y=0$, but in the Möbius band, there is no well-defined "bottom" edge. To wit, at the boundary: $r_t(0, y) = (0, y(1-t))$ and $r_t(1, y) = (1, y(1-t))$, but these points have to agree in the quotient! At $t=1$, $$ r_1(0, y) = (0, 0) $$ and $$ r_1(1, y) = (1, 0) $$ but $(0, 0) \not\sim (1, 0)$.
You can salvage this idea, but you need to shrink the square $[0, 1] \times [0, 1]$ towards the "equator" $[0, 1] \times \bigl\{ \tfrac12 \bigr\}$, since $$ \bigl( 0, \tfrac12 \bigr) \sim \bigl( 1, \tfrac12 \bigr). $$
Can you figure out the appropriate function $r_t$?