Show that the Möbius band has its central circle $C$ as a deformation retract

Question

I have started this problem by using the planar representation of the Möbius band and noted that a line down the middle is probably what is meant by the central circle, since travelling from top to bottom (where the left and right sides are identified in opposite directions) appears to trace out a circle.

The specifics of the deformation retraction are lost on me, I think I need a homotopy $f:M \times I \to S^1$ such that for all $m\in M$ and $x\in S^1$ we have $f(m,0)= m$, $ f(m,1)\in S^1$, $ f(x, 1) = x$.

Is there a parameterisation of the Möbius band that makes defining such a function explicitly obvious? Intuitively I can imagine pulling the sides of the Möbius band in continuously to the circle, is the idea of straight line homotopy useful here?

Answer 1

Yes, just “pulling the sides in” works and it’s not hard to describe explicitly.

Parameterize the planar representation $P$ of the band below, where $-\pi\le x \le \pi$ and $-1\le y\le1$, and where $e$ (which you need not give explicitly) is any embedding of $P$ into $\mathbb R^3$ that’s one-to-one except two-to-one on the left and right sides ($e(-\pi,y)=e(\pi,y)$). In other words, $M=e(P)$. Note that for $X$ equal to the $x$-axis, $e(X)$ is a circle $S^1$.

For a point $m\in M$, where $m=e(x,y)$, define $f(m,i)=e(x,y(1-i))$. You can check that this satisfies the requirements.

Answer 2

A deformation retract of a space $X$ to a subspace $A$ is a continuous map $F : X \times [0, 1] \to X$ such that $F(x, 0) = x$, $F(x, 1) \in A$ and $F(a, t) = a$ for all $a \in A$, $x \in X$ and $t \in [0, 1]$.

To prove that the moebius strip deformation retracts onto the center circle, consider the fundamental square $[0, 1] \times [0, 1]$ with the two sides $\{0\} \times [0, 1]$ and $\{1\} \times [0, 1]$ identified oppositely via identification $(0, t) \sim (1, 1-t)$.

Construct a deformation retract of this square onto the center interval $[0, 1] \times \{1/2\}$ via the map $F((x, y), t) = (x, t/2 + (1-t)y)$. Next, doing the identifications on the deformation retracted $[0, 1] \times \{1/2\}$ gives the center circle $[0, 1] \times \{1/2\}/(0, 1/2) \sim (1, 1/2) \cong S^1$.