6
$\begingroup$

Prove that there is no retraction (i.e. continuous function constant on the codomain) $r: M \rightarrow S^1 = \partial M$ where $M$ is the Möbius strip.

I've tried to find a contradiction using $r_*$ homomorphism between the fundamental groups, but they are both $\mathbb{Z}$ and nothing seems to go wrong...

3 Answers 3

10

If $\alpha\in\pi_1(\partial M)$ is a generator, its image $i_*(\alpha)\in\pi_1(M)$ under the inclusion $i:\partial M\to M$ is the square of an element of $\pi_1(M)$, so that if $r:M\to\partial M$ is a retraction, $\alpha=r_*i_*(\alpha)$ is also the square of an element of $\pi_1(\partial M)$. This is not so.

(For all this to work, one has to pick a basepoint $x_0\in\partial M$ and use it to compute both $\pi_1(M)$ and $\pi_1(\partial M)$)

  • 1
    why $i_* (\alpha) $ is the square of an element in $\pi_1 (M)$?2012-09-26
  • 2
    Essentially because it turns around the band twice, and a generator of the fundamental group of the band is a curve which turns around it only once. Notice that $i$ is a map one knows, so one can completely compute the morphism $i_*:\pi_1(\partial M,x_0)\to\pi_1(M,x_0)$.2012-09-26
  • 2
    Why is the following true: "if $r:M\to\partial M$ is a retraction, $\alpha=r_*i_*(\alpha)$ is also the square of an element of $\pi_1(\partial M)$"?2016-03-15
  • 0
    @AyushKhaitan this is because $i_*(\alpha) = \beta^2$ and as we are dealing with homomorphism this means $r_*i_*(\alpha)$ = r_*(\beta)^2$2018-02-22
2

For each $\alpha\in\partial M$, let $\gamma_\alpha$ be the closed loop in $M$ that starts at $\alpha$, goes directly across the strip to its antipode and then halfway around the boundary to its starting point in positive direction. Then $\alpha\mapsto\gamma_\alpha$ is a homotopy -- in particular every $\gamma_\alpha$ has the same homotopy class.

On the other hand, if $x$ and $y$ are antipodes, then when we form $\gamma_x+\gamma_y$, the "directly across" sections cancel out, and the concatenated curve is homotopic to a single turn around the entire boundary. So the homotopy class of $r(\gamma_x+\gamma_y)$ in $\partial M$ is $1$. On the other hand, $r$ ought to induce a homomorphism between the homotopy groups, but $1$ is not twice anything in $\mathbb Z$, which is a contradiction.

  • 0
    (Annoyingly, one has to do this with *based* loops :-/ )2012-09-25
  • 0
    @Mariano: Reparameterizing all of the loops to be based at $x$ and then throwing away the $\alpha$s where $\gamma_\alpha$ does not already pass through $x$ ought to take care of that.2012-09-25