# Retraction of the Möbius strip to its boundary

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...

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## 2 Answers

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 $\pi_1(\partial M)$)

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why $i_* (\alpha)$ is the square of an element in $\pi_1 (M)$? –  user32847 Sep 26 '12 at 7:31
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 $i_*:\pi_1(\partial M,x_0)\to\pi_1(M,x_0)$. –  Mariano Suárez-Alvarez Sep 26 '12 at 17:36

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.

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(Annoyingly, one has to do this with based loops :-/ ) –  Mariano Suárez-Alvarez Sep 25 '12 at 21:39
@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. –  Henning Makholm Sep 25 '12 at 21:43