Retraction of the Möbius strip to its boundary

Question

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...

Answer 1

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 $\pi_1(\partial M)$)

Answer 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.