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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...

5 Answers 5

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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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Suppose there were a retract $r:M\rightarrow \partial M$. By definition, this means that $r\circ i=\mathrm{id}\, _{\partial M}$, where $i:\partial M\rightarrow M$ is the inclusion. From functoriality, it follows that $r^*\circ i^*=\mathrm{id}\, _{\pi _1(\partial M)}$, where $f^*$ denotes the induced map of fundamental groups. Thus, $r^*:\pi _1(M)\rightarrow \pi _1(\partial M)$ is surjective. However, $\pi _1(M)\cong \mathbb{Z}\cong \pi _1(\partial M)$ and $r^*(n)=2n$, which is not surjective: a contradiction. Thus, there can be no such retract.

To see that $r^*(n)=2n$, I think it is easiest to view the Möbius strip as a quotient of the unit square in $\mathbb{R}^2$, obtained by identifying the left and right sides with the opposite orientation. Intuitively, if you go around the Möbius band once you, the projection onto the boundary goes around twice (draw a picture for yourself).

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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.