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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...

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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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    $\begingroup$ why $i_* (\alpha) $ is the square of an element in $\pi_1 (M)$? $\endgroup$ – user32847 Sep 26 '12 at 7:31
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    $\begingroup$ 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 $i_*:\pi_1(\partial M,x_0)\to\pi_1(M,x_0)$. $\endgroup$ – Mariano Suárez-Álvarez Sep 26 '12 at 17:36
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    $\begingroup$ 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 $\pi_1(\partial M)$"? $\endgroup$ – fierydemon Mar 15 '16 at 1:06
  • $\begingroup$ @AyushKhaitan this is because $i_*(\alpha) = \beta^2$ and as we are dealing with homomorphism this means $r_*i_*(\alpha)$ = r_*(\beta)^2$ $\endgroup$ – B.A Feb 22 '18 at 5:26
  • $\begingroup$ @MarianoSuárez-Álvarez what is the point of computing $\pi_1(M)$ and $\pi_1(\partial M)$? even if there is retraction, it is not a homotopy equivalence. Can we find a contradiction by these computations? Thanks! $\endgroup$ – PerelMan Feb 22 at 17:35
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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.

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  • $\begingroup$ (Annoyingly, one has to do this with based loops :-/ ) $\endgroup$ – Mariano Suárez-Álvarez Sep 25 '12 at 21:39
  • $\begingroup$ @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. $\endgroup$ – Henning Makholm Sep 25 '12 at 21:43
  • $\begingroup$ @HenningMakholm is $\pi_1(\partial M)=1$ or I misunderstood something? $\endgroup$ – PerelMan Feb 22 at 17:42
  • $\begingroup$ @PerelMan: $\partial M$ is a circle, so $\pi_1(\partial M)\cong\mathbb Z$. When I write $1$ I mean $1\in\mathbb Z$. $\endgroup$ – Henning Makholm Feb 22 at 18:42
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You can also prove this using homology, but it's somewhat more effort. The basic idea is that, if $B$ is the boundary circle and $r:M\rightarrow M$ is a retraction onto $B$ then the inclusion map also induces an injection $i_*:H_1(B)\rightarrow H_1(X)$ (to see this, apply $r_*$ to $i_*\alpha=i_*\beta$). Thus we have an exact sequence $$ 0\rightarrow H_1(B)\xrightarrow{i_*}H_1(M)\xrightarrow{q_*} H_1(M,B)\rightarrow 0 $$ coming from the reduced long exact sequence for the pair $(M,B)$. We know that $H_1(B)$ and $H_1(M)$ are both $\mathbb Z$ (because $B=S^1$ and $M$ deformation retracts onto its central circle) and, since $(M,B)$ is a good pair, $H_1(M,B)\cong H_1(M/B)$. But $M/B=\mathbb R\mathbb P^2$, as can be seen by their cell decompositions, where the pink indicates the boundary circle $B$:

enter image description here

thus $H_1(X,B)\cong \mathbb Z/2\mathbb Z$. The fact that $r$ is a retraction means that the above sequence splits, as $r_*:H_2(M)\rightarrow H_2(B)$ composed with $i_*$ is the identity. This is a contradiction.

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  • $\begingroup$ Thanks for this answer! I appreciated the unique perspective. But one question: Why does $r$ being a retraction mean that the above sequence splits? I understand that composing $r_*$ with $i_*$ is the identity. $\endgroup$ – Mathematical Mushroom Jun 11 at 22:33
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    $\begingroup$ This is just the splitting lemma, see en.wikipedia.org/wiki/Splitting_lemma. $\endgroup$ – Arbutus 2 days ago

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