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I would like a hint to solve the following problem:

Let $X$ be the Möbius band and $A$ its boundary (which is a circumference). Prove that there is not a retraction $r:X \rightarrow A$.

Each try I did gave no solution.

Any hint to solve this problem?

Thanks!

hbghlyj
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EQJ
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1 Answers1

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The fundamental group of a Moebius band is just $\mathbb{Z}$ (you can see that as it is homotopy equivalent to a circle). Pick a loop that generates it. If you don't know much about the fundamental group, just pick a circle that runs through the middle of the Moebius band; notice, it should go around once, not twice like the boundary.

Now, if there was a retraction this loop should map to a loop in the boundary. Is that possible? Why not? Can the class of the loop change in a retraction?

Edit: Extra hint:

What happens if you pre-compose with the inclusion $A \rightarrow X$ ? There are two maps induced $$\mathbb{Z}\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}$$ What do we know about the first, and what do we know about their composition?

Theo Douvropoulos
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  • I don't understand something, the intuitive idea is clear, but I don't know how to formalize this idea. If $r:X\rightarrow A$ is the retraction, what is the problem with the induces mapping $r_*:\pi(X)=\mathbb{Z}\rightarrow \pi(A)=\mathbb{Z}$? – EQJ Apr 08 '15 at 17:40
  • What happens if you compose with the inclusion $A \rightarrow X$ afterwards? – Theo Douvropoulos Apr 08 '15 at 17:54
  • Functoriality is the keyword here. – Daniel Valenzuela Apr 08 '15 at 18:05
  • Precompose means $ir$ or $ri$? Sorry for the silly question. – EQJ Apr 08 '15 at 18:15
  • $r\circ i: A\rightarrow X\rightarrow A$ – Theo Douvropoulos Apr 08 '15 at 18:18
  • I know that $r_$ must be surjective and $r_i_*$ has to be the identity map of $\mathbb{Z}$. Why is this a contradiction. Thanks for being patient. – EQJ Apr 08 '15 at 18:24
  • Because the map that is induced by $i$ is multiplication by 2 (can you see that?..) So now, your map $r:\mathbb{Z}\rightarrow \mathbb{Z}$ should satisfy $r(2n)=n$. But what do we know about any map $f:\mathbb{Z}\rightarrow \mathbb{Z}$? – Theo Douvropoulos Apr 08 '15 at 19:16
  • I am sure that $i_(n)=2n$, I get it intuitively, how can be done this formally? I think that since $r_$ has to be surjective then $r_(1)=\pm 1$ (indeed it has to be an isomorphism) and then $r_(2n)=2r_*(n)=\pm 2 n = n$ which implies $\pm 2=1$ a contradiction. Am I right? – EQJ Apr 08 '15 at 19:45
  • To do it formally you need to describe a loop that generates the fundamental group of the Moebius band, and see how many times you need to take that loop to get something homotopic to the boundary. To find such a loop the best way is to follow the proof that your $\pi_1$ is $\mathbb{Z}$. That is via the deformation retraction to the circle inside the Moebius strip. Now, you just need to show that the boundary of the Moebius strip is homotopic to twice the "median" circle. – Theo Douvropoulos Apr 08 '15 at 19:53
  • Let us continue this discussion in chat. – Theo Douvropoulos Apr 09 '15 at 00:49