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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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  • $\begingroup$ I reformatted the formulas. See math notation guide. $\endgroup$
    – user147263
    Commented Sep 27, 2014 at 17:15
  • $\begingroup$ The key here is functoriality of $\pi_1$. $\endgroup$ Commented Sep 27, 2014 at 18:30
  • $\begingroup$ Thank you for reformatting u.u $\endgroup$
    – Keith
    Commented Sep 28, 2014 at 12:27

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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    $\begingroup$ why $i_* (\alpha) $ is the square of an element in $\pi_1 (M)$? $\endgroup$
    – user32847
    Commented Sep 26, 2012 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 $i_*:\pi_1(\partial M,x_0)\to\pi_1(M,x_0)$. $\endgroup$ Commented Sep 26, 2012 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 $\pi_1(\partial M)$"? $\endgroup$
    – user67803
    Commented Mar 15, 2016 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
    Commented Feb 22, 2018 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! $\endgroup$
    – Conjecture
    Commented Feb 22, 2019 at 17:35
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Let $M$ be the Möbius strip defined by $M:=\frac{\displaystyle [0,1]\times [0,1]}{\displaystyle (0,t)\sim(1,1-t)}$ with quotient map $q\colon [0,1]\times [0,1]\to M$. Let $B:=q\big(\{(s,k):0\leq s\leq 1,k=0,1\}\big)$ be the boundary circle and $C:=q\left(\left\{\left(s,\frac{1}{2}\right):0\leq s\leq 1\right\}\right)$ be the central circle.

Consider the inclusion map $i\colon B\hookrightarrow M$. Also, consider the retraction $f\colon M\to C$ defined by $f:[(s,t)]\longmapsto\left[\left(s,\frac 12\right)\right].$

Now, the map $f\circ i\colon B\to C$ is a $2$-fold covering. Hence, $f_*\circ i_*\big(\pi_1(B)\big)$ is an index two subgroup of $\pi_1(C)$. See Theorem below.


Let $j\colon C\hookrightarrow M$ be the inclusion. Then, $f\circ j=\text{Id}_C$. So that, $f_*\circ j_*=\big(\text{Id}_C\big)_*=\text{Id}_{\pi_1(C)}$.

Next note that, $H\colon M\times[0,1]\to M$ defined by $$H:\big([(s,t)],t'\big)\longmapsto\left[\left(s,\frac{1}{2}t'+(1-t')t\right)\right]\text{ for } 0\leq s,t,t'\leq 1$$ is a homotopy between $H(-,0)=\text{Id}_M$ and $H(-,1)=j\circ f$. Hence, $\text{Id}_{\pi_1(M)}=\big(\text{Id}_M\big)_*=\big(j\circ f\big)_*=j_*\circ f_*$.

Therefore, $j_*\colon \pi_1(C)\to \pi_1(M)$ is an isomorphism.


Now, $i_*=\big(\text{Id}_M\big)_*\circ i_*=\big(j\circ f\big)_*\circ i_*=\big(j\circ f\circ i\big)_*=j_*\circ\big(f\circ i\big)_*$.

Hence, $i_*\big(\pi_1(B)\big)$ is an index two subgroup of $\pi_1(M)$.


Now, if possible assume, there is a retraction $r\colon M\to B$. Then $r\circ i=\text{Id}_B$. Then, $r_*\circ i_*=\text{Id}_{\pi_1(B)}$.

Note that both $B$ and $C$ are circles. So $\pi_1(B)$ and $\pi_1(C)$ are infinite cyclic groups. So, $\pi_1(M)$ is also an infinite cyclic group. Let $b$ be a generator of $\pi_1(B)$ and $m$ be a generator of $\pi_1(M)$.

Since, $i_*\big(\pi_1(B)\big)$ is an index two subgroup of $\pi_1(M)$, we have $i_*(b)$ equals to either $2m$ or $-2m$, here all group structure are written additive way. So, $$r_*\circ i_*=\text{Id}_{\pi_1(B)}\implies b=\text{Id}_{\pi_1(B)}(b)=r_*\big(i_*(b)\big)=r_*\big(\pm 2m\big)=\pm 2r_*(m).$$ Since, $r_*(m)\in \pi_1(B)$ we have some integer $n$ such that, $r_*(m)=nb$. Hence, $b=\pm2nb$, which is impossible.


$\textbf{Theorem:}$ Let $p\colon \widetilde X\to X$ be a covering map where $\widetilde X$ is a path-connected space, then for any $x_0\in X$ and $\widetilde {x_0}\in p^{-1}(x_0)$ we have $$\big|p^{-1}(x_0)\big|=\left[\pi_1(X,x_0):p_*\pi_1\left(\widetilde X,\widetilde{x_0}\right)\right]$$ For proof, see Theorem $10.9.$ here.

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  • $\begingroup$ I have not written with any new technique to solve this problem, I tried to write as explicit as possible. $\endgroup$
    – Sumanta
    Commented Sep 17, 2019 at 16:49
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    $\begingroup$ Wow it was a question posted long ago...I remember solviing it in a similar way you have done. Thank you for your answer anyway. $\endgroup$
    – Keith
    Commented Sep 17, 2019 at 19:01
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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 $i_*:H_1(B)\rightarrow H_1(M)$ (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 $H_1(M,B)\cong \mathbb Z/2\mathbb Z$. 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. $\endgroup$
    – user637978
    Commented Jun 11, 2019 at 22:33
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    $\begingroup$ This is just the splitting lemma, see en.wikipedia.org/wiki/Splitting_lemma. $\endgroup$
    – Arbutus
    Commented Jun 23, 2019 at 17:53
  • $\begingroup$ What is $H_1(X)$? $\endgroup$
    – kam
    Commented Feb 6, 2020 at 7:52
  • $\begingroup$ @kam, Fixed it - thanks. $\endgroup$
    – Arbutus
    Commented Feb 6, 2020 at 12:55
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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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    $\begingroup$ Shouldn't those $*$'s be lower? $\endgroup$
    – user5826
    Commented Feb 12, 2019 at 4:02
  • $\begingroup$ Isn't it $i^\ast (n)=2n$? I don't understand why going around the band once "projects" to going around the boundary twice $\endgroup$
    – VanD1206
    Commented Aug 10, 2019 at 16:47
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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 :-/ ) $\endgroup$ Commented Sep 25, 2012 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. $\endgroup$ Commented Sep 25, 2012 at 21:43
  • $\begingroup$ @HenningMakholm is $\pi_1(\partial M)=1$ or I misunderstood something? $\endgroup$
    – Conjecture
    Commented Feb 22, 2019 at 17:42
  • $\begingroup$ @PerelMan: $\partial M$ is a circle, so $\pi_1(\partial M)\cong\mathbb Z$. When I write $1$ I mean $1\in\mathbb Z$. $\endgroup$ Commented Feb 22, 2019 at 18:42

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