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

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.

Answer 3

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 $B$:

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.

Answer 4

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).

Answer 5

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.