Retraction of the Möbius strip to its boundary

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 $\pi_1(\partial M)$)

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 $\pi_1(C)$ both 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.