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 $M:=\frac{[0,1]\times [0,1]}{(0,t)\sim(1,1-t)}$ with quotient map $q:[0,1]\times [0,1]\to M$. Let $B:=q\big(\{(s,k):0\leq s\leq 1,k=0,1\}\big)$ be the boundary circle and $C:=q\big(\big\{\big(s,\frac{1}{2}\big):0\leq s\leq 1\big\}\big)$ be the central circle. Consider the inclusion map $i:B\hookrightarrow M$. Also, consider the retraction $f:M\to C$ defined by $f\big([(s,t)]\big)=\big[\big(s,\frac{1}{2}\big)\big]$.

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

Let $j:C\hookrightarrow M$ 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, $H:M\times[0,1]\to M$ defined by $H:\big([(s,t)],t'\big)\mapsto\big[\big(s,\frac{1}{2}t'+(1-t')t\big)\big];\forall 0\leq s,t,t'\leq 1$ 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, $j_*$ is an isomorphism between $\pi_1(C)$ and $\pi_1(M)$.

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 let, there is a retraction $r:M\to B$. 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 group. So that, $\pi_1(M)$ 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. But from $r_*\circ i_*=\text{Id}_{\pi_1(B)}$ we have, $b=\text{Id}_{\pi_1(B)}(b)=r_*\big(i_*(b)\big)=r_*\big(\pm 2m\big)=\pm 2r_*(m)$. Since, $r_*(m)\in \pi_1(B)$ we have some integer $n$ such that, $r_*(m)=nb$. Hence, $b=\pm2nb$, which is impossible.

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).