Let $X=([0,1]×[0,1])/\sim$ , where $\sim$ is the equivalence relation generated by $(0,y)\sim (1,1-y)$, and let $Y=([0,1]×\{ 0,1 \} )/\sim \ \subset X$. I must show that there are no retractions $r:X\to Y$.
First off, $X$ is homeomorphic to a Moebius strip ( the relation, in fact, glues the left side and the right side of the unitary square with opposite orientations); so $\pi_1(X,x)\simeq\langle\gamma\rangle$, since the Moebius strip retracts to $([0,1]×\frac 1 2)\simeq S^1$.
Now, $Y$ is a loop homotopically equivalent to $\gamma^2$, so it has winding number 2; I observe that $\operatorname{Id_X|_{\gamma([0,1])}} $ and $\iota \circ r |_{\gamma([0,1])}$ must be homotopic functions, since $r$ is a retraction. This actually means that $\gamma$ and $\iota \circ r \circ \gamma$ are homotopic loops, and so they must have equal winding number (namely$, \gamma$ has w.n. 1).
However, $\gamma(0)=\gamma(1)$ implies that $\iota \circ r \circ \gamma(0)=\iota \circ r \circ \gamma(1)$: starting from a point $y\in Y$, the only way to reach again $y$ moving continuously on $Y$ is to complete the whole loop $Y$ or to stop at a point and return back to $y$, because $Y$ doesn't self intersect in any point; in the fist case the winding number of $\iota \circ r \circ \gamma$ would be 2, in the second case 0. So a retraction $r$ from $X$ to $Y$ cannot exist.
I'm not sure about this reasoning, so I would be glad if someone told me if what I wrote is correct and if there is a more elegant way to do this exercise. Thank you in advance
Given an inclusion $i$ of a retract $r:X \rightarrow A$, we may write $0 \rightarrow \pi_1(A) \rightarrow \pi_1(X) \rightarrow \operatorname{coker} i_* \rightarrow 0$. Since $A$ is a retraction this short exact sequence splits which implies that $\pi_1(X)$ is a semidirect product of $\operatorname{coker}i_*$ and $\pi_1(A)$.
Since the cokernel of this inclusion is $\mathbb{Z}/2$, and the fundamental group of $X$ is $\mathbb{Z}$, this is clearly impossible.