If $X$ and $Y$ are subspaces of $Z$, $X \cong Y$ and $X$ is a retract of $Z$, is $Y$ also a retract of $Z$? If $X$ and $Y$ are subspaces of $Z$, $X \cong Y$ and $X$ is a retract of $Z$, is $Y$ also a retract of $Z$?
I think the answer is no, but I can't find a counterexample. Can anyone help me with this?
 A: You sure need some more conditions, otherwise there are trivial examples. Say you allow $Z$ be not connected (as you have). 
Then pick any spaces $Y\subseteq Z_0$ such that $Y$ is NOT a retract of $Z_0$. Pick such a space so that points are closed sets, that should not be a problem. Then take an isomorphic copy of $Y$ and call it $X$, and add it to the space $Z_0$ as a new component to obtain $Z$. 
Then $Y$ is not a retract of $Z$: assume otherwise, and let $f$ be a a retraction of $Z$ onto $Y$. The restriction of $f$ to $Z_0$ would be a retraction of $Z_0$ onto $Y$. 
But $X$ is a retract of $Z$: define the retraction as the identity on $X$ (obviously). Furthermore, pick any $x\in X$, and define the retraction on $Z_0$ as the constant $x$ function. This is continuous, as the pre-image of any closed set is closed.
A: Consider the Moebius strip $M$. It is homeomorphic to a rectangle with one pair of its opposite sides glued together.

Now note that the line connecting the centers of the identified sides is a retract of the Moebius strip. You can simply collapse the whole rectangle to it horizontally. This line is homeomorphic to $S^1$. Moreover, it is a deformation retract.
However, $M$ is not retractible to its boundary $\delta M$. On the picture the boundary is the left and right sides of the rectangle. Now assume that a retraction $r$ from $M$ onto its $\delta M$ exists. Let $id$ be the natural inclusion of $\delta M$ into $M$. The boundary is homeomorphic to $S^1$, hence $\pi_1(\delta M)=\mathbb{Z}.$ $id$ induces the group homeomorphism $id_*:\ \pi_1(\delta M)\to\pi_1(M)$ which sends $1$ to $2$ (the image of the loop generating $\pi_1(\delta M)$ under the deformation retraction onto the middle of the band circumvents the band twice). But $r$ also induces a group homeomorphism. Since $r|_{\delta M}=id$, $id_\ast\circ r_\ast=id:\ \pi_1(\delta M)\to\pi_1(\delta M)$ (here $id$ is the identity isomorphism), but we have already shown that $id_\ast\circ r_\ast(1)=2$. We arrived at a contradiction, so $\delta M$ is not a retract of $M$, even though $\delta M\simeq S^1$.
