One of the 2 examples of 2 spaces $X$ and $Y$ ($Y \subset X$) where $X$ and $Y$ are homotopy equivalent but $Y$ is not deformation retract of $X$ is the following:
$X$ is the annulus and $Y$ is a circle in $X$ that bounds a disk $D$ in $X$.
I can see why $Y$ is not a deformation retracts of $X$ (because of the non-retraction theorem, $Y$ is not a retract of the disk $D$) but I am having trouble to see why $X$ and $Y$ are homotopy equivalent. I always thought that $X$ is homotopy equivalent to the inner boundary circle but how can I see that $X$ is in fact homotopy equivalent to the circle $Y$.
Best Answer
Note that $Y$ is homeomorphic to the inner boundary circle which I'll denote $Z$. Since every homeomorphism is a homotopy equivalence, it follows that $Y$ is homotopy equivalent to $Z$.
And homotopy equivalence is a transitive relation (it is, in fact, an equivalence relation: reflexive, symmetric and transitive). Since $X$ is homotopy equivalent to $Z$, and since $Z$ is homotopy equivalent to $Y$, it follows that $X$ is homotopy equivalent to $Y$.
Keep in mind, just because two spaces like $X$ and $Y$ are homotopy equivalent to each other, it does not follow that every continuous map from $Y$ to $X$ is a homotopy equivalence. In particular, the inclusion map $i_Y : Y \to X$ is continuous, but it is not a homotopy equivalence. Instead, the composition of any homeomorphism $f : Y \to Z$ with the inclusion map $i_Z : Z \to X$ gives the needed homotopy equivalence.