I'm trying to prove that
$f: X \rightarrow Y$ is a homotopy equivalence $\iff$ $X$, $Y$ are both homeomorphic to a deformation retract of a space $Z$
The $\Leftarrow$ was not a problem. If both are deformation retracts it follows that $Z \simeq X$ and $Z \simeq Y$ and by transitivity $X \simeq Y$.
The first half of $\Rightarrow$ was also not a problem, because using the mapping cylinder $Z_f$ as their common super-space, $X \times I$ can be deformation-retracted down to $Y$ by the homotopy $(x,s,t)\mapsto (x,st)$, i.e. along a vertical line.
Now I'm stuck with the deformation retraction of $Z_f$ onto $X$. For example, do I construct a homotopy from $Z_f$ to $X \times I$ and then to $X$ or directly from $Z_f$ to $X$? I'm also not sure where to use $f \circ g \simeq id_Y$ and $g \circ f \simeq id_X$ and $id_{Z_f} \simeq i \circ r_Y$, where $r_Y$ is the retraction from $Z_f$ onto $Y$.
I'm tempted to do something like this but it doesn't seem to lead anywhere: $i \circ r_Y \simeq id_Y \simeq f \circ g$.
Can someone help me finish this proof? I'd much appreciate any help!