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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!

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    I believe this uses obstruction theory.2011-04-17
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    @Aaron: Are you sure? I just looked up what [it](http://en.wikipedia.org/wiki/Obstruction_theory) means but what I'm trying to prove doesn't need CW complexes.2011-04-17
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    Ah, you're right. Obstruction theory is supposed to give you a dimension-by-dimension complete obstruction to deforming a CW complex onto a CW subcomplex. You could possibly talk about how any space is weakly homotopy equivalent to a CW complex (and same for a pair), but that shouldn't be needed here. I'd like to say that it follows from the fact that the inclusion $X \hookrightarrow Z_f$ is a weak homotopy equivalence, although that route requires that you be working with CW-complexes too. It probably boils down to the same thing, anyways.2011-04-17

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