In this MO thread, the OP claimed that it is obvious that homotopy equivalence implies the mapping cone contractible, whereas the converse proposition is wrong. I hate to admit that it's not obvious to me at all.
I'm now putting this problem into a more conceptual or abstract setting, though I haven't learned anything on model category or homotopical algebra. Suppose $f\colon X\to Y$ is a continuous map of (pointed) topological spaces, and $Cf$ is the corresponding mapping cone. Then there's a long coexact sequence (Puppe sequence): $$X\to Y\to Cf\to\Sigma X\to\Sigma Y\to\Sigma Cf\to\dotsb$$ where $\Sigma Z$ is the reduced suspension of $Z$. In other words, for any topological space $Z$, we have a long exact sequence of pointed sets: $$\dotsb\to[\Sigma Cf,Z]\to[\Sigma Y,Z]\to[\Sigma X,Z]\to[Cf,Z]\to[Y,Z]\to[X,Z]$$ In addition, if $f$ is a homotopy equivalence, a fortiori $[\Sigma Y,Z]\to[\Sigma X,Z]$ and $[Y,Z]\to[X,Z]$ are isomorphisms, hence $[Cf,Z]=0$ for all $Z$. Take $Z=Cf$, we have $Cf$ is contractible (I'm not sure whether my argument is right). On the other hand, if $Cf$ is contractible, so is $\Sigma^{n-1}Cf$, therefore $(\Sigma^n f)^*\colon[\Sigma^n Y,Z]\to[\Sigma^n X,Z]$ is an isomorphism between groups for $n\ge1$, therefore $\Sigma^n Y$ and $\Sigma^n X$ are homotopically equivalent.
In contrast to space level version, the proposition for chain complexes is right. The preceding argument works when we replace $f$ by a cofibration, and the long exact sequence ends with an extra term $0$, therefore $X,Y$ are homotopically equivalent.
I don't know whether the topological version is right if we restrict to the category of some nice spaces, and it's still unclear what is the real obstruction in the topological case.
Any help? Thanks!