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!


1 Answer 1


Your argument for homotopy implying contractability is formally correct, but I am unsure how one would avoid circularity - from where do you get the fact that $[Cf, Z]=0$? The opposite direction is wrong: the isomorphisms of the groups you listed does not imply homotopy equivalence, and the very thread you linked gives an example of why it does not.

But this is not the right way to think about it: there is of course a geometric reason why this is true. If $f:X\to Y$, $g:Y\to X$ realize a homotopy equivalence, there is a homotopy of the identity on $Y$ to the map $fg$, which is of course in the base of the cone $Cf$. This can be extended to a homotopy on all of $Cf$ because the inclusion of the base is a cofibration. Concatenate this homotopy with the retraction to the point of the cone, and we have a contraction of $Cf$.

  • $\begingroup$ From where? I edited the post to clarify. Thanks, and I'll check your explanation later. $\endgroup$
    – Yai0Phah
    Oct 26, 2015 at 21:46
  • $\begingroup$ Oops yes I see. I'd still be suspicious of an argument that uses that machinery, though - chances are that either the statement that you're trying to prove, or something which is very similar, is involved in the construction. It's very overpowered. $\endgroup$
    – Peter Xu
    Oct 27, 2015 at 5:38

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