10

It is quite obvious that if a map is a homotopy equivalence, then its mapping cone is contractible, but is the converse true: mapping cone contractible => the map is a homotopy equivalence? I am thinking about both the topological category and the category of chain complexes.

Victor
  • 1,705
  • 4
    The mapping cylinder of $f:X\to Y$ is homotopy equivalent to $Y$. The mapping cone is contractible if (but not only if) $f$ is a homotopy equivalence. If $f$ is nullhomotopic then the mapping cone is homotopy equivalent to $Y\vee\Sigma X$. – Tom Goodwillie Aug 25 '11 at 18:38
  • 2
    It looks like maybe you made a typo when asking your question. My guess would be you switched "cylinder" for "cone" and "null homotopy" for "homotopy equivalence" ? I frequently say "cone" when I mean "cylinder" and I don't know why. – Ryan Budney Aug 25 '11 at 19:39
  • Thanks to Tom and Ryan for corrections. Sure I meant a completely different thing when asking my question; I meant mapping cone and I meant a homotopy equivalence. I am curious to see an example of a map which is not a homotopy equivalence, but its mapping cone is contractible. Especially in the category of chain complexes.

    Notice that a map is a quasi-isomorphism iff its mapping cone is acyclic.

     – Victor Aug 25 '11 at 20:09
  • 2
    Note that to get such an example you need to stay away from differential objects (i.e. $A$ with $d:A\rightarrow A$ s.t. $d\circ d = 0$) since for those $f$ is a homotopy equivalence iff the mapping cone is contractible. See: http://books.google.com/books?id=KaXmMjwBulgC&pg=PA74&lpg=PA74&dq=%22mapping+cone+is+contractible%22&source=bl&ots=kvxLiy84Yn&sig=xvW64C4vVYnogvfMtt6idpaf-Po&hl=en&ei=l7VWTqyDMIHs0gHO44itDA&sa=X&oi=book_result&ct=result&resnum=6&ved=0CDQQ6AEwBQ#v=onepage&q=%22mapping%20cone%20is%20contractible%22&f=false – David White Aug 25 '11 at 20:55
  • 1
    Another comment: if you end up getting an example from topology, stay away from the case when $X$ and $Y$ are simply connected CW complexes. According to exercise 9 in section 4.2 of Hatcher, that's a case where the mapping cone being contractible implies the map is a homotopy equivalence – David White Aug 25 '11 at 21:07
  • David, thank you for the 2 great comments! Chain complexes is a partial case of differential objects, thus one should have Contractible cone => homotopy equivalence in this case. but for spaces i would still like to see a counterexample. – Victor Aug 25 '11 at 21:30
  • @Eric. I don't quite understand your example. May be you mean $N_G(H)=H$? otherwise $H$ can be just the trivial subgroup, in which case $BH$ is also a point. – Victor Aug 25 '11 at 23:37
  • I don't think your example works. First, the higher homotopy groups of the mapping cone are not so easily related to the homotopy groups of the initial spaces, thus I doubt $\pi_n(BG\cup CBH)$, $n\geq 2$, are trivial. If your example was true then for any inclusion $H\subset G$ of groups with $H$ simple the induced map in the homology $H_BH\to H_BG$ must be an isomorphism (it is easy to show from the long exact sequence that a map is a quasi-isomorphism iff its mapping cone is acyclic), but now take any finite simple group and take its different Silov subgroups, you see this can not work. – Victor Aug 26 '11 at 05:05
  • Yeah, you're right, $(X, A)$ and $(X \cup CA, *)$ are different objects, so the long exact sequence of the pair doesn't say what I wanted it to. I don't follow your last argument about the homology of simple groups, but this is enough to break it anyway. I'll delete my comments in a short while to avoid confusing anyone who comes across this question later. – Eric Peterson Aug 27 '11 at 20:40
  • Ryan: I, too, am prone to saying "mapping cone" for "mapping cylinder". I think that, for me at least, the reason is that in the special case when $Y$ is a point the mapping cylinder of $f:X\to Y$ is called the cone on $X$. And that in the general case if you think of $X$ as an object in the category of spaces over $Y$ and think of the mapping cone of $f$ as another such object then the latter is a "cone" in that category: the mapping cylinder is a fiberwise cone. – Tom Goodwillie Feb 07 '12 at 14:45

3 Answers3

17

Ok, going off my second comment from above, in exercise 9 from section 4.2 Hatcher gives a hint that solves your problem. Let $X$ be an acyclic CW-complex which isn’t contractible (I'll give an example below to be complete). Let $f: X \rightarrow *$. The mapping cone of this is $SX$, the suspension of $X$. Exercise 8 of section 4.2 proves that $SX$ is contractible because $X$ is acyclic. But the map $X \rightarrow *$ is not a homotopy equivalence.

Solution to Exercise 8: Suppose $\tilde{H}_*(X)=0$. From $H_0$ we see $X$ is path connected, so $SX$ is simply connected. Thus $H_* (SX) = 0 $ . By the Hurewicz theorem $\pi_*(SX)=0$ so $SX$ is contractible as desired.

An acyclic CW-complex which is not contractible (thanks to link)... Let $a$ and $b$ be the two loops in $X=S^1 \vee S^1$. Glue in two 2-cells along the words $a^5b^{-3}$ and $b^3(ab)^{-2}$. Then $\pi_1(X) = \langle a,b|a^5b^{-3},b^3(ab)^{-2}\rangle$ and this surjects onto the symmetry group of a regular dodecahedron (so $\pi_1(X)\not \cong 0$) while $H_1(X)$ is trivial because it's the abelianization and playing with symbols lets you reduce those relations to $H_1(X) \cong \langle a,b | a,b\rangle \cong 0$. It's clear that $H_n(X)\cong 0$ for $n>1$ because of the dimensions of the cells involved.

David White
  • 29,779
3

The accepted answer addresses the version of the question for topological spaces. But there is also a positive result for triangulated categories: a morphism $f \colon X \to Y$ is an isomorphism if and only if any distinguished triangle $X \to Y \to Z \to X[1]$ has $Z \cong 0$. See Tag 05QR.

This in particular applies to the homotopy category $K(\mathcal A)$ of chain complexes with values in an additive category $\mathcal A$, which is a triangulated category by Tag 014S. Applying this to the mapping cone in $K(\mathcal A)$ shows that $C(f)$ is contractible if and only if $f$ is a homotopy equivalence.

This also applies to the stable homotopy category of spectra, where it implies that a map of topological spaces $f \colon X \to Y$ with contractible mapping cone is a stable homotopy equivalence.

In examples like David White's answer, you only need to go up to the first suspension $\Sigma X$, by construction (but see also this answer). I'm not sure if something like this is supposed to be true in general, but I'm guessing it's more complicated.

R. van Dobben de Bruyn
  • 27,218
1

Weibel: An Introduction to Homological Algebra, Exercise 1.5.10 (1994 edition) asks you to show that for $C,C'$ split complexes with splittings $s,s'$ the cone on a map $f: C \to C'$ is split by the map $(c,c') \mapsto (-s(c),s'(c')-s'fs(c))$ if and only if the induced map on homology $H(f): H(C) \to H(C')$ is zero.

rob mckemey
  • 11