Timeline for when mapping cone is contractible
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 26, 2020 at 22:38 | answer | added | R. van Dobben de Bruyn | timeline score: 3 | |
| Feb 7, 2012 at 14:45 | comment | added | Tom Goodwillie | 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. | |
| Feb 7, 2012 at 13:19 | answer | added | rob mckemey | timeline score: 1 | |
| Aug 27, 2011 at 20:40 | comment | added | Eric Peterson | 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. | |
| Aug 26, 2011 at 5:05 | comment | added | Victor | 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. | |
| Aug 25, 2011 at 23:37 | comment | added | Victor | @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. | |
| Aug 25, 2011 at 22:02 | vote | accept | Victor | ||
| Aug 25, 2011 at 21:30 | comment | added | Victor | 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. | |
| Aug 25, 2011 at 21:25 | answer | added | David White | timeline score: 17 | |
| Aug 25, 2011 at 21:07 | comment | added | David White | 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 | |
| Aug 25, 2011 at 20:55 | comment | added | David White | 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: books.google.com/… | |
| Aug 25, 2011 at 20:09 | comment | added | Victor | 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. | |
| Aug 25, 2011 at 20:02 | history | edited | Victor | CC BY-SA 3.0 |
changed cylinder -> cone; added 16 characters in body
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| Aug 25, 2011 at 19:39 | comment | added | Ryan Budney | 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. | |
| Aug 25, 2011 at 18:38 | comment | added | Tom Goodwillie | 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 $Y\vee\Sigma X$. | |
| Aug 25, 2011 at 18:27 | history | asked | Victor | CC BY-SA 3.0 |