Timeline for Is there a (discrete) monoid M injecting into its group completion G for which BM is not homotopy equivalent to BG?
Current License: CC BY-SA 3.0
19 events
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| S Jul 23, 2020 at 4:07 | history | bounty ended | CommunityBot | ||
| S Jul 23, 2020 at 4:07 | history | notice removed | CommunityBot | ||
| Jul 18, 2020 at 21:41 | review | Suggested edits | |||
| Jul 18, 2020 at 23:27 | |||||
| S Jul 15, 2020 at 3:01 | history | bounty started | Omar Antolín-Camarena | ||
| S Jul 15, 2020 at 3:01 | history | notice added | Omar Antolín-Camarena | Authoritative reference needed | |
| Feb 19, 2013 at 22:44 | comment | added | Benjamin Steinberg | In semigroup jargon the 5-element example with classifying space $S^2$ is a 2x2 rectangular band with adjoined identity. | |
| Feb 19, 2013 at 13:48 | comment | added | Omar Antolín-Camarena | Yes, that is correct. Alternatively, instead of using the universal cover argument, I think you can use Quillen's theorem B to show the slice category is simply connected, and thus contractible iff it has trivial homology, in which case Theorem B implies the canonical map is an equivalence BM->BG. | |
| Feb 19, 2013 at 12:23 | comment | added | Benjamin Steinberg | If I understood correctly the classifying space of the slice category is the universal cover of BM. So it is contractible iff BM is a K(G,1). The homology of this slice category is the homology of M with coefficients in ZG. In particular BM is homotopy equivalent to BG iff the canonical map is an equivalence. | |
| Feb 19, 2013 at 3:52 | comment | added | Omar Antolín-Camarena | That's a nice paper, thanks for the reference, @BenjaminSteinberg! | |
| Feb 19, 2013 at 3:14 | comment | added | Benjamin Steinberg | In Fiedorowicz, Z. Classifying spaces of topological monoids and categories. Amer. J. Math. 106 (1984), no. 2, 301–350 it is shown that the natural map $BM\to BG$ is a homotopy equivalence if and ony if $H_n(M,\mathbb ZG)=0$ for $n\geq 1$. | |
| Feb 18, 2013 at 23:49 | comment | added | Omar Antolín-Camarena | Plus, it id's not true that just adding fillers for the outer horns of NM gets you NKM, since you also have compositions of elements of KM\M. | |
| Feb 18, 2013 at 23:44 | comment | added | Omar Antolín-Camarena | I'm also confused by talk of "minimal fibrations" in Spice the bird's answer since there is no map NZ->NN and the inclusion NN->NZ is not a fibration, of course. I think Spice might have meant NKM is a minimal simplicial set or something like that instead... | |
| Feb 18, 2013 at 23:38 | comment | added | Omar Antolín-Camarena | I don't understand Spice the bird's argument. He or she seems to be saying that for a "cancelable" M, NKM is obtained from NM by attaching files for outer horns, in particular, by only adding simplices, so it would seem the argument would imply that M injects into KM, but Malcev's example shows that's not always true. (Maybe "cancelable" means "injects into KM" rather than "has left and right cancellation" in Spice's answer?) | |
| Feb 18, 2013 at 22:28 | comment | added | Benjamin Steinberg | Spice the bird seems to answer that no examples exist in his/her answer to mathoverflow.net/questions/94017/…. I never completely understood his/her answer so would be grateful if someone could. | |
| Feb 18, 2013 at 22:22 | history | edited | Omar Antolín-Camarena | CC BY-SA 3.0 |
remove LaTeX from title, to improve appearence; edited title
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| Feb 18, 2013 at 21:10 | comment | added | Benjamin Steinberg | Yes Ore=calculus of fractions. | |
| Feb 18, 2013 at 20:52 | comment | added | Omar Antolín-Camarena | Is an Ore condition something like having a right calculus of fractions? If so, the proof sketched in 2 covers this case as well. I forgot to mention that, thanks Benjamin. | |
| Feb 18, 2013 at 20:18 | comment | added | Benjamin Steinberg | If the monoid satisfies an Ore condition then the classifying spaces are equivalent. | |
| Feb 18, 2013 at 19:34 | history | asked | Omar Antolín-Camarena | CC BY-SA 3.0 |