Timeline for Do finite simplicial sets jointly detect isomorphisms in the homotopy category? [duplicate]
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Oct 15, 2018 at 5:53 | vote | accept | Edoardo Lanari | ||
| Oct 14, 2018 at 0:52 | history | closed | CommunityBot | Duplicate of Counterexamples in algebraic topology? | |
| Oct 12, 2018 at 8:15 | review | Close votes | |||
| Oct 14, 2018 at 0:55 | |||||
| Oct 12, 2018 at 8:02 | comment | added | Kevin Carlson | @SimonHenry That there is no such set consisting of finite complexes follows immediately from the linked counterexample, while the generalization is new, due to myself and Dan Christensen, and should appear soon. But as I say, if you believe the counterexample with restricted countable symmetric group you might not find the generalization too suspicious, as it just does something very similar with bigger groups. | |
| Oct 12, 2018 at 8:00 | answer | added | Karol Szumiło | timeline score: 7 | |
| Oct 12, 2018 at 7:59 | comment | added | Simon Henry | @kevin carlson : I have doubt about your second claim, do you have a references ? there is Freyd's result that shows that there is no small such familly which have no phantom maps, but detecting weak equivalences is a very different question. | |
| Oct 12, 2018 at 7:55 | comment | added | Kevin Carlson | Oops, I guess Lawson didn’t post the counterexample, but sufficient conditions to get around it. | |
| Oct 12, 2018 at 7:47 | comment | added | Kevin Carlson | Cont’d : In fact there are no small families of objects in the homotopy category for which the analogous claim is true. Similar counterexamples go through for classifying spaces of arbitrarily big restricted symmetric groups. This is a way in which the unpointed homotopy category is unavoidably terrible compared to the pointed or stable categories. | |
| Oct 12, 2018 at 7:41 | comment | added | Kevin Carlson | Lawson’s counterexample of the shift map $s$ on the classifying space $X$of the restricted infinite symmetric group for spaces also applies to simplicial sets, since the equivalence between the two categories sends finite simplicial sets to CW complexes and vice versa, up to weak equivalence. It is indeed true, as Simon points out, that any finite family of circles in $X$ is sent by $s$ to a family of circles representing simultaneously conjugate classes in the fundamental group. But this isn’t enough-there are infinite families not simultaneously conjugate to their shifts. | |
| Oct 12, 2018 at 6:36 | comment | added | Simon Henry | In fact looking at unbased homotopy class of maps from a bouquet of $k$ $n$-sphere to your space gives you the quotient of the products of $k$ copy of $\pi_n$ by the action of $\pi_1$. Which should be enoug to show that $f$ is a bijection on $\pi_n$. I can't write the details today sadly. | |
| Oct 12, 2018 at 6:20 | comment | added | Simon Henry | You're right that was stupid I realised just after writing it... What you get from the unbased map maping set though are the quotient of the $\pi_n$ by the action of $\pi_1$. So it at least works in cases where the $\pi_1$ action is trivial | |
| Oct 12, 2018 at 6:13 | answer | added | Tim Porter | timeline score: 3 | |
| Oct 12, 2018 at 6:11 | comment | added | Edoardo Lanari | @Gasterbiter Interesting, indeed Tyler Lawson's reply is related to the idea that I had of relating pointed and unpointed mapping spaces (I remember there is some fiber sequence of some sort somewhere in Strom's book) | |
| Oct 12, 2018 at 6:07 | comment | added | Edoardo Lanari | @SimonHenry but $\pi_0$ does not commute with (homotopy) pullbacks in general | |
| Oct 12, 2018 at 5:58 | comment | added | user123627 | I don’t think this is true though, see eg mathoverflow.net/questions/55365/… for a counterexample to the CW version. | |
| Oct 12, 2018 at 5:39 | comment | added | Harry Gindi | yes that was my concern | |
| Oct 12, 2018 at 5:38 | comment | added | Edoardo Lanari | Hi Harry. Yes, surely that is the intuition but these are unpointed mapping spaces, so how do you recover that? I was looking for a "conceptual" proof whatever that might mean, but I couldn't come up with a combinatorics one either. | |
| Oct 12, 2018 at 5:36 | comment | added | Harry Gindi | Might be dumb, but there are finite models for all of the simplicial spheres, so shouldn't this imply that the map induces isomorphisms on all homotopy groups at all basepoints? Not sure if considering basepoints is kosher after going to the homotopy category. | |
| Oct 12, 2018 at 4:46 | history | asked | Edoardo Lanari | CC BY-SA 4.0 |