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Apr 13, 2017 at 12:58 history edited CommunityBot
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Dec 9, 2015 at 5:45 vote accept user46652
Mar 31, 2015 at 21:29 comment added David Roberts @DanRamras The conditions do imply they are good simplicial spaces.
Mar 31, 2015 at 20:04 answer added archipelago timeline score: 6
Mar 30, 2015 at 23:24 comment added archipelago I somehow disagree. Life isn't always as beautiful as the homotopy type of a CW complex especially if one is not interested in algebraic topological questions for its own sake.
Mar 29, 2015 at 3:31 comment added Qiaochu Yuan $BG$ is fundamentally an object which only makes sense up to weak homotopy equivalence; asking questions like this means getting bogged down in the technicalities of spaces not having the homotopy type of a CW complex and there's just no reason to torture yourself like that.
Mar 29, 2015 at 1:49 comment added John Klein @user46652: you need to specify which classifying space functor $B$ you are using for the question to make any sense.
Mar 29, 2015 at 1:43 comment added Denis Nardin @user46652 If I understand what you're asking, yes if $G$ has the homotopy type of a CW complex so does $BG$ (because after all it can be constructed using the bar construction).
Mar 29, 2015 at 0:48 comment added Dan Ramras (cont'd) I think your conditions on $G$ and $H$ ought to imply that these nerves are good simplicial spaces. But I'm not certain if that's ensure that a level-wise homotopy equivalence induces a homotopy equivalence of realizations. Goerss and Jardine's book would be one place to look.
Mar 29, 2015 at 0:46 comment added Dan Ramras You might try looking at Segal's paper Classifying Spaces and Spectral Sequences, where he explains that Milnor's infinite join construction of $BG$ can be built as the classifying space of a certain topological category $G_N$. Then induced functor $G_N \to H_N$ will induce a level-wise homotopy equivalence between the nerves of these cateogies (whose realizations give $BG$ and $BH$)...
Mar 29, 2015 at 0:43 comment added user46652 Thanks for the comment! This means in order for my very last statement to work, $G$ would have to be a lot nicer, so I guess showing $BG$ has the homotopy type of a CW complex is not the right way to show that $Bf$ is a homotopy equivalence. Nonetheless, do you think the converse to your comment hold somehow?
Mar 29, 2015 at 0:01 comment added Denis Nardin If $BG$ has the homotopy type of a CW complex then $G=\Omega BG$ has it too.
Mar 28, 2015 at 23:32 review First posts
Mar 28, 2015 at 23:56
Mar 28, 2015 at 23:32 history asked user46652 CC BY-SA 3.0