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Aug 11, 2012 at 14:30 answer added Peter May timeline score: 3
Aug 10, 2012 at 22:43 comment added Dan Ramras Oscar, thanks. I deleted my comment, since yours addresses the relationship with the infinite join correctly...
Aug 10, 2012 at 22:15 comment added Oscar Randal-Williams @Dan: The infinite join of copies of $G$, $\mathcal{B} G$, has a map down to the infinite join of copies of $*$, which is $\Delta^\infty$. It is different from $BG$, with either the fat or thin realisation (this is all in Segal's paper).
Aug 10, 2012 at 15:09 comment added André Henriques The example I would try to work out in detail is the compact group given by an infinite product of copies of $\mathbb Z/2$. You should also keep in mind that there are two possible ways of interpreting $B_\bullet(pt,G,pt)$: one using the usual geometric realization, and one using the fat geometric realization.
Aug 10, 2012 at 13:15 comment added Mark Grant Looking at Segal's "Classifying spaces and spectral sequences" (section 3) it seems that the bundle map you describe is not locally trivial in general. But perhaps it is when $G$ is compactly-generated weak Hausdorff (a standing assumption in Rudyak's book)?
Aug 10, 2012 at 13:09 comment added some guy on the street Not that I know how to answer in that case, but just to be clear, by "topological group", does it mean strict group?
Aug 10, 2012 at 12:44 comment added Mark Grant I guess you mean theorem 1.65 in chapter IV?
Aug 10, 2012 at 12:21 history asked Ulrich Pennig CC BY-SA 3.0