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Aug 21, 2021 at 1:20 vote accept geometricK
Aug 20, 2021 at 15:58 comment added Z. M It seems to me that you are interested in the case that $G$ is endowed with a topology. If $G$ is a condensed group (given by, say, a Lie group), then $BG=[*/G]$ is a condensed anima and the usual "classifying space" is referring to its homotopy type. See Peter Scholze's Lecture. This point of view is heavily adopted in Fargues-Scholze.
Aug 20, 2021 at 14:36 answer added IJL timeline score: 2
Aug 13, 2021 at 17:10 comment added Fernando Muro 1. Yes, you have different functorial constructions for both spaces. Not the one indicated above, though. 2. Yes, the product total space clearly satisfies the properties characterizing the universal space for proper actions so you get a homotopy equivalence. I don't know if you get a homeomorphism for some functorial construction, though.
Aug 13, 2021 at 15:42 comment added archipelago $\underline{B}G\not\simeq BG$
Aug 13, 2021 at 8:45 comment added Konrad Waldorf This is model-dependent. If you construct $BG$ and $EG$ via geometric realization of the groupoids $G \Rightarrow \ast$ and $G \times G \Rightarrow G$, respecitvely, then everything should be fine. See Segal's paper "Classifying spaces and spectral sequences".
Aug 13, 2021 at 6:56 review Close votes
Aug 18, 2021 at 3:03
Aug 13, 2021 at 2:00 comment added Kapil The constructions are not functorial. However, given $f:G\to H$ we can always replace $EG$ with $EG\times EH$ with the diagonal action. Now, there will be a projection map etc.
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Aug 13, 2021 at 0:06 history asked geometricK CC BY-SA 4.0