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Nov 25, 2019 at 21:25 vote accept pyroscepter
Nov 25, 2019 at 0:53 comment added Student I cannot see.. I cannot even prove $B\mathbb{Z}$ is $S^1$..
Nov 24, 2019 at 14:48 history edited Paolo Perrone CC BY-SA 4.0
added higher simplices, thanks to the commenters!
Nov 24, 2019 at 14:42 comment added Paolo Perrone @AndreasBlass oops, the higher simplices. Let me correct. Thanks!
Nov 23, 2019 at 23:05 comment added JK34 To expand the point of @AndreasBlass, $BG$ is in fact the nerve as I described above, and thus contains simplecies of arbitrary dimension.
Nov 23, 2019 at 21:04 comment added user44191 "Given $g \in G$, also $g^{-1}$ is in $G$. Don't add an extra path for $g^{-1}$, just imagine that your loops can be walked either way." It might be helpful to instead include those paths separately, but point out that by the later construction, you can construct a homotopy between $gg^{-1}$ and the trivial loop - and so the new object will "look like" the two paths ($g$ reversed and $g^{-1}$) are the same.
Nov 23, 2019 at 20:55 comment added Andreas Blass Your description of $BG$ includes only simplices of dimension $\leq2$. In general, you'll need some higher-dimensional simplices to make all the higher homotopy vanish.
Nov 23, 2019 at 20:38 history answered Paolo Perrone CC BY-SA 4.0