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Sep 13, 2016 at 20:16 history closed Franz Lemmermeyer
Wolfgang
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john mangual
Ryan Budney
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Sep 13, 2016 at 16:27 comment added Qiaochu Yuan The claim is slightly incorrect as stated; either you should take pointed homotopy or you should quotient by conjugation by $G$.
Sep 13, 2016 at 13:08 comment added S. Carnahan Could you fix your link? Even after inserting a colon after the http, Numdam said that the document wasn't found.
Sep 13, 2016 at 13:05 answer added Jens Reinhold timeline score: 2
Sep 13, 2016 at 12:48 comment added Uri Bader Note that the set of pairs of commuting elements in $G$ is in bijections with $\text{Hom}(\mathbb{Z}^2,G)$.
Sep 13, 2016 at 6:29 comment added SashaP This is true only if $G$ is discrete, see math.stackexchange.com/questions/36488/… and use that $G\sim \Omega BG$, so $\pi_1(BG)=\pi_0(G),\pi_2(BG)=\pi_1(G)$.
Sep 13, 2016 at 4:12 review Close votes
Sep 13, 2016 at 20:16
Sep 13, 2016 at 3:43 history asked user88649 CC BY-SA 3.0