Timeline for What is π_1(BG) for an arbitrary topological group $G$?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 25, 2011 at 17:28 | answer | added | Ronnie Brown | timeline score: 2 | |
| Oct 12, 2010 at 14:25 | answer | added | John Rognes | timeline score: 15 | |
| Oct 11, 2010 at 13:49 | vote | accept | Jeremy Brazas | ||
| Oct 9, 2010 at 22:27 | comment | added | André Henriques | In general, there is no map $\pi_1(BG) \to \pi_0(G)$. See my post below. | |
| Oct 9, 2010 at 22:05 | answer | added | André Henriques | timeline score: 33 | |
| Oct 9, 2010 at 21:42 | comment | added | some guy on the street | @Jeremy, that it's a homomorphism looks fairly obvious to me --- but don't ask me why... @Fei, following on Jeremy's remark, the proofs (e.g. of long exact sequence) are usually presented for topological groups of the homotopy type of a CW complex, which is a specialization explicitly excluded by the question as posed. Put another way, it's not clear to me even that the sequence is exact. Jeremy, just how arbitrary can these groups be? Is a real algebraic group with the Zariski topology a case you want to consider? | |
| Oct 9, 2010 at 21:05 | comment | added | Jeremy Brazas | Is it obvious that the function $\pi_{1}(BG)\rightarrow \pi_{0}(G)$ from the LES is a homomorphism? | |
| Oct 9, 2010 at 20:19 | comment | added | Fei YE | Should that be $\pi_0(G)$? Since the universal bundle $EG$ is contractible, then the long exact sequence of homotopy groups break into [1=\pi_1(EG)\to\pi_1(BG)\to \pi_0{G}\to\pi_0(EG)=1.] | |
| Oct 9, 2010 at 17:52 | history | edited | Harry Gindi | CC BY-SA 2.5 |
edited title
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| Oct 9, 2010 at 17:29 | history | asked | Jeremy Brazas | CC BY-SA 2.5 |