Timeline for Restriction of "$\pi_{1}$" to topological groups
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 28, 2014 at 11:15 | comment | added | nsrt | If G and H are topological abelian groups, then they are products of Eilenberg-MacLane spaces. So G has a "projection to the $\pi_1$ factor" and H has the "inclusion of the $\pi_1$ factor" i.e. the map you are asking for exists as a composition $G\to K(\pi_1G,1)\stackrel{K(\phi,1)}{\to} K(\pi_1H,1)\to H$. | |
| Feb 27, 2014 at 20:23 | comment | added | Ali Taghavi | @nsrt A group homomorphism between product group is not necessarily a product of group homomorphism. Is not this statement an obstruction for your argument? | |
| Feb 26, 2014 at 11:30 | comment | added | nsrt | Topological abelian groups are products of Eilenberg-MacLane spaces so the answer is yes. | |
| Feb 26, 2014 at 6:47 | comment | added | Mark Grant | For path connected abelian Lie groups the answer is yes (because they are all $K(\pi,1)$s). I don't know enough about general topological groups to answer your further question. In fact, I can't think of an example of a path connected abelian topological group which isn't a $K(\pi,1)$. | |
| Feb 25, 2014 at 18:21 | comment | added | Ali Taghavi | thank you very much for your interesting answer. what about for further assumption "G and H are path connected abelian topological group"? | |
| Feb 25, 2014 at 18:17 | vote | accept | Ali Taghavi | ||
| Feb 24, 2014 at 15:43 | history | answered | Mark Grant | CC BY-SA 3.0 |