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Jun 9, 2011 at 14:52 comment added mikhail skopenkov Maybe even more amazing wrong belief in this field: $\dim(E/G)\le\dim E$ (there are counterexamples by A.N. Kolmogorov)
Mar 6, 2011 at 21:06 comment added Dan Ramras Yes, precisely. It's an odd little example, but helpful when people forget to include the proper conditions...
Mar 6, 2011 at 20:01 comment added Autumn Kent Oh! You're saying that a point is not a classifying space for G with some other topology. I thought you were saying that $G^i/G$ wasn't $BG^i$. Thanks for the clarification!
Mar 6, 2011 at 19:58 history edited Dan Ramras CC BY-SA 2.5
clarified the topology on the group G
Mar 6, 2011 at 19:57 comment added Dan Ramras The group in this example starts out with some topology. (I guess I didn't specify that I was thinking of a topological group.) If G started with the indiscrete topology, then your commment makes sense, and we would have a principal bundle for this indiscrete group. But if G is not indiscrete, then the map $(e_1, e_2) \mapsto g$ is not continuous as a map into the topological group G. The proof that continuity of the translation map forces this to be a principal bundle can be found in Husemoller's book on fiber bundles (it's not hard). Let me know if this didn't answer your questions.
Mar 6, 2011 at 17:52 comment added Autumn Kent I'm a little confused. How does requiring that $(e_1, e_2) \mapsto g$ be continuous fix things? In the indiscrete case, this map is continuous (since every map to the group is). And why isn't $G^i \to G^i/G$ a principal $G^i$--bundle?
Sep 27, 2010 at 1:41 history answered Dan Ramras CC BY-SA 2.5