Timeline for Stable triviality of fiber bundles
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Feb 20, 2012 at 5:52 | vote | accept | Craig Westerland | ||
| Feb 20, 2012 at 2:24 | answer | added | Tyler Lawson | timeline score: 3 | |
| Feb 19, 2012 at 2:22 | comment | added | Craig Westerland | Ah, ok. If I put some finite generation assumption on $B$, then I can rule that out... | |
| Feb 17, 2012 at 2:22 | comment | added | Tyler Lawson |
I think you can let $E = B = colim (Y \times Y \times \cdots \times Y)$ for some space $Y$ with a choice of point that you use to define the colimit, and let $X = \Sigma^\infty_+ Y$. Then the smash product has the desired isomorphism, but the desired conclusion doesn't hold unless $\Sigma^\infty_+ Y \simeq S^0$.
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| Feb 16, 2012 at 21:17 | comment | added | Craig Westerland | Ah, cool! I'm having trouble seeing how you use the Eilenberg swindle here, though... | |
| Feb 16, 2012 at 2:46 | comment | added | Tyler Lawson | Also, if $B$ is nonempty then there are no $\Sigma^\infty B_+$-acyclics, because it has $S^0$ as a retract; in fact, I think that the existence of the map $f$ implies that if $E$ and $B$ are nonempty (and $X$ is connective) you might already be able to find a retract from $X$ to $S^0$. | |
| Feb 16, 2012 at 2:37 | comment | added | Tyler Lawson | I think that if you have $F = *$ you can hit it with an Eilenberg swindle. | |
| Feb 15, 2012 at 20:57 | comment | added | Craig Westerland | Yup! Good point, John. | |
| Feb 15, 2012 at 12:17 | comment | added | John Klein | I think you're also missing a condition on $X$: perhaps $X$ should have $S^0$ as a retract? Otherwise, I don't see how you're going to conclude, as you write above, that $p$ has a stable section. | |
| Feb 15, 2012 at 7:02 | history | edited | Craig Westerland | CC BY-SA 3.0 |
Added locality condition to avoid "obvious" counterexample.
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| Feb 15, 2012 at 6:10 | history | asked | Craig Westerland | CC BY-SA 3.0 |