Timeline for Is the stable homotopy category idempotent complete?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 21, 2012 at 22:04 | comment | added | Jonathan Beardsley | Aha of course, that's clever! And yeah, I guess it is equivalent, by precisely your argument above. Thanks again. :-) | |
| Oct 21, 2012 at 21:23 | comment | added | Akhil Mathew | I suppose, but the latter can be proved purely algebraically. Namely, given an idempotent $e: h \to h$ on a cohomology theory $h$, you can define a new cohomology theory which sends any space (or spectrum) $Y$ to the image of $e$ in $h^*(Y)$. | |
| Oct 21, 2012 at 17:51 | comment | added | Jonathan Beardsley | Hi @Akhil, thanks! However, the statement that the category of cohomology theories is idempotent complete seems to be equivalent to saying that the category of spectra is idempotent complete? | |
| Oct 21, 2012 at 16:25 | vote | accept | Jonathan Beardsley | ||
| Oct 21, 2012 at 2:09 | comment | added | Akhil Mathew | Hi Jon. Another argument for this result (besides the one Neil Strickland gives) is to use Brown representability: given an idempotent $e: X \to X$, then $e$ defines an idempotent in $X$-cohomology and therefore gives another cohomology theory on spaces (or spectra), since the category of cohomology theories is idempotent complete. Now represent this theory by a spectrum. | |
| Oct 20, 2012 at 19:48 | answer | added | Neil Strickland | timeline score: 14 | |
| Oct 20, 2012 at 19:09 | history | asked | Jonathan Beardsley | CC BY-SA 3.0 |