Is the stable homotopy category idempotent complete? I have not been able to prove it, and the proof for abelian groups seems to strongly rely on looking at elements.
Thanks, Jon
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asked Oct 20, 2012 at 19:09
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8$\begingroup$ 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. $\endgroup$– Akhil MathewOct 21, 2012 at 2:09
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$\begingroup$ 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? $\endgroup$– Jonathan BeardsleyOct 21, 2012 at 17:51
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3$\begingroup$ 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 $h^*(Y)$. $\endgroup$– Akhil MathewOct 21, 2012 at 21:23
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$\begingroup$ Aha of course, that's clever! And yeah, I guess it is equivalent, by precisely your argument above. Thanks again. :-) $\endgroup$– Jonathan BeardsleyOct 21, 2012 at 22:04
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1 Answer
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Yes, this is a standard fact. Given a self-map $e\colon X\to X$, we write $e^{-1}X$ for the telescope of the sequence $X\xrightarrow{e}X\xrightarrow{e}X\xrightarrow{e}\dotsb$ (constructed as the cofibre of a suitable self-map of $\bigvee_{i=0}^\infty X$). If $e$ is idempotent, one can check that the natural map $X\to e^{-1}X\vee (1-e)^{-1}X$ is an equivalence, and that $e$ acts as the identity on the first factor and as zero on the second; in other words, we have a splitting of $e$.
answered Oct 20, 2012 at 19:48
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$\begingroup$ Awesome! Thanks so much. I figured it must be true. That's immensely helpful (since I can apply to recent work on idempotent complete triangulated categories!). $\endgroup$ Oct 21, 2012 at 16:28