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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

flag	asked Oct 20 at 19:09 Jon Beardsley 1,732●3●11

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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. – Akhil Mathew Oct 21 at 2:09

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? – Jon Beardsley Oct 21 at 17:51

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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 $h^*(Y)$. – Akhil Mathew Oct 21 at 21:23

Aha of course, that's clever! And yeah, I guess it is equivalent, by precisely your argument above. Thanks again. :-) – Jon Beardsley Oct 21 at 22:04

Neil Strickland · Accepted Answer · 2012-10-20 19:48:49Z UTC

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 $e$.

Is the stable homotopy category idempotent complete?

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