MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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 '12 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 '12 at 17:51
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 '12 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 '12 at 22:04
up vote 13 down vote accepted

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 $e$.

share|cite|improve this answer
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!). – Jon Beardsley Oct 21 '12 at 16:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.