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I have added to the entry split idempotent the statement (here) that in a triangulated category in which the direct sum of two triangles is a triangle, then idempotents split.
(Maybe that should rather go into the entry Cauchy complete category?)
I have used this to write out a full proof (here) of the “Adams universal coefficient theorem”, namely the statement that for $E$ a homotopy ring spectrum such that $E_\bullet(E)$ is flat over $\pi_\bullet(E)$ and $X$ a spectrum such that $E_\bullet(X)$ is projective over $\pi_\bullet(E)$, then
$[X, E \wedge Y]_\bullet \overset{\simeq}{\longrightarrow} Hom^\bullet_{E_\bullet(E)}( E_\bullet(X), E_\bullet(Y) )$is an isomorphism, where on the right we have the graded hom of comodule homomorphisms over the dual $E$-Steenrod algebra.
The proof in Adams 74, p. 323 of this fact invokes the stronger statement that under these conditions and for $Z$ any homotopy $E$-module spectrum then also
$[X, Z]_\bullet \simeq Hom_{\pi_\bullet(E)}( E_\bullet(X), \pi_\bullet(Z) )$is an isomorphism (where what is needed for the above is only the case that $Z$ is a free module spectrum $E \wedge Y$ ), but this stronger statement (which is really what Adams calles the UCT) holds only under much more restrictive conditions. The general proof of the above I am deducing from Schwede 12, chapter II, prop. 6.20, which however at the end is vague about how the conclusion via idempotent splitting in $Ho(Spectra)$ follows. (It looks like there may have been a copy-and-pasting mismatch in the compilation of that proof.) It’s precisely that result by Bökstedt-Neemn in #1 above which concludes that proof.
The proof in Bokstedt-Neeman of idempotent completeness for triangulated categories appeals to the existence and exactness of arbitrary direct sums, not just binary direct sums, since it uses totalizations to construct the splittings. The proposition is false under the weaker hypothesis. Remark 3.2 in Schnurer’s “Homotopy categories and idempotent completeness, weight structures and weight complex functors” gives a counter-example.
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