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Jul 19, 2021 at 9:19 comment added Lao-tzu I think since projective abelian groups are just free, you mention about filtered colimits is not about projectives, but to view infinite direct sums as a filtered colimit of finite direct sums.
Feb 2, 2012 at 23:39 comment added Marc Hoyois Yes, but it suffices to know that $H$ has these properties: $(colim MA_i)\wedge HZ=colim (MA_i\wedge HZ)=colim HA_i=H(colim A_i)$, hence $colim MA_i=M(colim A_i)$.
Feb 2, 2012 at 21:28 comment added Jonathan Beardsley Been thinking about this a little more. You seem to imply that what you say following "For example..." follows from the fact that $H$ preserves sums and filtered colimits. It seems that what we really need is the functor $M:A\to M(A,0)$ which takes $A$ to its Moore spectrum, to commute with filtered colimits, at least in this case, since we're replacing $\mathbb{Z}$ with $\mathbb{S}$, not $H\mathbb{Z}$. Is this correct?
Feb 2, 2012 at 13:41 history edited Marc Hoyois CC BY-SA 3.0
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Feb 2, 2012 at 5:52 comment added Marc Hoyois The fact about tensor products (which I just edited, I was being a bit hasty) is in EKMM, Theorem 2.1. I can't think of a reference for the others, but the fact about sums and filtered colimits is seen just by looking at homotopy groups, and the fact about exact sequences follows from the fact that $HA$ represents ordinary (co)homology which transforms s.e.s. of coefficients into l.e.s.
Feb 2, 2012 at 5:35 history edited Marc Hoyois CC BY-SA 3.0
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Feb 2, 2012 at 3:26 vote accept Jonathan Beardsley
Feb 2, 2012 at 3:25 comment added Jonathan Beardsley Wow thanks so much. Ultimately my question really concerned the behavior of the functor H, and what you told me makes that really clear! Is there a good reference that proves that fact?
Feb 2, 2012 at 2:31 history answered Marc Hoyois CC BY-SA 3.0