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An extension of the Dold-Kan equivalence gives an adjunction between the stable homotopy category and the (unbounded) derived category of abelian groups $SH \rightleftarrows D(Ab)$.

Question 1: Is the right adjoint $D(Ab) \to SH$ faithful?

Question 2: If not, is there a class of objects on which it is faithful (for example compact objects).

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2 Answers 2

up vote 8 down vote accepted

I've found the following somewhat intricate way of answering Q1 in the affirmative. Any complex in $D(Ab)$ quasi-isomorphic to a graded abelian group. Hence, it is enough to consider complexes concentrated in a single degree. Given an abelian group $A$ and $n\in\mathbb Z$, let $(A,n)$ be the abelian group $A$ concentrated in degree $n$. For simplicity, I will use the same notation for the Eilenberg-MacLane spectrum $\Sigma^nHA$. In the derived category we have, $$D(Ab)((A,n),(B,n))=\operatorname{Hom}(A,B),$$ $$D(Ab)((A,n),(B,n+1))=\operatorname{Ext}(A,B),$$ $$D(Ab)((A,n),(B,m))=0\text{ otherwise}.$$ In the stable homotopy category we have the stable Eilenberg-MacLane groups $$SH((A,n),(B,m))=H^{m+k}(A,n+k;B),\quad k>>0.$$ It is well known, since E-ML's "On the groups..." (Annals) that $$SH((A,n),(B,n))=\operatorname{Hom}(A,B),$$ $$SH((A,n),(B,n+1))=\operatorname{Ext}(A,B),$$ and that the functor $D(Ab)\rightarrow SH$ is the identity on the previous morphism sets. Hence we are done. The groups $SH((A,n),(B,m))$ are however non-trivial for $m>n+1$, in general.

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Thanks for your answer Fernando. I was actually initially trying to find a counter-example using this method. Something I don't understand is "Any complex is quasi-isomorphic to a graded abelian group". What do you mean by this? I don't see how to reduce to complexes concentrated in a single degree. –  name Jan 30 '13 at 15:11
A graded abelian group need not be concentrated in a single degree. It's just a complex with trivial differentials. This holds for abelian groups and more generally for complexes of modules over a hereditary ring. The proof is straightforward, see 1.6 in homepages.math.uic.edu/~bshipley/krause.chicago.pdf –  Fernando Muro Jan 30 '13 at 15:40

I think the answer to Question 1 is positive. Think of $SH$ as the homotopy category of modules over the sphere spectrum $S$. The category $D(Ab)$ is equivalent to the homotopy category of modules over the Eilenberg-Mac Lane spectrum $HZ$. Your adjunction is equivalent to the adjunction between $S$-modules and $HZ$ modules, where the right adjoint is pullback along the natural map of ring spectra $S\to HZ$, and the left adjoint is the functor $M\mapsto HZ\wedge M$.

Your question is equivalent to this: given $HZ$-modules $M, N$, is the map

  • $ [M, N]_{HZ}\to [M, N]_S$

injective? By adjunction

$[M,N] = [HZ \wedge M, N]_{HZ}$

and the map * is induced by the map $HZ\wedge M \to M$. I claim that the last map is a split surjection in the homotopy category of $HZ$-modules.

Edited to account for Fernando's comment.

Since every $HZ$-module splits (non-naturally) as a wedge sum of Eilenberg Maclane modules It is enough to check this claim when $M=HA$, in which case it is an easy calculation. The homotopy groups of $HZ\wedge HA$ are isomorphic to the homology groups of $HA$. By Huriewicz theorem, this is $A$ in dimension zero. Using the general splitting result again, it follows that $HA$ is a summand of $HZ\wedge HA$ in the category of $HZ$-modules.

Therefore * is injective.

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Why is it enough to check the case $M=HZ$? Somehow, you're deducing from this case that any $HZ$-module is a retract of an induced $HZ$-module, along the ring spectrum morphism $S\rightarrow HZ$. This may be true for this ring spectrum map (I don't know), but it is false in general (consider simply ring homomorphisms, e.g. from a field $k$ to a $k$-algebra with non-projective modules). Hence, if true, there should be a good reason for $S\rightarrow HZ$ to have this property. –  Fernando Muro Jan 29 '13 at 22:55
+1 You are right. I was originally going to say that it is enough to check it for $M$ an Eilenberg-Maclane spectrum, using the same splitting argument as you did (honest). Then somehow convinced myself in a hurry that I could get away with a categorical argument. But it does not work. The map $HZ \wedge M \to M$ splits, but not naturally. I will edit. –  Gregory Arone Jan 29 '13 at 23:28
Essentially, the problem with making a general categorical argument for this is that the unit map $M\to HZ\wedge M$ is not an $HZ$-module map. Or, to put it another way, the map $HZ\wedge HZ\to HZ$ splits as a map of $HZ$-modules but not as a map of $HZ$-bimodules. –  Eric Wofsey Jan 29 '13 at 23:42
Indeed, if a general categorical argument worked then you could replace $S \to HZ$ by $HZ \to HZ/p^2$ and "prove" that the derived category of $Z/p^2$ embeds into the derived category of $HZ$. –  Tyler Lawson Jan 30 '13 at 5:07

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