Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).

share|improve this question
add comment

2 Answers

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.

share|improve this answer
    
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
add comment

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.

share|improve this answer
3  
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
2  
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
add comment

Your Answer

 
discard

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.