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

Let $\mathscr{C}$ be a DG category (not much will be lost if you assume that $\mathscr{C}$ has one object, i.e. is a DG algebra). One way to construct the unbounded derived category of $\mathscr{C}$-modules is by using semi-free resolutions. Recall that a DG module $F$ is free if it is a sum of shifts of corepresentable modules, or semi-free if it has an exhaustive filtration $0 = F_0 \subset F_1 \subset \cdots \subset F$ such that the subquotients $F_{n+1}/F_n$ are free.

I've seen the existence of semi-free resolutions stated as follows: for any DG module $M$, there is a semi-free module $F$ and a surjective quasi-isomorphism $F \to M$. But how does one actually construct $F$? I'm especially confused about how to make $F \to M$ surjective, since generators of free modules are cycles.

Edit: I found a construction of $F \to M$ in Drinfeld's paper "DG quotients of DG categories." Start by choosing a free module $F_1$ and a morphism $F_1 \to M$ which is surjective on cohomology. Now induct: given $F_n \to M$, let $C_n = \text{Cone}(F_n \to M)$ then find a free module $P_n$ and a morphism $P_n \to C_n$ which is surjective on cohomology. Using the map $C_n \to F_n[1]$ we get $P_n \to F_n[1]$, and Drinfeld takes $F_{n+1} = \text{Cone}(P_n[-1] \to F_n)$. Of course $F$ is the direct limit of the $F_n$.

This is nice, but I still don't understand how to make $F \to M$ surjective. Does this happen automatically for the construction given above, or do we have to do something extra?

share|improve this question
    
Do you mean the references where you've seen that statement do not provide proofs nor refer to other papers where the statement is proven? –  Fernando Muro Dec 4 '13 at 21:46
    
@Fernando: That's right. –  Justin Campbell Dec 4 '13 at 21:54
    
The keyword is bar-resolution. –  Sasha Dec 5 '13 at 9:12
1  
Sasha, could you elaborate? I don't see how to write down a semi-free resolution using the bar construction. –  Justin Campbell Dec 5 '13 at 18:54
    
What do you need surjectivity for anyway? –  Qiaochu Yuan Dec 5 '13 at 22:47

1 Answer 1

up vote 4 down vote accepted

There is a recent paper of Tobi Barthel, Emily Riehl, and myself that answers this question in a model categorical framework. We were lazy and only considered the one object case, although we believe our work generalizes to DG categories. In another respect we were not lazy: we work over a general commutative ring $R$, not just a field, and this introduces interesting subtleties. The reference is

http://front.math.ucdavis.edu/1310.1159

The standard model structure on (unbounded) DG modules over a DG $R$-algebra $A$ takes quasi-isomorphisms as weak equivalences. Cofibrant approximations are more general than semi-free resolutions, but they are retracts of semi-free resolutions given by model theoretic cellular DG modules. There is an early construction by my adviser, John Moore, in the 1959-60 Cartan seminar, which we modernize. This uses bicomplexes, as usual in differential homological algebra. There is another construction, originally due to Gugenheim and myself, dating from 1974, which we also modernize. Qiaochu, you will be interested that we drop surjectivity in that construction, meaning that we do not have model theoretic fibrations. We use multicomplexes, which are bigraded but have differentials that are sums of pieces that mimic differentials in spectral sequences. The drop of surjectivity and the use of multicomplexes gives a great gain of computability, as we illustrate by modernizing 1974 applications to the computation of the cohomology of many homogeneous spaces.

We also explain the role of the bar construction. When R is a field it constructs cofibrant approximations in the standard model structure. In general, it constructs cofibrant approximations in a relative model structure for which the weak equivalences are the maps of DG $A$-modules which are chain homotopy equivalences of underlying $R$-modules, and in that generality it is not semi-free.

share|improve this answer
    
Thanks for the reference, I'll have a look at that paper and probably talk to Tobi too. I have a question about your answer: "Cofibrant approximations are more general than semi-free resolutions, but they are retracts of semi-free resolutions given by model theoretic cellular DG modules." But the category of semi-free modules has all homotopy colimits, including retracts, right? –  Justin Campbell Dec 6 '13 at 4:35
    
No, just as projective modules are retracts of free modules but not necessarily free, retracts of semi-free need not be semi-free. –  Peter May Dec 6 '13 at 5:02
    
Ah, I guess I was thinking that a retraction of a semi-free module is quasi-isomorphic to a semi-free module, but duh, every module is quasi-isomorphic to a semi-free module. –  Justin Campbell Dec 6 '13 at 5:28
    
Cool, the construction I need is contained in the proof of Proposition 8.8. –  Justin Campbell Dec 6 '13 at 6:40
    
Just to emphasize, that result and that section modernize the paper of Moore that I cited originally and the classic memoir ``Foundations of relative homological algebra'' by Eilenberg and Moore, from 1966. –  Peter May Dec 6 '13 at 15:25

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.