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


This probably just technical, but anyway:

In "Simplicial Homotopy Theory" by Goerss and Jardine, chap. III, par. 2, after cor. 2.12, they describe a model structure on $Ch^{+}$, the category of chain complexes in non-negative degress. Equivalences are quasi-isom., fibrations are those maps which are surjective in degrees $n \ge 1$ (notice the $1$!).

Then they note: "After the fact, it turns out that the cofibrations are those monomorphisms of chain complexes having degreewise projective cokernels".

It seems this is not correct as stated. Maybe if it would say $n \ge 0$ upstairs it would be correct, but we do have $n \ge 1$ (corresponding to the picture of simplicial abelian groups).

Is it a typo or I don't understand something? Is there a nice description of cofibrations in this case?

Thank you, Sasha

Update: I was just confused with homological/cohmological indexing conventions. One needs either chains in positive degrees, or cochains in negative degrees. I by mistake used positive degrees as they, but with cochains - by habit.

share|improve this question
add comment

2 Answers

up vote 5 down vote accepted

The definition is correct. (That is, it correctly describes a model structure, and this is a commonly used model structure.)

Note that for nonnegatively graded chain complexes there is no chance of requiring all fibrations to be surjective in degree zero because (given the requirement that every map can be factored as fibration composed with weak equivalence) this would force all maps to induce surjections in $H_0$.

For chain complexes graded by $\mathbb Z$ there is indeed a model structure with surjections as the fibrations. Again, the cofibrations work out to be the monomorphisms with degreewise projective cokernel.

All of this generalizes to chain complexes of $R$-modules, except that in the $\mathbb Z$-graded case it turns out that for general $R$ not every monomorphism with projective cokernel is a cofibration.

Your mention of simplicial abelian groups reminds me to point out that Dold-Kan gives an equivalence of categories, so that a model structure on simplicial groups corresponds precisely to such a structure on nonnegatively graded chain complexes. This also generalizes to $R$-modules rather than abelian groups. The relevant model structure on simplicial modules is such that a map is a fibration [resp. weak equivalence] if and only if the forgetful functor to simplicial sets takes it to a fibration [resp. weak equivalence].

share|improve this answer
I understand that the model in which the fibrations are surjections in degree $n\ge 1$ is the correct one. What I wanted to ask is that the consequent claim that cofibrant objects are complexes with projective terms seems to be wrong. For example, complex with P in degrees 0,1 and 0 otherwise, is not cofibrant; we have a fibration from complex with P in degrees 1,2 onto it, and this fibration does not admit a section. Maybe I am wrong in something? –  Sasha Apr 1 '12 at 22:44
I believe that the "fibration" of one complex "onto" another that you are thinking of is not a chain map. Or maybe I am misunderstanding you. Certainly a complex $D$ that is free of rank one in degrees $0$ and $1$ (and boundary map an isomorphism) and $0$ in other degrees has the left lifting property with respect to maps that are surjective in degree one. Because maps $D\to X$ correspond naturally with elements of $X_1$. –  Tom Goodwillie Apr 1 '12 at 23:03
Oh, very sorry for the confusion! I was thinking about cohomologically indexed complexes in positive degrees - the wrong thing. A stupid mistake. –  Sasha Apr 2 '12 at 7:44
add comment

This is the same way that the fibrations are described in this paper by Dwyer and Spalinski. Section 7 contains the relevant material. Here the cofibrations are monomorphisms in degrees $n \geq 0$ with projective cokernal.

share|improve this answer
Thank you for the nice reference. –  Sasha Apr 2 '12 at 7:52
add comment

Your Answer


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.