MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


The Quillen model structure on spaces has weak equivalences given by the weak homotopy equivalences and the fibrations are the Serre fibrations. The cofibrations are characterized by the lifting property, but in the end they are those inclusions which are built up by cell attachments (or are retracts of such things).

The Strøm model structure on spaces has weak equivalences = the homotopy equivalences and fibrations the Hurewicz fibrations. Cofibrations the closed inclusions which satisfy the homotopy extension property.

The projective model structure on non-negatively graded chain complexes over $\Bbb Z$ has fibrations given by the degreewise surjections and weak equivalences given by the quasi-isomorphisms. Cofibrations are given by the degree-wise split inclusions such that the quotient complex is degree-wise free.

From the above, it would appear to me that the projective model structure on chain complexes is analogous to the Quillen model structure on spaces.


Is there a model structure on (non-negatively graded) chain complexes over $\Bbb Z$ for which the weak equivalences are the chain homotopy equivalences?

(Extra wish: I want cofibrations in the projective model structure to be a sub-class of the cofibrations in the model structure answering the question. Conjecturally, they should be the inclusions satisfying the chain homotopy extension property.)

share|cite|improve this question
There is the paper "The homotopy category of chain complexes is a homotopy category" by Golasiński, Marek; Gromadzki, Grzegorz, It treats all complexes, not necessarily bounded on any side but should give what you want. Unfortunately, the paper seems not to be available online, so I cannot check whether your extra wish is fulfilled in that structure without doing some thinking. – Theo Buehler Jan 19 '11 at 12:54
This seems to be the same as this question:… Denis-Charles Cisinski provided a positive answer. – Neil Strickland Jan 19 '11 at 12:55
Thanks Neil, that does what I want. – John Klein Jan 19 '11 at 13:00
+1! Very good question and answer! – Harry Gindi Jan 19 '11 at 14:27
up vote 8 down vote accepted

There are several other papers, I think earlier ones, that cover this.

[32] M. Cole. The homotopy category of chain complexes is a homotopy category. Preprint (1990's)

[29] J. Daniel Christensen and Mark Hovey. Quillen model structures for relative homological algebra. Math. Proc. Cambridge Philos. Soc., 133(2):261–293, 2002.

[121] R. Schw¨anzl and R. M. Vogt. Strong cofibrations and fibrations in enriched categories. Arch. Math. (Basel), 79(6):449–462, 2002.

The references are from the book ``More concise algebraic topology: localizations, completions, and model categories'', by Kate Ponto and myself. It will be published this year by the University of Chicago Press. It includes and compares three natural model structures on spaces and chain complexes, the Strom model structure, the Quillen model structure, and Cole's mixed model structures:

33] Michael Cole. Mixing model structures. Topology Appl., 153(7):1016–1032, 2006.

share|cite|improve this answer
+1 mainly out of excitement for the new book! – Dylan Wilson Feb 4 '11 at 4:06
Thanks Peter. That's probably a definitive list. – John Klein Feb 4 '11 at 11:58

Very recently a paper appeared on the arxiv by Barthel-May-Riehl which addresses this question in a very complete way. It discusses the three model structures on DG-algebras (answering the OPs question and covering the mixed model structure as well), then goes on to give six model structures on DG modules over a DGA. This paper generalizing the references in Peter May's answer above and gives model category foundations for some classical constructions in differential graded algebra.

share|cite|improve this answer

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.