I'm a student learning the basics of working with the unbounded derived category $D(\mathcal{A})$. I arrived at the natural question, "is every K-projective complex formed out of projective objects?" (having learned that this isn't sufficient to guarantee K-projectivity), and a literature hunt led to the introduction of the paper "The Direct Limit Closure of Perfect Complexes" by Christensen and Holm (http://arxiv.org/pdf/1301.0731.pdf), where the authors mention K-projective and formed out of projective objects is equivalent to DG-projective (in the first paragraph).

I think I managed to prove this, but I still have two questions:

i. I gather that this last statement can be found in Avramov-Foxby-Halperin's Differential Graded Homological Algebra, but I couldn't locate a copy of this on the web. Does anyone know where I could obtain a copy of this (preferably digital)?

ii. With the result, it seems like there should be complexes which are K-projective but not DG-projective, but I have been unable to build one. Is there a reference where one is built/does anyone know of such an example (my hunch is that Avramov-Foxby-Halperin should be said reference)?

Many thanks!


3 Answers 3


K-projectivity of a complex is a property of its homotopy equivalence class, i.e., any complex homotopy equivalent to a K-projective complex is K-projective. In particular, any contractible complex is K-projective. Taking the cone of the identity endomorphism of any complex that is not formed out of projective objects thus provides an example of a K-projective complex that is not formed out of projective objects.


In response to your search for the paper by Avramov Foxby and Halperin, as of 9.01.2003 it is still a preprint entitled "differential graded homological algebra", and the second page reads:

FRIENDLY REMINDER What follows is a partially proofread and completely preliminary version of a paper, which has been in preparation for a number of years. Any user of these notes is expected to adhere to the following conditions: • No further circulation will be made without an author’s consent. • Corrections, comments, and suggestions will be forwarded to an author. • Results will be used at one’s own risk.

I would recommend finding a copy however as it does have a lot of information on the kind of issue you asked about. I'm happy to say a bit more on who you could (respectuflly!) ask for a copy, if you email me at smp12tbs@sheffield.ac.uk




Here is an example (which is due to A. Dold) Consider the complex $A$ of free $\mathbb{Z}/4$-modules: $$\ldots\stackrel{2}{\rightarrow}\mathbb{Z}/4\stackrel{2}{\rightarrow}\mathbb{Z}/4\stackrel{2}{\rightarrow}\mathbb{Z}/4\stackrel{2}{\rightarrow}\mathbb{Z}/4\stackrel{2}{\rightarrow}\ldots $$ It is acyclic, thus the morphism $A\rightarrow 0$ into the zero chain complex is a quasi-isomorphism if this complex were $\mathbb{Z}/4$-DG-projective it would be a homotopy equivalence. But when tensoring with $\mathbb{Z}/2$ we get the complex $$\ldots\stackrel{0}{\rightarrow}\mathbb{Z}/2\stackrel{0}{\rightarrow}\mathbb{Z}/2\stackrel{0}{\rightarrow}\mathbb{Z}/2\stackrel{0}{\rightarrow}\mathbb{Z}/2\stackrel{0}{\rightarrow}\ldots $$ which is not acyclic, thus $A$ is not DG-projective.

  • $\begingroup$ Hey thanks for the response! I'm a beginner, so it's very possible I'm goofing something up, but: I don't think $A$ is even $K$ projective ($Hom_{\mathcal{K}(\mathbb{Z}/4)}(A,A) \simeq \mathbb{Z}/2 \neq 0$, or alternatively use your $\otimes \mathbb{Z}/2$ along with $K$ projective implies $K$ flat) $\endgroup$ Jan 28, 2013 at 14:36
  • $\begingroup$ You are right! It does not answer your question. $\endgroup$
    – David C
    Jan 28, 2013 at 15:56

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