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1. • CommentRowNumber1.
   • CommentAuthorEmily Riehl
   • CommentTimeDec 30th 2011
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   Author: Emily Riehl
   Format: MarkdownItex(First, an apology: I accidentally posted this in the wrong place, under "MathForge general discussions." I tried to delete it from there but couldn't figure out how to do so. Could someone help me?) My question relates to the entry http://ncatlab.org/nlab/show/derived+functor#InHomologicalAlgebra I'd like to say that $RCh_*(F)$ is a point-set right derived functor of $Ch_*(F)$ but I don't know how one knows that there is a fibrant replacement *functor* on $Ch_*(A)$ (here meaning the category of non-negatively graded cochain complexes), assuming $A$ has enough injectives. Naively of course, one can construct an injective resolution $I_*(A)$ of an object of $A$. Given a map $A \to B$ one can build a map $I_*(A) \to I_*(B)$ but this is only unique up to chain homotopy equivalence and so this construction is not functorial on the point-set level (though of course suffices for defining total derived functors). Presumably this naive construction can be generalized to construct, given a non-negative graded cochain complex $A_*$, a cochain complex $I_*(A_*)$ of injectives together with a quasi-isomorphism $A_* \to I_*(A_*)$ but will similarly fail to be strictly functorial. In the dual situation (non-negatively graded chain complexes) if we take the abelian category $A$ to be modules over some ring, then the appropriate model structure is cofibrantly generated so it's clear there is a functorial cofibrant replacement. But I don't know that the injective model structure is cofibrantly generated, even in this case.

   (First, an apology: I accidentally posted this in the wrong place, under “MathForge general discussions.” I tried to delete it from there but couldn’t figure out how to do so. Could someone help me?)

   My question relates to the entry

   http://ncatlab.org/nlab/show/derived+functor#InHomologicalAlgebra

   I’d like to say that $RCh_*(F)$ is a point-set right derived functor of $Ch_*(F)$ but I don’t know how one knows that there is a fibrant replacement functor on $Ch_*(A)$ (here meaning the category of non-negatively graded cochain complexes), assuming $A$ has enough injectives.

   Naively of course, one can construct an injective resolution $I_*(A)$ of an object of $A$. Given a map $A \to B$ one can build a map $I_*(A) \to I_*(B)$ but this is only unique up to chain homotopy equivalence and so this construction is not functorial on the point-set level (though of course suffices for defining total derived functors). Presumably this naive construction can be generalized to construct, given a non-negative graded cochain complex $A_*$, a cochain complex $I_*(A_*)$ of injectives together with a quasi-isomorphism $A_* \to I_*(A_*)$ but will similarly fail to be strictly functorial.

   In the dual situation (non-negatively graded chain complexes) if we take the abelian category $A$ to be modules over some ring, then the appropriate model structure is cofibrantly generated so it’s clear there is a functorial cofibrant replacement. But I don’t know that the injective model structure is cofibrantly generated, even in this case.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeDec 30th 2011
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   Author: Urs
   Format: MarkdownItexJust a quick comment: in section 5.4 of * Christensen, Hovey, _Quillen model structures for relative homological algebra_ ([pdf](http://jdc.math.uwo.ca/papers/relative.pdf)) are some results on model structures on chain complexes that are provably _not_ cofibrantly generated.

   Just a quick comment: in section 5.4 of

   • Christensen, Hovey, Quillen model structures for relative homological algebra (pdf)

   are some results on model structures on chain complexes that are provably not cofibrantly generated.

3. • CommentRowNumber3.
   • CommentAuthorAndrew Stacey
   • CommentTimeDec 30th 2011
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   Author: Andrew Stacey
   Format: MarkdownItex> First, an apology: I accidentally posted this in the wrong place, under "MathForge general discussions." I tried to delete it from there but couldn't figure out how to do so. Could someone help me? No apology necessary. I know it's confusing to sometimes wind up there, but it is useful for me to have that place. I've deleted the original post.

   First, an apology: I accidentally posted this in the wrong place, under “MathForge general discussions.” I tried to delete it from there but couldn’t figure out how to do so. Could someone help me?

   No apology necessary. I know it’s confusing to sometimes wind up there, but it is useful for me to have that place. I’ve deleted the original post.

4. • CommentRowNumber4.
   • CommentAuthorTobyBartels
   • CommentTimeDec 31st 2011
   • (edited Dec 31st 2011)
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   Author: TobyBartels
   Format: MarkdownItexAnother technical tip: since you're already using a Markdown formatting method, put ASCII angle brackets around a URI to make it live: <http://ncatlab.org/nlab/show/derived+functor#InHomologicalAlgebra>

   Another technical tip: since you’re already using a Markdown formatting method, put ASCII angle brackets around a URI to make it live: http://ncatlab.org/nlab/show/derived+functor#InHomologicalAlgebra

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeJan 1st 2012
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   Author: Mike Shulman
   Format: MarkdownItexIn Hovey's book *Model Categories*, Theorem 2.3.13 says that the injective model structure on chain complexes of modules over a ring is cofibrantly generated. For an arbitrary abelian category with enough injectives, I sure don't see any way to obtain a fibrant replacement functor. However, I'd expect that in many cases arising in practice (e.g. abelian sheaves), you could find a concrete injective-hull functor and thereby also a fibrant replacement functor, without necessarily needing to go through the machinery of cofibrant generation.

   In Hovey’s book Model Categories, Theorem 2.3.13 says that the injective model structure on chain complexes of modules over a ring is cofibrantly generated.

   For an arbitrary abelian category with enough injectives, I sure don’t see any way to obtain a fibrant replacement functor. However, I’d expect that in many cases arising in practice (e.g. abelian sheaves), you could find a concrete injective-hull functor and thereby also a fibrant replacement functor, without necessarily needing to go through the machinery of cofibrant generation.

6. • CommentRowNumber6.
   • CommentAuthorEmily Riehl
   • CommentTimeJan 1st 2012
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   Author: Emily Riehl
   Format: MarkdownItexThanks for these references and formatting tips. And for confirming what I'd suspected. Happy new year!

   Thanks for these references and formatting tips. And for confirming what I’d suspected. Happy new year!

7. • CommentRowNumber7.
   • CommentAuthorDylan Wilson
   • CommentTimeJan 10th 2012
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   Author: Dylan Wilson
   Format: TextI may be imagining this, but it feels like the category of chain complexes on any Grothendieck category should admit a combinatorial model structure that allows for fibrant replacement... This should cover a lot of the examples (if it's true!). This "feeling" extends to a feeling that the proofs of the relevant results should be scattered across the appendix to HTT and the section in HA on Grothendieck abelian categories. Of course, there's always the possibility that I'm missing something.

   I may be imagining this, but it feels like the category of chain complexes on any Grothendieck category should admit a combinatorial model structure that allows for fibrant replacement... This should cover a lot of the examples (if it's true!).
   
   This "feeling" extends to a feeling that the proofs of the relevant results should be scattered across the appendix to HTT and the section in HA on Grothendieck abelian categories.
   
   Of course, there's always the possibility that I'm missing something.