I am trying to reproduce to me familiar methods of homological algebra using the language of model categories, but I run into a few small problems. Consider a left exact functor $F: \mathcal{A} \to \mathcal{B}$ between Grothendieck abelian categories and the induced functor $Ch(F): Ch(\mathcal{A}) \to Ch(\mathcal{B})$ on unbounded (or maybe bounded above) chain complex categories.

  • To define $RF$ via model categories, I need $Ch(F)$ to be a right Quillen functor. For this, I seem to need a left adjoint $G: Ch(\mathcal{B}) \to Ch(\mathcal{A})$ which I don't know how to get. Can one somehow pass to chain complexes up to homotopy and use Brown representability? Also, even if I had $G$, wouldn't I need $F$ to preserve injectives to get a Quillen adjunction?

  • In usual homological algebra, one often wants to use acyclic resolutions instead of injective ones (e.g. to prove results like $R(G \circ F) \cong RG \circ RF$ if $F$ sends injectives to not-necessarily injective $G$-acyclic objects). Is there a way to do this in model category language? E.g. is there a model structure on $Ch(\mathcal(A))$ where fibrations are degreewise epimorphisms with $F$-acyclic kernel?

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    $\begingroup$ For the first question, I think the entering wedge you want to exploit is that any such $\operatorname{Ch}(F)$ automatically preserves chain homotopies. On the second, my instincts are always to pass to the category of bicomplexes to have more room to maneuver, rather than stick to just complexes. I assume there is actually a Quillen equivalence between the categories of complexes and bicomplexes, with one direction being the total complex functor? I never think of things in those terms. $\endgroup$ – Hurkyl Sep 9 '17 at 9:50
  • $\begingroup$ If I were you, I would try math.overflow. $\endgroup$ – fauxefox Mar 19 at 1:33

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