Derived functors of abelian categories via model categories.How are injective model structures cofibrantly...
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Derived functors of abelian categories via model categories.
How are injective model structures cofibrantly generated?The projective model structure on chain complexesThe hyper-derived functors $mathbb L_bullet F$ are just derived functors of $H_0F$?Is the Quillen Injective Model Structure on the category of positive cochain complexes of R-modules (co)fibrantly generated?Where to learn about model categories?Are the model structure(s) on chain complexes and the triangulated structure interchangeable, or complementary?Composition of three right derived functorsDerived functors and induced functorsWhy are there not $4$ kinds of derived functors/Enough injectives in the category of chain complexes
$begingroup$
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?
category-theory homological-algebra abelian-categories model-categories
edited Mar 14 at 12:55
Arnaud D.
16.2k52444
asked Sep 8 '17 at 10:46
Johann HaasJohann Haas
70329
$endgroup$
add a comment |
$begingroup$
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?
category-theory homological-algebra abelian-categories model-categories
edited Mar 14 at 12:55
Arnaud D.
16.2k52444
asked Sep 8 '17 at 10:46
Johann HaasJohann Haas
70329
$endgroup$
1
$begingroup$
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
add a comment |
$begingroup$
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?
category-theory homological-algebra abelian-categories model-categories
edited Mar 14 at 12:55
Arnaud D.
16.2k52444
asked Sep 8 '17 at 10:46
Johann HaasJohann Haas
70329
$endgroup$
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?
category-theory homological-algebra abelian-categories model-categories
category-theory homological-algebra abelian-categories model-categories
edited Mar 14 at 12:55
Arnaud D.
16.2k52444
asked Sep 8 '17 at 10:46
Johann HaasJohann Haas
70329
edited Mar 14 at 12:55
Arnaud D.
16.2k52444
asked Sep 8 '17 at 10:46
Johann HaasJohann Haas
70329
edited Mar 14 at 12:55
Arnaud D.
16.2k52444
edited Mar 14 at 12:55
Arnaud D.
16.2k52444
edited Mar 14 at 12:55
Arnaud D.
16.2k52444
16.2k52444
asked Sep 8 '17 at 10:46
Johann HaasJohann Haas
70329
asked Sep 8 '17 at 10:46
Johann HaasJohann Haas
70329
asked Sep 8 '17 at 10:46
Johann HaasJohann Haas
70329
70329
1
$begingroup$
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
add a comment |
1
$begingroup$
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
1
1
$begingroup$
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
$begingroup$
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
$begingroup$
If I were you, I would try math.overflow.
$endgroup$
– fauxefox
Mar 19 at 1:33
add a comment |
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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