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constructing a right adjoint to i:eff C ---> mod C
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In "Deriving Auslander's Formula (Theorem 2.2)", I'm trying to construct a right adjoint to inclusion functor $mathsfi:eff~C to mod~C$. I constructed it as follows:
Now, my problem is: "If $Hom_mathsfC(c,-)$ is exact, the latter assignment (above) sends each object to zero object in $mathsfeff~C$. In this case, zero object in $mathsfeff~C$ has many presentations." Is abelianness of $mathsfC$ sufficient for $Hom_mathsfC(c,-)$ to be exact? If yes, how can I resolve this difficulty?
The functor $mathsfj:mod~C to eff~C$ (right adjoint of $mathsfi$) sends each object $f$ to itself if $f$ is in $mathsfeff~C$, and otherwise it sends $f$ $($with presentation $(-,A) to (-,B) to f to 0$$)$ to $mathsfcoker((-,B) to (-,coker(A to B)))$.
Notes and definitions:
$mathsfC$: an abelian category
$mathsf(C^op,Ab)$: the category of functors from $mathsfC^op$ to $mathsfAb$
Finitely presented object: An object $F$ in $mathsf(C^op,Ab)$ is called finitely presented if it has an exact presentation $(-,A) to (-,B) to F to 0$
$mathsfmod~C$ : the full subcategory of $mathsf(C^op,Ab)$ consisting of finitely presented objects
$mathsfeff~C$: the full subcategory of $mathsfmod~C$ consisting of objects $F$ (in definition of "finitely presented object") with $A to B$ an epimorphism in $mathsfC$
category-theory representation-theory abelian-categories adjoint-functors functors
asked Jul 28 at 14:27
math16
465
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In "Deriving Auslander's Formula (Theorem 2.2)", I'm trying to construct a right adjoint to inclusion functor $mathsfi:eff~C to mod~C$. I constructed it as follows:
Now, my problem is: "If $Hom_mathsfC(c,-)$ is exact, the latter assignment (above) sends each object to zero object in $mathsfeff~C$. In this case, zero object in $mathsfeff~C$ has many presentations." Is abelianness of $mathsfC$ sufficient for $Hom_mathsfC(c,-)$ to be exact? If yes, how can I resolve this difficulty?
The functor $mathsfj:mod~C to eff~C$ (right adjoint of $mathsfi$) sends each object $f$ to itself if $f$ is in $mathsfeff~C$, and otherwise it sends $f$ $($with presentation $(-,A) to (-,B) to f to 0$$)$ to $mathsfcoker((-,B) to (-,coker(A to B)))$.
Notes and definitions:
$mathsfC$: an abelian category
$mathsf(C^op,Ab)$: the category of functors from $mathsfC^op$ to $mathsfAb$
Finitely presented object: An object $F$ in $mathsf(C^op,Ab)$ is called finitely presented if it has an exact presentation $(-,A) to (-,B) to F to 0$
$mathsfmod~C$ : the full subcategory of $mathsf(C^op,Ab)$ consisting of finitely presented objects
$mathsfeff~C$: the full subcategory of $mathsfmod~C$ consisting of objects $F$ (in definition of "finitely presented object") with $A to B$ an epimorphism in $mathsfC$
category-theory representation-theory abelian-categories adjoint-functors functors
asked Jul 28 at 14:27
math16
465
What difficulty are you referring to? Also, $hom_C(c, -) : C^op to Ab$ generally isn't exact; presumably the quoted passage (which I can't find in your link) is only meant to apply when it is.
– Hurkyl
Jul 29 at 8:32
@Hurkyl The quoted passage is my problem (it's mine) and it's not in the paper. If Hom_C(c, -) : C^op ---> Ab is exaxt, Obj(eff C) is just zero object of mod C. In this case, I think zero object has many (non isomorphic) presentations.
– math16
Jul 29 at 10:07
The zero object has many nonisomorphic presentations regardless; e.g. $(-, A) to (-, A) to 0 to 0$ is a presentation of zero for every object $A$, or more generally $(-, A) to (-, B) to 0 to 0$ whenever $A to B$ is a split epimorphism. I still don't understand what you refer to by "the difficulty"; the vibe I get from your writing is that you think what you have written has an obvious consequence that is an (apparent) obstacle or paradox or contradiction... but I can't figure out what obvious consequence you have in mind.
– Hurkyl
Jul 29 at 10:14
@Hurkyl You mean that having different (non isomorphic) presentations is not a problem?
– math16
Jul 29 at 10:50
Not in of itself; that's the usual and expected state of affairs.
– Hurkyl
Jul 29 at 14:06
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In "Deriving Auslander's Formula (Theorem 2.2)", I'm trying to construct a right adjoint to inclusion functor $mathsfi:eff~C to mod~C$. I constructed it as follows:
Now, my problem is: "If $Hom_mathsfC(c,-)$ is exact, the latter assignment (above) sends each object to zero object in $mathsfeff~C$. In this case, zero object in $mathsfeff~C$ has many presentations." Is abelianness of $mathsfC$ sufficient for $Hom_mathsfC(c,-)$ to be exact? If yes, how can I resolve this difficulty?
The functor $mathsfj:mod~C to eff~C$ (right adjoint of $mathsfi$) sends each object $f$ to itself if $f$ is in $mathsfeff~C$, and otherwise it sends $f$ $($with presentation $(-,A) to (-,B) to f to 0$$)$ to $mathsfcoker((-,B) to (-,coker(A to B)))$.
Notes and definitions:
$mathsfC$: an abelian category
$mathsf(C^op,Ab)$: the category of functors from $mathsfC^op$ to $mathsfAb$
Finitely presented object: An object $F$ in $mathsf(C^op,Ab)$ is called finitely presented if it has an exact presentation $(-,A) to (-,B) to F to 0$
$mathsfmod~C$ : the full subcategory of $mathsf(C^op,Ab)$ consisting of finitely presented objects
$mathsfeff~C$: the full subcategory of $mathsfmod~C$ consisting of objects $F$ (in definition of "finitely presented object") with $A to B$ an epimorphism in $mathsfC$
category-theory representation-theory abelian-categories adjoint-functors functors
asked Jul 28 at 14:27
math16
465
In "Deriving Auslander's Formula (Theorem 2.2)", I'm trying to construct a right adjoint to inclusion functor $mathsfi:eff~C to mod~C$. I constructed it as follows:
Now, my problem is: "If $Hom_mathsfC(c,-)$ is exact, the latter assignment (above) sends each object to zero object in $mathsfeff~C$. In this case, zero object in $mathsfeff~C$ has many presentations." Is abelianness of $mathsfC$ sufficient for $Hom_mathsfC(c,-)$ to be exact? If yes, how can I resolve this difficulty?
The functor $mathsfj:mod~C to eff~C$ (right adjoint of $mathsfi$) sends each object $f$ to itself if $f$ is in $mathsfeff~C$, and otherwise it sends $f$ $($with presentation $(-,A) to (-,B) to f to 0$$)$ to $mathsfcoker((-,B) to (-,coker(A to B)))$.
Notes and definitions:
$mathsfC$: an abelian category
$mathsf(C^op,Ab)$: the category of functors from $mathsfC^op$ to $mathsfAb$
Finitely presented object: An object $F$ in $mathsf(C^op,Ab)$ is called finitely presented if it has an exact presentation $(-,A) to (-,B) to F to 0$
$mathsfmod~C$ : the full subcategory of $mathsf(C^op,Ab)$ consisting of finitely presented objects
$mathsfeff~C$: the full subcategory of $mathsfmod~C$ consisting of objects $F$ (in definition of "finitely presented object") with $A to B$ an epimorphism in $mathsfC$
category-theory representation-theory abelian-categories adjoint-functors functors
asked Jul 28 at 14:27
math16
465
edited Jul 28 at 15:16
asked Jul 28 at 14:27
math16
465
asked Jul 28 at 14:27
math16
465
asked Jul 28 at 14:27
math16
465
465
What difficulty are you referring to? Also, $hom_C(c, -) : C^op to Ab$ generally isn't exact; presumably the quoted passage (which I can't find in your link) is only meant to apply when it is.
– Hurkyl
Jul 29 at 8:32
@Hurkyl The quoted passage is my problem (it's mine) and it's not in the paper. If Hom_C(c, -) : C^op ---> Ab is exaxt, Obj(eff C) is just zero object of mod C. In this case, I think zero object has many (non isomorphic) presentations.
– math16
Jul 29 at 10:07
The zero object has many nonisomorphic presentations regardless; e.g. $(-, A) to (-, A) to 0 to 0$ is a presentation of zero for every object $A$, or more generally $(-, A) to (-, B) to 0 to 0$ whenever $A to B$ is a split epimorphism. I still don't understand what you refer to by "the difficulty"; the vibe I get from your writing is that you think what you have written has an obvious consequence that is an (apparent) obstacle or paradox or contradiction... but I can't figure out what obvious consequence you have in mind.
– Hurkyl
Jul 29 at 10:14
@Hurkyl You mean that having different (non isomorphic) presentations is not a problem?
– math16
Jul 29 at 10:50
Not in of itself; that's the usual and expected state of affairs.
– Hurkyl
Jul 29 at 14:06
add a comment |Â
What difficulty are you referring to? Also, $hom_C(c, -) : C^op to Ab$ generally isn't exact; presumably the quoted passage (which I can't find in your link) is only meant to apply when it is.
– Hurkyl
Jul 29 at 8:32
@Hurkyl The quoted passage is my problem (it's mine) and it's not in the paper. If Hom_C(c, -) : C^op ---> Ab is exaxt, Obj(eff C) is just zero object of mod C. In this case, I think zero object has many (non isomorphic) presentations.
– math16
Jul 29 at 10:07
The zero object has many nonisomorphic presentations regardless; e.g. $(-, A) to (-, A) to 0 to 0$ is a presentation of zero for every object $A$, or more generally $(-, A) to (-, B) to 0 to 0$ whenever $A to B$ is a split epimorphism. I still don't understand what you refer to by "the difficulty"; the vibe I get from your writing is that you think what you have written has an obvious consequence that is an (apparent) obstacle or paradox or contradiction... but I can't figure out what obvious consequence you have in mind.
– Hurkyl
Jul 29 at 10:14
@Hurkyl You mean that having different (non isomorphic) presentations is not a problem?
– math16
Jul 29 at 10:50
Not in of itself; that's the usual and expected state of affairs.
– Hurkyl
Jul 29 at 14:06
What difficulty are you referring to? Also, $hom_C(c, -) : C^op to Ab$ generally isn't exact; presumably the quoted passage (which I can't find in your link) is only meant to apply when it is.
– Hurkyl
Jul 29 at 8:32
What difficulty are you referring to? Also, $hom_C(c, -) : C^op to Ab$ generally isn't exact; presumably the quoted passage (which I can't find in your link) is only meant to apply when it is.
– Hurkyl
Jul 29 at 8:32
@Hurkyl The quoted passage is my problem (it's mine) and it's not in the paper. If Hom_C(c, -) : C^op ---> Ab is exaxt, Obj(eff C) is just zero object of mod C. In this case, I think zero object has many (non isomorphic) presentations.
– math16
Jul 29 at 10:07
@Hurkyl The quoted passage is my problem (it's mine) and it's not in the paper. If Hom_C(c, -) : C^op ---> Ab is exaxt, Obj(eff C) is just zero object of mod C. In this case, I think zero object has many (non isomorphic) presentations.
– math16
Jul 29 at 10:07
The zero object has many nonisomorphic presentations regardless; e.g. $(-, A) to (-, A) to 0 to 0$ is a presentation of zero for every object $A$, or more generally $(-, A) to (-, B) to 0 to 0$ whenever $A to B$ is a split epimorphism. I still don't understand what you refer to by "the difficulty"; the vibe I get from your writing is that you think what you have written has an obvious consequence that is an (apparent) obstacle or paradox or contradiction... but I can't figure out what obvious consequence you have in mind.
– Hurkyl
Jul 29 at 10:14
The zero object has many nonisomorphic presentations regardless; e.g. $(-, A) to (-, A) to 0 to 0$ is a presentation of zero for every object $A$, or more generally $(-, A) to (-, B) to 0 to 0$ whenever $A to B$ is a split epimorphism. I still don't understand what you refer to by "the difficulty"; the vibe I get from your writing is that you think what you have written has an obvious consequence that is an (apparent) obstacle or paradox or contradiction... but I can't figure out what obvious consequence you have in mind.
– Hurkyl
Jul 29 at 10:14
@Hurkyl You mean that having different (non isomorphic) presentations is not a problem?
– math16
Jul 29 at 10:50
@Hurkyl You mean that having different (non isomorphic) presentations is not a problem?
– math16
Jul 29 at 10:50
Not in of itself; that's the usual and expected state of affairs.
– Hurkyl
Jul 29 at 14:06
Not in of itself; that's the usual and expected state of affairs.
– Hurkyl
Jul 29 at 14:06
add a comment |Â
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What difficulty are you referring to? Also, $hom_C(c, -) : C^op to Ab$ generally isn't exact; presumably the quoted passage (which I can't find in your link) is only meant to apply when it is.
– Hurkyl
Jul 29 at 8:32
@Hurkyl The quoted passage is my problem (it's mine) and it's not in the paper. If Hom_C(c, -) : C^op ---> Ab is exaxt, Obj(eff C) is just zero object of mod C. In this case, I think zero object has many (non isomorphic) presentations.
– math16
Jul 29 at 10:07
The zero object has many nonisomorphic presentations regardless; e.g. $(-, A) to (-, A) to 0 to 0$ is a presentation of zero for every object $A$, or more generally $(-, A) to (-, B) to 0 to 0$ whenever $A to B$ is a split epimorphism. I still don't understand what you refer to by "the difficulty"; the vibe I get from your writing is that you think what you have written has an obvious consequence that is an (apparent) obstacle or paradox or contradiction... but I can't figure out what obvious consequence you have in mind.
– Hurkyl
Jul 29 at 10:14
@Hurkyl You mean that having different (non isomorphic) presentations is not a problem?
– math16
Jul 29 at 10:50
Not in of itself; that's the usual and expected state of affairs.
– Hurkyl
Jul 29 at 14:06