Mixing
Defining Category of Adjunctions
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$newcommandAmathcalA
newcommandBmathcalB
newcommandCmathcalC
newcommandADJmathsfADJ
newcommandidmathrmId$Morphism between a pair of adjunctions $(F : A leftrightarrows B : G)$ and
$(F' : A' leftrightarrows B' : G')$ is a pair of functors $H : A to A'$ and $F : B to B'$ such that $FK = HF'$ and $GH = KG'$, and either
(I) $H(eta_(cdot)) = eta_H(cdot)' $, where $eta$ and $eta'$ are unit natural transforms corresponding to the adjunctions.
(II) $K(epsilon_(cdot)) = epsilon'_K(cdot)$, where $epsilon$ and $epsilon'$ are coint natural transforms corresponding to the adjunctions.
(III) for all $A in A, B in B,f : F(A) to B$, it holds that $H(f^top) = (Kf)^top$, where by $(cdot)^top$ I denote transposition associated with both adjunctions.
Conditions (I),(II),(III) are equivalent as I proved in the exercise recently.
However it seems to me that, as adjunctions have morphisms, they should have their own category:
Consider category $mathsfADJ$ with objects being pairs of categories supplied each with an appropriate adjunction i. e. $(F : A leftrightarrows B : G) in mathsfADJ$ and morphism defined as above. Then, composition is defined by a pairwise functor composition, and identity is a pair of identity functors. Associativity of composition is easily seen from version (III) of definition.
So It seems that $ADJ$ is a category. But in fact it is not as it contains itself in form $(id : ADJ rightleftarrows ADJ : id) in ADJ$. Moreover every category $C$ has a canonical representative $(id : C rightleftarrows C : id) in ADJ$, which means that $ADJ$ contains all categories in certain sense. Both of these are bas things for a set theoretical reasons, thus $ADJ$ should not be defined as a category.
This situation is simillar to the idea of category of categories here so I see two pathways:
1) Restrict $ADJ$ to small categories or locally small ones.
2) Define $ADJ$ as a 2-category by introducing 'nutaral transfoms' as morphisms of morphims.
I personally prefer the latter solution.
Are categories of adjunctons used somewhere? If so, how are these categories defined?
P.S.
While writing this question I took liberty to define command A -> $mathcalA$, B -> $mathcalB$, C -> $mathcalC$, ADJ -> $mathsfADJ$; id ->$mathrmId$. You can use them in your answers and comments.
category-theory definition adjoint-functors higher-category-theory
edited Jul 23 at 12:02
Fabio Lucchini
5,62411025
asked Jul 14 at 22:28
Nik Pronko
795717
add a comment |Â
up vote
3
down vote
favorite
$newcommandAmathcalA
newcommandBmathcalB
newcommandCmathcalC
newcommandADJmathsfADJ
newcommandidmathrmId$Morphism between a pair of adjunctions $(F : A leftrightarrows B : G)$ and
$(F' : A' leftrightarrows B' : G')$ is a pair of functors $H : A to A'$ and $F : B to B'$ such that $FK = HF'$ and $GH = KG'$, and either
(I) $H(eta_(cdot)) = eta_H(cdot)' $, where $eta$ and $eta'$ are unit natural transforms corresponding to the adjunctions.
(II) $K(epsilon_(cdot)) = epsilon'_K(cdot)$, where $epsilon$ and $epsilon'$ are coint natural transforms corresponding to the adjunctions.
(III) for all $A in A, B in B,f : F(A) to B$, it holds that $H(f^top) = (Kf)^top$, where by $(cdot)^top$ I denote transposition associated with both adjunctions.
Conditions (I),(II),(III) are equivalent as I proved in the exercise recently.
However it seems to me that, as adjunctions have morphisms, they should have their own category:
Consider category $mathsfADJ$ with objects being pairs of categories supplied each with an appropriate adjunction i. e. $(F : A leftrightarrows B : G) in mathsfADJ$ and morphism defined as above. Then, composition is defined by a pairwise functor composition, and identity is a pair of identity functors. Associativity of composition is easily seen from version (III) of definition.
So It seems that $ADJ$ is a category. But in fact it is not as it contains itself in form $(id : ADJ rightleftarrows ADJ : id) in ADJ$. Moreover every category $C$ has a canonical representative $(id : C rightleftarrows C : id) in ADJ$, which means that $ADJ$ contains all categories in certain sense. Both of these are bas things for a set theoretical reasons, thus $ADJ$ should not be defined as a category.
This situation is simillar to the idea of category of categories here so I see two pathways:
1) Restrict $ADJ$ to small categories or locally small ones.
2) Define $ADJ$ as a 2-category by introducing 'nutaral transfoms' as morphisms of morphims.
I personally prefer the latter solution.
Are categories of adjunctons used somewhere? If so, how are these categories defined?
P.S.
While writing this question I took liberty to define command A -> $mathcalA$, B -> $mathcalB$, C -> $mathcalC$, ADJ -> $mathsfADJ$; id ->$mathrmId$. You can use them in your answers and comments.
category-theory definition adjoint-functors higher-category-theory
edited Jul 23 at 12:02
Fabio Lucchini
5,62411025
asked Jul 14 at 22:28
Nik Pronko
795717
2
ncatlab.org/nlab/show/2-category+of+adjunctions
– Derek Elkins
Jul 14 at 23:23
1
Working with a category of small categories isn't as restrictive as you might think; when necessary you can switch which universe of sets you are taking to be the universe of small sets. (assuming you take foundations with enough universes)
– Hurkyl
Jul 14 at 23:44
Remark that 2) is no solution to your set-theoretic issue.
– Pece
Jul 15 at 9:40
@DerekElkins Thanks, This is interesting. As I understood In my context, underlying 2-category would be $mathsfCAT$ with categories as object, Adjunctions (of categories) as horizontal arrows, and morphims of adjunctions (as defined above) as vertical arrows.
– Nik Pronko
Jul 15 at 20:52
@Pece I understand that this is not a solution, as 1-skeleton of the category will contain itself. I understand that this is incorrect, that I was thinking about, is that If 0-skeleton of $mathsfSET$ is not a set itself, why can't I have (In certain specific foundations) a 2-category those 1-skeleton is not a category. What I'm trying to Introduce here is some type of structure that are like categories, but not count as categories then you quantify statements other categories. If I could do something like this with k-skeletons of n-categories I would get something akin Russells's hierarchy.
– Nik Pronko
Jul 15 at 21:14
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$newcommandAmathcalA
newcommandBmathcalB
newcommandCmathcalC
newcommandADJmathsfADJ
newcommandidmathrmId$Morphism between a pair of adjunctions $(F : A leftrightarrows B : G)$ and
$(F' : A' leftrightarrows B' : G')$ is a pair of functors $H : A to A'$ and $F : B to B'$ such that $FK = HF'$ and $GH = KG'$, and either
(I) $H(eta_(cdot)) = eta_H(cdot)' $, where $eta$ and $eta'$ are unit natural transforms corresponding to the adjunctions.
(II) $K(epsilon_(cdot)) = epsilon'_K(cdot)$, where $epsilon$ and $epsilon'$ are coint natural transforms corresponding to the adjunctions.
(III) for all $A in A, B in B,f : F(A) to B$, it holds that $H(f^top) = (Kf)^top$, where by $(cdot)^top$ I denote transposition associated with both adjunctions.
Conditions (I),(II),(III) are equivalent as I proved in the exercise recently.
However it seems to me that, as adjunctions have morphisms, they should have their own category:
Consider category $mathsfADJ$ with objects being pairs of categories supplied each with an appropriate adjunction i. e. $(F : A leftrightarrows B : G) in mathsfADJ$ and morphism defined as above. Then, composition is defined by a pairwise functor composition, and identity is a pair of identity functors. Associativity of composition is easily seen from version (III) of definition.
So It seems that $ADJ$ is a category. But in fact it is not as it contains itself in form $(id : ADJ rightleftarrows ADJ : id) in ADJ$. Moreover every category $C$ has a canonical representative $(id : C rightleftarrows C : id) in ADJ$, which means that $ADJ$ contains all categories in certain sense. Both of these are bas things for a set theoretical reasons, thus $ADJ$ should not be defined as a category.
This situation is simillar to the idea of category of categories here so I see two pathways:
1) Restrict $ADJ$ to small categories or locally small ones.
2) Define $ADJ$ as a 2-category by introducing 'nutaral transfoms' as morphisms of morphims.
I personally prefer the latter solution.
Are categories of adjunctons used somewhere? If so, how are these categories defined?
P.S.
While writing this question I took liberty to define command A -> $mathcalA$, B -> $mathcalB$, C -> $mathcalC$, ADJ -> $mathsfADJ$; id ->$mathrmId$. You can use them in your answers and comments.
category-theory definition adjoint-functors higher-category-theory
edited Jul 23 at 12:02
Fabio Lucchini
5,62411025
asked Jul 14 at 22:28
Nik Pronko
795717
$newcommandAmathcalA
newcommandBmathcalB
newcommandCmathcalC
newcommandADJmathsfADJ
newcommandidmathrmId$Morphism between a pair of adjunctions $(F : A leftrightarrows B : G)$ and
$(F' : A' leftrightarrows B' : G')$ is a pair of functors $H : A to A'$ and $F : B to B'$ such that $FK = HF'$ and $GH = KG'$, and either
(I) $H(eta_(cdot)) = eta_H(cdot)' $, where $eta$ and $eta'$ are unit natural transforms corresponding to the adjunctions.
(II) $K(epsilon_(cdot)) = epsilon'_K(cdot)$, where $epsilon$ and $epsilon'$ are coint natural transforms corresponding to the adjunctions.
(III) for all $A in A, B in B,f : F(A) to B$, it holds that $H(f^top) = (Kf)^top$, where by $(cdot)^top$ I denote transposition associated with both adjunctions.
Conditions (I),(II),(III) are equivalent as I proved in the exercise recently.
However it seems to me that, as adjunctions have morphisms, they should have their own category:
Consider category $mathsfADJ$ with objects being pairs of categories supplied each with an appropriate adjunction i. e. $(F : A leftrightarrows B : G) in mathsfADJ$ and morphism defined as above. Then, composition is defined by a pairwise functor composition, and identity is a pair of identity functors. Associativity of composition is easily seen from version (III) of definition.
So It seems that $ADJ$ is a category. But in fact it is not as it contains itself in form $(id : ADJ rightleftarrows ADJ : id) in ADJ$. Moreover every category $C$ has a canonical representative $(id : C rightleftarrows C : id) in ADJ$, which means that $ADJ$ contains all categories in certain sense. Both of these are bas things for a set theoretical reasons, thus $ADJ$ should not be defined as a category.
This situation is simillar to the idea of category of categories here so I see two pathways:
1) Restrict $ADJ$ to small categories or locally small ones.
2) Define $ADJ$ as a 2-category by introducing 'nutaral transfoms' as morphisms of morphims.
I personally prefer the latter solution.
Are categories of adjunctons used somewhere? If so, how are these categories defined?
P.S.
While writing this question I took liberty to define command A -> $mathcalA$, B -> $mathcalB$, C -> $mathcalC$, ADJ -> $mathsfADJ$; id ->$mathrmId$. You can use them in your answers and comments.
category-theory definition adjoint-functors higher-category-theory
edited Jul 23 at 12:02
Fabio Lucchini
5,62411025
asked Jul 14 at 22:28
Nik Pronko
795717
edited Jul 23 at 12:02
Fabio Lucchini
5,62411025
edited Jul 23 at 12:02
Fabio Lucchini
5,62411025
edited Jul 23 at 12:02
Fabio Lucchini
5,62411025
5,62411025
asked Jul 14 at 22:28
Nik Pronko
795717
asked Jul 14 at 22:28
Nik Pronko
795717
asked Jul 14 at 22:28
Nik Pronko
795717
795717
2
ncatlab.org/nlab/show/2-category+of+adjunctions
– Derek Elkins
Jul 14 at 23:23
1
Working with a category of small categories isn't as restrictive as you might think; when necessary you can switch which universe of sets you are taking to be the universe of small sets. (assuming you take foundations with enough universes)
– Hurkyl
Jul 14 at 23:44
Remark that 2) is no solution to your set-theoretic issue.
– Pece
Jul 15 at 9:40
@DerekElkins Thanks, This is interesting. As I understood In my context, underlying 2-category would be $mathsfCAT$ with categories as object, Adjunctions (of categories) as horizontal arrows, and morphims of adjunctions (as defined above) as vertical arrows.
– Nik Pronko
Jul 15 at 20:52
@Pece I understand that this is not a solution, as 1-skeleton of the category will contain itself. I understand that this is incorrect, that I was thinking about, is that If 0-skeleton of $mathsfSET$ is not a set itself, why can't I have (In certain specific foundations) a 2-category those 1-skeleton is not a category. What I'm trying to Introduce here is some type of structure that are like categories, but not count as categories then you quantify statements other categories. If I could do something like this with k-skeletons of n-categories I would get something akin Russells's hierarchy.
– Nik Pronko
Jul 15 at 21:14
add a comment |Â
2
ncatlab.org/nlab/show/2-category+of+adjunctions
– Derek Elkins
Jul 14 at 23:23
1
Working with a category of small categories isn't as restrictive as you might think; when necessary you can switch which universe of sets you are taking to be the universe of small sets. (assuming you take foundations with enough universes)
– Hurkyl
Jul 14 at 23:44
Remark that 2) is no solution to your set-theoretic issue.
– Pece
Jul 15 at 9:40
@DerekElkins Thanks, This is interesting. As I understood In my context, underlying 2-category would be $mathsfCAT$ with categories as object, Adjunctions (of categories) as horizontal arrows, and morphims of adjunctions (as defined above) as vertical arrows.
– Nik Pronko
Jul 15 at 20:52
@Pece I understand that this is not a solution, as 1-skeleton of the category will contain itself. I understand that this is incorrect, that I was thinking about, is that If 0-skeleton of $mathsfSET$ is not a set itself, why can't I have (In certain specific foundations) a 2-category those 1-skeleton is not a category. What I'm trying to Introduce here is some type of structure that are like categories, but not count as categories then you quantify statements other categories. If I could do something like this with k-skeletons of n-categories I would get something akin Russells's hierarchy.
– Nik Pronko
Jul 15 at 21:14
2
2
ncatlab.org/nlab/show/2-category+of+adjunctions
– Derek Elkins
Jul 14 at 23:23
ncatlab.org/nlab/show/2-category+of+adjunctions
– Derek Elkins
Jul 14 at 23:23
1
1
Working with a category of small categories isn't as restrictive as you might think; when necessary you can switch which universe of sets you are taking to be the universe of small sets. (assuming you take foundations with enough universes)
– Hurkyl
Jul 14 at 23:44
Working with a category of small categories isn't as restrictive as you might think; when necessary you can switch which universe of sets you are taking to be the universe of small sets. (assuming you take foundations with enough universes)
– Hurkyl
Jul 14 at 23:44
Remark that 2) is no solution to your set-theoretic issue.
– Pece
Jul 15 at 9:40
Remark that 2) is no solution to your set-theoretic issue.
– Pece
Jul 15 at 9:40
@DerekElkins Thanks, This is interesting. As I understood In my context, underlying 2-category would be $mathsfCAT$ with categories as object, Adjunctions (of categories) as horizontal arrows, and morphims of adjunctions (as defined above) as vertical arrows.
– Nik Pronko
Jul 15 at 20:52
@DerekElkins Thanks, This is interesting. As I understood In my context, underlying 2-category would be $mathsfCAT$ with categories as object, Adjunctions (of categories) as horizontal arrows, and morphims of adjunctions (as defined above) as vertical arrows.
– Nik Pronko
Jul 15 at 20:52
@Pece I understand that this is not a solution, as 1-skeleton of the category will contain itself. I understand that this is incorrect, that I was thinking about, is that If 0-skeleton of $mathsfSET$ is not a set itself, why can't I have (In certain specific foundations) a 2-category those 1-skeleton is not a category. What I'm trying to Introduce here is some type of structure that are like categories, but not count as categories then you quantify statements other categories. If I could do something like this with k-skeletons of n-categories I would get something akin Russells's hierarchy.
– Nik Pronko
Jul 15 at 21:14
@Pece I understand that this is not a solution, as 1-skeleton of the category will contain itself. I understand that this is incorrect, that I was thinking about, is that If 0-skeleton of $mathsfSET$ is not a set itself, why can't I have (In certain specific foundations) a 2-category those 1-skeleton is not a category. What I'm trying to Introduce here is some type of structure that are like categories, but not count as categories then you quantify statements other categories. If I could do something like this with k-skeletons of n-categories I would get something akin Russells's hierarchy.
– Nik Pronko
Jul 15 at 21:14
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
(note: I had misread the construction in the OP when writing this)
You can't get away from the size issues; an easy way to see definition 2 doesn't help is that you can extract from it the paradoxical 1-category simply by taking its 0 and 1-cells.
It is standard to define Adj on the same 'level' as Cat; e.g. given a cardinal $kappa$, if you are working with the 2-category $mathbfCat_kappa$ of $kappa$-small categories, functors, and natural transformations, then you would normally take $mathbfAdj_kappa$ to be the 2-category of $kappa$-small categories, adjunctions, and natural transformations. Or if you are working with Cat as a 1-category, then you do the same with Adj.
As Derek Elkins points out in comments, Cat isn't special here; given any 2-category $mathcalC$ you can construct the 2-category of adjunctions in $mathcalC$.
answered Jul 14 at 23:42
Hurkyl
108k9112253
2
I think it's not exactly the 2-category defined by the OP, rather it is the full subcategory on adjunctions of the vertical category of the double category $CAT$ (with functors as vertical, and profunctors as horizontal arrows).
– Berci
Jul 15 at 0:02
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
(note: I had misread the construction in the OP when writing this)
You can't get away from the size issues; an easy way to see definition 2 doesn't help is that you can extract from it the paradoxical 1-category simply by taking its 0 and 1-cells.
It is standard to define Adj on the same 'level' as Cat; e.g. given a cardinal $kappa$, if you are working with the 2-category $mathbfCat_kappa$ of $kappa$-small categories, functors, and natural transformations, then you would normally take $mathbfAdj_kappa$ to be the 2-category of $kappa$-small categories, adjunctions, and natural transformations. Or if you are working with Cat as a 1-category, then you do the same with Adj.
As Derek Elkins points out in comments, Cat isn't special here; given any 2-category $mathcalC$ you can construct the 2-category of adjunctions in $mathcalC$.
answered Jul 14 at 23:42
Hurkyl
108k9112253
2
I think it's not exactly the 2-category defined by the OP, rather it is the full subcategory on adjunctions of the vertical category of the double category $CAT$ (with functors as vertical, and profunctors as horizontal arrows).
– Berci
Jul 15 at 0:02
add a comment |Â
up vote
2
down vote
accepted
(note: I had misread the construction in the OP when writing this)
You can't get away from the size issues; an easy way to see definition 2 doesn't help is that you can extract from it the paradoxical 1-category simply by taking its 0 and 1-cells.
It is standard to define Adj on the same 'level' as Cat; e.g. given a cardinal $kappa$, if you are working with the 2-category $mathbfCat_kappa$ of $kappa$-small categories, functors, and natural transformations, then you would normally take $mathbfAdj_kappa$ to be the 2-category of $kappa$-small categories, adjunctions, and natural transformations. Or if you are working with Cat as a 1-category, then you do the same with Adj.
As Derek Elkins points out in comments, Cat isn't special here; given any 2-category $mathcalC$ you can construct the 2-category of adjunctions in $mathcalC$.
answered Jul 14 at 23:42
Hurkyl
108k9112253
2
I think it's not exactly the 2-category defined by the OP, rather it is the full subcategory on adjunctions of the vertical category of the double category $CAT$ (with functors as vertical, and profunctors as horizontal arrows).
– Berci
Jul 15 at 0:02
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
(note: I had misread the construction in the OP when writing this)
You can't get away from the size issues; an easy way to see definition 2 doesn't help is that you can extract from it the paradoxical 1-category simply by taking its 0 and 1-cells.
It is standard to define Adj on the same 'level' as Cat; e.g. given a cardinal $kappa$, if you are working with the 2-category $mathbfCat_kappa$ of $kappa$-small categories, functors, and natural transformations, then you would normally take $mathbfAdj_kappa$ to be the 2-category of $kappa$-small categories, adjunctions, and natural transformations. Or if you are working with Cat as a 1-category, then you do the same with Adj.
As Derek Elkins points out in comments, Cat isn't special here; given any 2-category $mathcalC$ you can construct the 2-category of adjunctions in $mathcalC$.
answered Jul 14 at 23:42
Hurkyl
108k9112253
(note: I had misread the construction in the OP when writing this)
You can't get away from the size issues; an easy way to see definition 2 doesn't help is that you can extract from it the paradoxical 1-category simply by taking its 0 and 1-cells.
It is standard to define Adj on the same 'level' as Cat; e.g. given a cardinal $kappa$, if you are working with the 2-category $mathbfCat_kappa$ of $kappa$-small categories, functors, and natural transformations, then you would normally take $mathbfAdj_kappa$ to be the 2-category of $kappa$-small categories, adjunctions, and natural transformations. Or if you are working with Cat as a 1-category, then you do the same with Adj.
As Derek Elkins points out in comments, Cat isn't special here; given any 2-category $mathcalC$ you can construct the 2-category of adjunctions in $mathcalC$.
answered Jul 14 at 23:42
Hurkyl
108k9112253
edited Jul 15 at 21:58
answered Jul 14 at 23:42
Hurkyl
108k9112253
answered Jul 14 at 23:42
Hurkyl
108k9112253
answered Jul 14 at 23:42
Hurkyl
108k9112253
108k9112253
2
I think it's not exactly the 2-category defined by the OP, rather it is the full subcategory on adjunctions of the vertical category of the double category $CAT$ (with functors as vertical, and profunctors as horizontal arrows).
– Berci
Jul 15 at 0:02
add a comment |Â
2
I think it's not exactly the 2-category defined by the OP, rather it is the full subcategory on adjunctions of the vertical category of the double category $CAT$ (with functors as vertical, and profunctors as horizontal arrows).
– Berci
Jul 15 at 0:02
2
2
I think it's not exactly the 2-category defined by the OP, rather it is the full subcategory on adjunctions of the vertical category of the double category $CAT$ (with functors as vertical, and profunctors as horizontal arrows).
– Berci
Jul 15 at 0:02
I think it's not exactly the 2-category defined by the OP, rather it is the full subcategory on adjunctions of the vertical category of the double category $CAT$ (with functors as vertical, and profunctors as horizontal arrows).
– Berci
Jul 15 at 0:02
add a comment |Â
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2
ncatlab.org/nlab/show/2-category+of+adjunctions
– Derek Elkins
Jul 14 at 23:23
1
Working with a category of small categories isn't as restrictive as you might think; when necessary you can switch which universe of sets you are taking to be the universe of small sets. (assuming you take foundations with enough universes)
– Hurkyl
Jul 14 at 23:44
Remark that 2) is no solution to your set-theoretic issue.
– Pece
Jul 15 at 9:40
@DerekElkins Thanks, This is interesting. As I understood In my context, underlying 2-category would be $mathsfCAT$ with categories as object, Adjunctions (of categories) as horizontal arrows, and morphims of adjunctions (as defined above) as vertical arrows.
– Nik Pronko
Jul 15 at 20:52
@Pece I understand that this is not a solution, as 1-skeleton of the category will contain itself. I understand that this is incorrect, that I was thinking about, is that If 0-skeleton of $mathsfSET$ is not a set itself, why can't I have (In certain specific foundations) a 2-category those 1-skeleton is not a category. What I'm trying to Introduce here is some type of structure that are like categories, but not count as categories then you quantify statements other categories. If I could do something like this with k-skeletons of n-categories I would get something akin Russells's hierarchy.
– Nik Pronko
Jul 15 at 21:14