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Algebra operations as natural transformations
The Next CEO of Stack OverflowMorphisms in the category of natural transformations?Understanding associators as natural transformationsWhat's the name of a morphism the morphism category of the category of categories?Natural transformations and the definition of Monoidal lax functorsWhat are the group objects in the category of finite sets and bijections, and its functor category?“Alternatives” to Natural TransformationsRepresenting natural transformations with diagramsBasic category theory: applying natural transformationsDefinition of category and natural transformationsCan natural transformations be viewed as functors between images of functors?
$begingroup$
Apologies in advance if the following makes little to no sense, but here goes ..
Denote $m_G : Gtimes Gto G$ the multiplication of a group $G$. Does it make sense to think of the map $m_G$ as some kind of morphism in a (at the moment unspecified) category?
Even more, could we think of the family $m_G$ (indexed by the class of groups) as a natural transformation of some or other functors?
Let's call our mystery categories $mathcal A$ and $mathcal B$ with mystery functors $X,Y :mathcal Atomathcal B$ such that the natural transformation condition holds:
$$ m_HX(f) = Y(f)m_G $$
where $f$ is a morphism in $mathcal A$ and $G,H$ are groups.
For this to make sense, we need a category with objects $Gtimes G$ and $G$. So, we extend (?) the category of groups by $mathcal B_0 := mboxGrp_0 cup Gtimes G mid GinmboxGrp_0$. Morphisms in 'separate components' remain as they are in $mboxGrp$ and $mboxGrptimesmboxGrp$ respectively. There would be no morphisms of the form $Gto Htimes H$ and
$$mathcal B(Gtimes G,H) := varphi m_G mid varphi in mboxHom(G,H) $$
where for every $x,yin G$ $varphi m_G (x,y) := varphi (xy) = varphi (x)varphi (y) =: m_H(varphi,varphi)(x,y)$. The identities are $1_G$ or $(1_G,1_G)$ depending on the component and composition of the morphisms would happen naturally.
- Is it guaranteed $(A,B) neq (A',B') implies mathcal B(A,B)capmathcal B(A',B') =emptyset, A,A',B,B'inmathcal B_0$?
- Taking $mathcal A = mboxGrp$ with $X :mathcal Atomathcal B$ given such that $X(G) = Gtimes G$ and for every morphism $f:Gto H$, $X(f) = (f,f)$. Put $Y:mathcal Atomathcal B$ as the embedding, then we would have $m_G$ as a natural transformation $XRightarrow Y$.
I omit the routine checks here, for they aren't important for this discussion. I am interested in whether this idea of regarding families of operations as natural transformations is, call it, well-founded
Questionnaire.
- Would such an approach be the only one? How else (if at all) would we regard the family $m_G$ as a natural transformation?
- Is this a more general thing in universal algebra? Given a class of algebras with certain operations of various arities, could we regard every family of operations as a natural transformation? (For instance, inverse operation or unit element operation of groups)
group-theory category-theory universal-algebra natural-transformations
$endgroup$
add a comment |
$begingroup$
Apologies in advance if the following makes little to no sense, but here goes ..
Denote $m_G : Gtimes Gto G$ the multiplication of a group $G$. Does it make sense to think of the map $m_G$ as some kind of morphism in a (at the moment unspecified) category?
Even more, could we think of the family $m_G$ (indexed by the class of groups) as a natural transformation of some or other functors?
Let's call our mystery categories $mathcal A$ and $mathcal B$ with mystery functors $X,Y :mathcal Atomathcal B$ such that the natural transformation condition holds:
$$ m_HX(f) = Y(f)m_G $$
where $f$ is a morphism in $mathcal A$ and $G,H$ are groups.
For this to make sense, we need a category with objects $Gtimes G$ and $G$. So, we extend (?) the category of groups by $mathcal B_0 := mboxGrp_0 cup Gtimes G mid GinmboxGrp_0$. Morphisms in 'separate components' remain as they are in $mboxGrp$ and $mboxGrptimesmboxGrp$ respectively. There would be no morphisms of the form $Gto Htimes H$ and
$$mathcal B(Gtimes G,H) := varphi m_G mid varphi in mboxHom(G,H) $$
where for every $x,yin G$ $varphi m_G (x,y) := varphi (xy) = varphi (x)varphi (y) =: m_H(varphi,varphi)(x,y)$. The identities are $1_G$ or $(1_G,1_G)$ depending on the component and composition of the morphisms would happen naturally.
- Is it guaranteed $(A,B) neq (A',B') implies mathcal B(A,B)capmathcal B(A',B') =emptyset, A,A',B,B'inmathcal B_0$?
- Taking $mathcal A = mboxGrp$ with $X :mathcal Atomathcal B$ given such that $X(G) = Gtimes G$ and for every morphism $f:Gto H$, $X(f) = (f,f)$. Put $Y:mathcal Atomathcal B$ as the embedding, then we would have $m_G$ as a natural transformation $XRightarrow Y$.
I omit the routine checks here, for they aren't important for this discussion. I am interested in whether this idea of regarding families of operations as natural transformations is, call it, well-founded
Questionnaire.
- Would such an approach be the only one? How else (if at all) would we regard the family $m_G$ as a natural transformation?
- Is this a more general thing in universal algebra? Given a class of algebras with certain operations of various arities, could we regard every family of operations as a natural transformation? (For instance, inverse operation or unit element operation of groups)
group-theory category-theory universal-algebra natural-transformations
$endgroup$
1
$begingroup$
@H.H. the components must also be morphisms themselves in the destination category so if it were to work, then that must involve some 'weird' categories. I'm concerned about the way I defined $mathcal B$ yet I can't find any rule that says a category can't contain objects of different 'type'.
$endgroup$
– Alvin Lepik
Mar 18 at 11:27
add a comment |
$begingroup$
Apologies in advance if the following makes little to no sense, but here goes ..
Denote $m_G : Gtimes Gto G$ the multiplication of a group $G$. Does it make sense to think of the map $m_G$ as some kind of morphism in a (at the moment unspecified) category?
Even more, could we think of the family $m_G$ (indexed by the class of groups) as a natural transformation of some or other functors?
Let's call our mystery categories $mathcal A$ and $mathcal B$ with mystery functors $X,Y :mathcal Atomathcal B$ such that the natural transformation condition holds:
$$ m_HX(f) = Y(f)m_G $$
where $f$ is a morphism in $mathcal A$ and $G,H$ are groups.
For this to make sense, we need a category with objects $Gtimes G$ and $G$. So, we extend (?) the category of groups by $mathcal B_0 := mboxGrp_0 cup Gtimes G mid GinmboxGrp_0$. Morphisms in 'separate components' remain as they are in $mboxGrp$ and $mboxGrptimesmboxGrp$ respectively. There would be no morphisms of the form $Gto Htimes H$ and
$$mathcal B(Gtimes G,H) := varphi m_G mid varphi in mboxHom(G,H) $$
where for every $x,yin G$ $varphi m_G (x,y) := varphi (xy) = varphi (x)varphi (y) =: m_H(varphi,varphi)(x,y)$. The identities are $1_G$ or $(1_G,1_G)$ depending on the component and composition of the morphisms would happen naturally.
- Is it guaranteed $(A,B) neq (A',B') implies mathcal B(A,B)capmathcal B(A',B') =emptyset, A,A',B,B'inmathcal B_0$?
- Taking $mathcal A = mboxGrp$ with $X :mathcal Atomathcal B$ given such that $X(G) = Gtimes G$ and for every morphism $f:Gto H$, $X(f) = (f,f)$. Put $Y:mathcal Atomathcal B$ as the embedding, then we would have $m_G$ as a natural transformation $XRightarrow Y$.
I omit the routine checks here, for they aren't important for this discussion. I am interested in whether this idea of regarding families of operations as natural transformations is, call it, well-founded
Questionnaire.
- Would such an approach be the only one? How else (if at all) would we regard the family $m_G$ as a natural transformation?
- Is this a more general thing in universal algebra? Given a class of algebras with certain operations of various arities, could we regard every family of operations as a natural transformation? (For instance, inverse operation or unit element operation of groups)
group-theory category-theory universal-algebra natural-transformations
$endgroup$
Apologies in advance if the following makes little to no sense, but here goes ..
Denote $m_G : Gtimes Gto G$ the multiplication of a group $G$. Does it make sense to think of the map $m_G$ as some kind of morphism in a (at the moment unspecified) category?
Even more, could we think of the family $m_G$ (indexed by the class of groups) as a natural transformation of some or other functors?
Let's call our mystery categories $mathcal A$ and $mathcal B$ with mystery functors $X,Y :mathcal Atomathcal B$ such that the natural transformation condition holds:
$$ m_HX(f) = Y(f)m_G $$
where $f$ is a morphism in $mathcal A$ and $G,H$ are groups.
For this to make sense, we need a category with objects $Gtimes G$ and $G$. So, we extend (?) the category of groups by $mathcal B_0 := mboxGrp_0 cup Gtimes G mid GinmboxGrp_0$. Morphisms in 'separate components' remain as they are in $mboxGrp$ and $mboxGrptimesmboxGrp$ respectively. There would be no morphisms of the form $Gto Htimes H$ and
$$mathcal B(Gtimes G,H) := varphi m_G mid varphi in mboxHom(G,H) $$
where for every $x,yin G$ $varphi m_G (x,y) := varphi (xy) = varphi (x)varphi (y) =: m_H(varphi,varphi)(x,y)$. The identities are $1_G$ or $(1_G,1_G)$ depending on the component and composition of the morphisms would happen naturally.
- Is it guaranteed $(A,B) neq (A',B') implies mathcal B(A,B)capmathcal B(A',B') =emptyset, A,A',B,B'inmathcal B_0$?
- Taking $mathcal A = mboxGrp$ with $X :mathcal Atomathcal B$ given such that $X(G) = Gtimes G$ and for every morphism $f:Gto H$, $X(f) = (f,f)$. Put $Y:mathcal Atomathcal B$ as the embedding, then we would have $m_G$ as a natural transformation $XRightarrow Y$.
I omit the routine checks here, for they aren't important for this discussion. I am interested in whether this idea of regarding families of operations as natural transformations is, call it, well-founded
Questionnaire.
- Would such an approach be the only one? How else (if at all) would we regard the family $m_G$ as a natural transformation?
- Is this a more general thing in universal algebra? Given a class of algebras with certain operations of various arities, could we regard every family of operations as a natural transformation? (For instance, inverse operation or unit element operation of groups)
group-theory category-theory universal-algebra natural-transformations
group-theory category-theory universal-algebra natural-transformations
edited Mar 18 at 11:13
Alvin Lepik
asked Mar 18 at 10:54
Alvin LepikAlvin Lepik
2,7961924
2,7961924
1
$begingroup$
@H.H. the components must also be morphisms themselves in the destination category so if it were to work, then that must involve some 'weird' categories. I'm concerned about the way I defined $mathcal B$ yet I can't find any rule that says a category can't contain objects of different 'type'.
$endgroup$
– Alvin Lepik
Mar 18 at 11:27
add a comment |
1
$begingroup$
@H.H. the components must also be morphisms themselves in the destination category so if it were to work, then that must involve some 'weird' categories. I'm concerned about the way I defined $mathcal B$ yet I can't find any rule that says a category can't contain objects of different 'type'.
$endgroup$
– Alvin Lepik
Mar 18 at 11:27
1
1
$begingroup$
@H.H. the components must also be morphisms themselves in the destination category so if it were to work, then that must involve some 'weird' categories. I'm concerned about the way I defined $mathcal B$ yet I can't find any rule that says a category can't contain objects of different 'type'.
$endgroup$
– Alvin Lepik
Mar 18 at 11:27
$begingroup$
@H.H. the components must also be morphisms themselves in the destination category so if it were to work, then that must involve some 'weird' categories. I'm concerned about the way I defined $mathcal B$ yet I can't find any rule that says a category can't contain objects of different 'type'.
$endgroup$
– Alvin Lepik
Mar 18 at 11:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $DeclareMathOperatorSetSetSet$ denote the category of sets and $DeclareMathOperatorGrpGrpGrp$ denote the category of groups.
Let $Upsilon:GrptoSet$ denote the forgetful functor and $Delta:SettoSet$ be the diagonal functor $Delta(X)=Xtimes X$.
Then you are looking for a natural transformation $mu:DeltacircUpsilontoUpsilon$.
If $X$ is a group, then $Upsilon(X)$ is its underlying set.
For each group $X$ let $mu_X:(DeltacircUpsilon)(X)toUpsilon(X)$ be its composition law.
If $f:Xto Y$ is a group homomorphism, then we have a commutative diagram$requireAMScd$:
beginCD
Upsilon(X)timesUpsilon(X)@=(DeltacircUpsilon)(X)@>mu_X>>Upsilon(X)\
@VUpsilon(f)timesUpsilon(f)VV@V(DeltacircUpsilon)(f)VV@VVUpsilon(f)V\
Upsilon(Y)timesUpsilon(Y)@=(DeltacircUpsilon)(Y)@>>mu_Y >Upsilon(Y)
endCD
which proves naturalness of $mu_X$.
$endgroup$
1
$begingroup$
This looks much neater and cleaner than my attempt. Thanks, I have a much better understanding of this idea now.
$endgroup$
– Alvin Lepik
Mar 18 at 12:31
$begingroup$
Also thanks for demonstrating the amscd package, looks much more convenient than tikz syntax-wise. Can't believe I didn't know about it!
$endgroup$
– Alvin Lepik
Mar 18 at 12:44
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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active
oldest
votes
$begingroup$
Let $DeclareMathOperatorSetSetSet$ denote the category of sets and $DeclareMathOperatorGrpGrpGrp$ denote the category of groups.
Let $Upsilon:GrptoSet$ denote the forgetful functor and $Delta:SettoSet$ be the diagonal functor $Delta(X)=Xtimes X$.
Then you are looking for a natural transformation $mu:DeltacircUpsilontoUpsilon$.
If $X$ is a group, then $Upsilon(X)$ is its underlying set.
For each group $X$ let $mu_X:(DeltacircUpsilon)(X)toUpsilon(X)$ be its composition law.
If $f:Xto Y$ is a group homomorphism, then we have a commutative diagram$requireAMScd$:
beginCD
Upsilon(X)timesUpsilon(X)@=(DeltacircUpsilon)(X)@>mu_X>>Upsilon(X)\
@VUpsilon(f)timesUpsilon(f)VV@V(DeltacircUpsilon)(f)VV@VVUpsilon(f)V\
Upsilon(Y)timesUpsilon(Y)@=(DeltacircUpsilon)(Y)@>>mu_Y >Upsilon(Y)
endCD
which proves naturalness of $mu_X$.
$endgroup$
1
$begingroup$
This looks much neater and cleaner than my attempt. Thanks, I have a much better understanding of this idea now.
$endgroup$
– Alvin Lepik
Mar 18 at 12:31
$begingroup$
Also thanks for demonstrating the amscd package, looks much more convenient than tikz syntax-wise. Can't believe I didn't know about it!
$endgroup$
– Alvin Lepik
Mar 18 at 12:44
add a comment |
$begingroup$
Let $DeclareMathOperatorSetSetSet$ denote the category of sets and $DeclareMathOperatorGrpGrpGrp$ denote the category of groups.
Let $Upsilon:GrptoSet$ denote the forgetful functor and $Delta:SettoSet$ be the diagonal functor $Delta(X)=Xtimes X$.
Then you are looking for a natural transformation $mu:DeltacircUpsilontoUpsilon$.
If $X$ is a group, then $Upsilon(X)$ is its underlying set.
For each group $X$ let $mu_X:(DeltacircUpsilon)(X)toUpsilon(X)$ be its composition law.
If $f:Xto Y$ is a group homomorphism, then we have a commutative diagram$requireAMScd$:
beginCD
Upsilon(X)timesUpsilon(X)@=(DeltacircUpsilon)(X)@>mu_X>>Upsilon(X)\
@VUpsilon(f)timesUpsilon(f)VV@V(DeltacircUpsilon)(f)VV@VVUpsilon(f)V\
Upsilon(Y)timesUpsilon(Y)@=(DeltacircUpsilon)(Y)@>>mu_Y >Upsilon(Y)
endCD
which proves naturalness of $mu_X$.
$endgroup$
1
$begingroup$
This looks much neater and cleaner than my attempt. Thanks, I have a much better understanding of this idea now.
$endgroup$
– Alvin Lepik
Mar 18 at 12:31
$begingroup$
Also thanks for demonstrating the amscd package, looks much more convenient than tikz syntax-wise. Can't believe I didn't know about it!
$endgroup$
– Alvin Lepik
Mar 18 at 12:44
add a comment |
$begingroup$
Let $DeclareMathOperatorSetSetSet$ denote the category of sets and $DeclareMathOperatorGrpGrpGrp$ denote the category of groups.
Let $Upsilon:GrptoSet$ denote the forgetful functor and $Delta:SettoSet$ be the diagonal functor $Delta(X)=Xtimes X$.
Then you are looking for a natural transformation $mu:DeltacircUpsilontoUpsilon$.
If $X$ is a group, then $Upsilon(X)$ is its underlying set.
For each group $X$ let $mu_X:(DeltacircUpsilon)(X)toUpsilon(X)$ be its composition law.
If $f:Xto Y$ is a group homomorphism, then we have a commutative diagram$requireAMScd$:
beginCD
Upsilon(X)timesUpsilon(X)@=(DeltacircUpsilon)(X)@>mu_X>>Upsilon(X)\
@VUpsilon(f)timesUpsilon(f)VV@V(DeltacircUpsilon)(f)VV@VVUpsilon(f)V\
Upsilon(Y)timesUpsilon(Y)@=(DeltacircUpsilon)(Y)@>>mu_Y >Upsilon(Y)
endCD
which proves naturalness of $mu_X$.
$endgroup$
Let $DeclareMathOperatorSetSetSet$ denote the category of sets and $DeclareMathOperatorGrpGrpGrp$ denote the category of groups.
Let $Upsilon:GrptoSet$ denote the forgetful functor and $Delta:SettoSet$ be the diagonal functor $Delta(X)=Xtimes X$.
Then you are looking for a natural transformation $mu:DeltacircUpsilontoUpsilon$.
If $X$ is a group, then $Upsilon(X)$ is its underlying set.
For each group $X$ let $mu_X:(DeltacircUpsilon)(X)toUpsilon(X)$ be its composition law.
If $f:Xto Y$ is a group homomorphism, then we have a commutative diagram$requireAMScd$:
beginCD
Upsilon(X)timesUpsilon(X)@=(DeltacircUpsilon)(X)@>mu_X>>Upsilon(X)\
@VUpsilon(f)timesUpsilon(f)VV@V(DeltacircUpsilon)(f)VV@VVUpsilon(f)V\
Upsilon(Y)timesUpsilon(Y)@=(DeltacircUpsilon)(Y)@>>mu_Y >Upsilon(Y)
endCD
which proves naturalness of $mu_X$.
edited Mar 19 at 16:23
Litho
3,4671716
3,4671716
answered Mar 18 at 12:18
Fabio LucchiniFabio Lucchini
9,56611426
9,56611426
1
$begingroup$
This looks much neater and cleaner than my attempt. Thanks, I have a much better understanding of this idea now.
$endgroup$
– Alvin Lepik
Mar 18 at 12:31
$begingroup$
Also thanks for demonstrating the amscd package, looks much more convenient than tikz syntax-wise. Can't believe I didn't know about it!
$endgroup$
– Alvin Lepik
Mar 18 at 12:44
add a comment |
1
$begingroup$
This looks much neater and cleaner than my attempt. Thanks, I have a much better understanding of this idea now.
$endgroup$
– Alvin Lepik
Mar 18 at 12:31
$begingroup$
Also thanks for demonstrating the amscd package, looks much more convenient than tikz syntax-wise. Can't believe I didn't know about it!
$endgroup$
– Alvin Lepik
Mar 18 at 12:44
1
1
$begingroup$
This looks much neater and cleaner than my attempt. Thanks, I have a much better understanding of this idea now.
$endgroup$
– Alvin Lepik
Mar 18 at 12:31
$begingroup$
This looks much neater and cleaner than my attempt. Thanks, I have a much better understanding of this idea now.
$endgroup$
– Alvin Lepik
Mar 18 at 12:31
$begingroup$
Also thanks for demonstrating the amscd package, looks much more convenient than tikz syntax-wise. Can't believe I didn't know about it!
$endgroup$
– Alvin Lepik
Mar 18 at 12:44
$begingroup$
Also thanks for demonstrating the amscd package, looks much more convenient than tikz syntax-wise. Can't believe I didn't know about it!
$endgroup$
– Alvin Lepik
Mar 18 at 12:44
add a comment |
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$begingroup$
@H.H. the components must also be morphisms themselves in the destination category so if it were to work, then that must involve some 'weird' categories. I'm concerned about the way I defined $mathcal B$ yet I can't find any rule that says a category can't contain objects of different 'type'.
$endgroup$
– Alvin Lepik
Mar 18 at 11:27