Two monoidal structures and copowering The Next CEO of Stack OverflowDefinition of enriched caterories or internal homs without using monoidal categories.Unitalization internal to monoidal categoriesCorrespondence between operads and monads requires tensor distribute over coproduct?Making additive envelopes of monoidal categories monoidalEnriching categories and equivalencesSeeking more information regarding the “rigoidal category” of $mathbbN$-graded setsIs there a monoidal category that coclassifies enriched category structures for a given set?Biased vs unbiased lax monoidal categoriesDefinitions of enriched monoidal categoryEnrichment of lax monoidal functors between closed monoidal categories
Two monoidal structures and copowering
The Next CEO of Stack OverflowDefinition of enriched caterories or internal homs without using monoidal categories.Unitalization internal to monoidal categoriesCorrespondence between operads and monads requires tensor distribute over coproduct?Making additive envelopes of monoidal categories monoidalEnriching categories and equivalencesSeeking more information regarding the “rigoidal category” of $mathbbN$-graded setsIs there a monoidal category that coclassifies enriched category structures for a given set?Biased vs unbiased lax monoidal categoriesDefinitions of enriched monoidal categoryEnrichment of lax monoidal functors between closed monoidal categories
$begingroup$
Let $(mathbfM,otimes,1)$ be a closed monoidal category and $(mathbfC,oplus,0)$ an $mathbfM$-enriched monoidal category. Furthermore, assume that we have a copowering $odot:mathbfMtimesmathbfCto mathbfC$. Is there a canonical morphism
$$(Aodot X)oplus (Bodot Y)to (Aotimes B)odot (Xoplus Y)$$
The question came to my mind because in order to spell out the axioms (in one of the definitions) for an algebra over an operad $mathcalO$ in the above setting, we need for the associativity axiom a morphism
$$mathcalO(r)odot left(bigoplus_i (mathcalO(k_i)odot X^oplus k_i)right)toleft(mathcalO(r)otimesbigotimes_imathcalO(k_i)right)odot left(bigoplus_iX^oplus k_iright)$$
Or the other direction. If $mathbfM$ is considered to be enriched over itself, everything is fine because then $otimes=odot=oplus$, but in general?
ct.category-theory monoidal-categories operads enriched-category-theory
asked 2 days ago
FKranholdFKranhold
3236
$endgroup$
add a comment |
$begingroup$
Let $(mathbfM,otimes,1)$ be a closed monoidal category and $(mathbfC,oplus,0)$ an $mathbfM$-enriched monoidal category. Furthermore, assume that we have a copowering $odot:mathbfMtimesmathbfCto mathbfC$. Is there a canonical morphism
$$(Aodot X)oplus (Bodot Y)to (Aotimes B)odot (Xoplus Y)$$
The question came to my mind because in order to spell out the axioms (in one of the definitions) for an algebra over an operad $mathcalO$ in the above setting, we need for the associativity axiom a morphism
$$mathcalO(r)odot left(bigoplus_i (mathcalO(k_i)odot X^oplus k_i)right)toleft(mathcalO(r)otimesbigotimes_imathcalO(k_i)right)odot left(bigoplus_iX^oplus k_iright)$$
Or the other direction. If $mathbfM$ is considered to be enriched over itself, everything is fine because then $otimes=odot=oplus$, but in general?
ct.category-theory monoidal-categories operads enriched-category-theory
asked 2 days ago
FKranholdFKranhold
3236
$endgroup$
1
$begingroup$
Okay, it seems to be equivalent to the formulation: “The copowering is a monoidal functor with respect to the component-wise monoidal structure.”
$endgroup$
– FKranhold
2 days ago
add a comment |
$begingroup$
Let $(mathbfM,otimes,1)$ be a closed monoidal category and $(mathbfC,oplus,0)$ an $mathbfM$-enriched monoidal category. Furthermore, assume that we have a copowering $odot:mathbfMtimesmathbfCto mathbfC$. Is there a canonical morphism
$$(Aodot X)oplus (Bodot Y)to (Aotimes B)odot (Xoplus Y)$$
The question came to my mind because in order to spell out the axioms (in one of the definitions) for an algebra over an operad $mathcalO$ in the above setting, we need for the associativity axiom a morphism
$$mathcalO(r)odot left(bigoplus_i (mathcalO(k_i)odot X^oplus k_i)right)toleft(mathcalO(r)otimesbigotimes_imathcalO(k_i)right)odot left(bigoplus_iX^oplus k_iright)$$
Or the other direction. If $mathbfM$ is considered to be enriched over itself, everything is fine because then $otimes=odot=oplus$, but in general?
ct.category-theory monoidal-categories operads enriched-category-theory
asked 2 days ago
FKranholdFKranhold
3236
$endgroup$
Let $(mathbfM,otimes,1)$ be a closed monoidal category and $(mathbfC,oplus,0)$ an $mathbfM$-enriched monoidal category. Furthermore, assume that we have a copowering $odot:mathbfMtimesmathbfCto mathbfC$. Is there a canonical morphism
$$(Aodot X)oplus (Bodot Y)to (Aotimes B)odot (Xoplus Y)$$
The question came to my mind because in order to spell out the axioms (in one of the definitions) for an algebra over an operad $mathcalO$ in the above setting, we need for the associativity axiom a morphism
$$mathcalO(r)odot left(bigoplus_i (mathcalO(k_i)odot X^oplus k_i)right)toleft(mathcalO(r)otimesbigotimes_imathcalO(k_i)right)odot left(bigoplus_iX^oplus k_iright)$$
Or the other direction. If $mathbfM$ is considered to be enriched over itself, everything is fine because then $otimes=odot=oplus$, but in general?
ct.category-theory monoidal-categories operads enriched-category-theory
ct.category-theory monoidal-categories operads enriched-category-theory
asked 2 days ago
FKranholdFKranhold
3236
asked 2 days ago
FKranholdFKranhold
3236
asked 2 days ago
FKranholdFKranhold
3236
asked 2 days ago
FKranholdFKranhold
3236
asked 2 days ago
FKranholdFKranhold
3236
3236
1
$begingroup$
Okay, it seems to be equivalent to the formulation: “The copowering is a monoidal functor with respect to the component-wise monoidal structure.”
$endgroup$
– FKranhold
2 days ago
add a comment |
1
$begingroup$
Okay, it seems to be equivalent to the formulation: “The copowering is a monoidal functor with respect to the component-wise monoidal structure.”
$endgroup$
– FKranhold
2 days ago
1
1
$begingroup$
Okay, it seems to be equivalent to the formulation: “The copowering is a monoidal functor with respect to the component-wise monoidal structure.”
$endgroup$
– FKranhold
2 days ago
$begingroup$
Okay, it seems to be equivalent to the formulation: “The copowering is a monoidal functor with respect to the component-wise monoidal structure.”
$endgroup$
– FKranhold
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No. Consider the case where $(M,otimes,1)$ is $(mathbfSet,times,1)$, so the enrichment is vacuous, and $(C,oplus,0)$ is $(mathbfSet,+,0)$, with copowering $odot$ given by $times$.
Then the morphism you ask for would give a map
$$(A times X) + (B times Y) longrightarrow (A times B) times (X + Y) $$
which doesn’t exist in general: consider $A = X = Y = 1$, $B = 0$.
However, there is a natural map in the other direction. There are natural maps $A to C(X,A odot X)$ and $B to C(Y,B odot Y)$, the structure maps of the copowering. Also, the definition of enriched monoidal category includes the condition that $oplus$ is an enriched bifunctor, so there’s a general map $C(X,X') otimes C(Y,Y') to C(X oplus Y, X' oplus Y')$. Putting these together, we get a map
$$ A otimes B longrightarrow C(X, A odot X) otimes C(Y, B odot Y) longrightarrow C(X oplus Y, (A odot X) oplus (B odot Y)) $$
which corresponds under copowering to a map $(A otimes B) odot (X oplus Y) to (A odot X) oplus (B odot Y)$.
answered 2 days ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87113871
$endgroup$
$begingroup$
Good counterexample! Then maybe there is a canonical morphism in the other direction? Otherwise, I have the above problem with the associativity axiom for algebras over operads, as long as we do not assume that the copowering is a monoidal functor …
$endgroup$
– FKranhold
2 days ago
$begingroup$
@FKranhold: Yes, there is a natural map in the converse direction — I’ll add the description of that in my answer.
$endgroup$
– Peter LeFanu Lumsdaine
2 days ago
$begingroup$
I don’t see the maps $Ato C(X,Aodot X)$ and $Bto C(Y,Bodot Y)$. It is clear that we have $1to C(X,X)$ and the only thing I know from the copower is that $C(Aodot X,Y)cong C(X,Y)^A$, right? How can I get $Ato C(X,Aodot X)$? The rest of your explanation is well understandable.
$endgroup$
– FKranhold
2 days ago
1
$begingroup$
In the universal property of the copower, as you state it in your comment, take $Y=Aodot X$. This gives $C(Aodot X,Aodot X)cong C(X,Aodot X)^A$. Combining this with the identity map you mention gives $1to C(X,Aodot X)^A$, which transposes to give the desired map $Ato C(X,Aodot X)$.
$endgroup$
– Peter LeFanu Lumsdaine
yesterday
$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
yesterday
add a comment |
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1 Answer
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$begingroup$
No. Consider the case where $(M,otimes,1)$ is $(mathbfSet,times,1)$, so the enrichment is vacuous, and $(C,oplus,0)$ is $(mathbfSet,+,0)$, with copowering $odot$ given by $times$.
Then the morphism you ask for would give a map
$$(A times X) + (B times Y) longrightarrow (A times B) times (X + Y) $$
which doesn’t exist in general: consider $A = X = Y = 1$, $B = 0$.
However, there is a natural map in the other direction. There are natural maps $A to C(X,A odot X)$ and $B to C(Y,B odot Y)$, the structure maps of the copowering. Also, the definition of enriched monoidal category includes the condition that $oplus$ is an enriched bifunctor, so there’s a general map $C(X,X') otimes C(Y,Y') to C(X oplus Y, X' oplus Y')$. Putting these together, we get a map
$$ A otimes B longrightarrow C(X, A odot X) otimes C(Y, B odot Y) longrightarrow C(X oplus Y, (A odot X) oplus (B odot Y)) $$
which corresponds under copowering to a map $(A otimes B) odot (X oplus Y) to (A odot X) oplus (B odot Y)$.
answered 2 days ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87113871
$endgroup$
$begingroup$
Good counterexample! Then maybe there is a canonical morphism in the other direction? Otherwise, I have the above problem with the associativity axiom for algebras over operads, as long as we do not assume that the copowering is a monoidal functor …
$endgroup$
– FKranhold
2 days ago
$begingroup$
@FKranhold: Yes, there is a natural map in the converse direction — I’ll add the description of that in my answer.
$endgroup$
– Peter LeFanu Lumsdaine
2 days ago
$begingroup$
I don’t see the maps $Ato C(X,Aodot X)$ and $Bto C(Y,Bodot Y)$. It is clear that we have $1to C(X,X)$ and the only thing I know from the copower is that $C(Aodot X,Y)cong C(X,Y)^A$, right? How can I get $Ato C(X,Aodot X)$? The rest of your explanation is well understandable.
$endgroup$
– FKranhold
2 days ago
1
$begingroup$
In the universal property of the copower, as you state it in your comment, take $Y=Aodot X$. This gives $C(Aodot X,Aodot X)cong C(X,Aodot X)^A$. Combining this with the identity map you mention gives $1to C(X,Aodot X)^A$, which transposes to give the desired map $Ato C(X,Aodot X)$.
$endgroup$
– Peter LeFanu Lumsdaine
yesterday
$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
yesterday
add a comment |
$begingroup$
No. Consider the case where $(M,otimes,1)$ is $(mathbfSet,times,1)$, so the enrichment is vacuous, and $(C,oplus,0)$ is $(mathbfSet,+,0)$, with copowering $odot$ given by $times$.
Then the morphism you ask for would give a map
$$(A times X) + (B times Y) longrightarrow (A times B) times (X + Y) $$
which doesn’t exist in general: consider $A = X = Y = 1$, $B = 0$.
However, there is a natural map in the other direction. There are natural maps $A to C(X,A odot X)$ and $B to C(Y,B odot Y)$, the structure maps of the copowering. Also, the definition of enriched monoidal category includes the condition that $oplus$ is an enriched bifunctor, so there’s a general map $C(X,X') otimes C(Y,Y') to C(X oplus Y, X' oplus Y')$. Putting these together, we get a map
$$ A otimes B longrightarrow C(X, A odot X) otimes C(Y, B odot Y) longrightarrow C(X oplus Y, (A odot X) oplus (B odot Y)) $$
which corresponds under copowering to a map $(A otimes B) odot (X oplus Y) to (A odot X) oplus (B odot Y)$.
answered 2 days ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87113871
$endgroup$
$begingroup$
Good counterexample! Then maybe there is a canonical morphism in the other direction? Otherwise, I have the above problem with the associativity axiom for algebras over operads, as long as we do not assume that the copowering is a monoidal functor …
$endgroup$
– FKranhold
2 days ago
$begingroup$
@FKranhold: Yes, there is a natural map in the converse direction — I’ll add the description of that in my answer.
$endgroup$
– Peter LeFanu Lumsdaine
2 days ago
$begingroup$
I don’t see the maps $Ato C(X,Aodot X)$ and $Bto C(Y,Bodot Y)$. It is clear that we have $1to C(X,X)$ and the only thing I know from the copower is that $C(Aodot X,Y)cong C(X,Y)^A$, right? How can I get $Ato C(X,Aodot X)$? The rest of your explanation is well understandable.
$endgroup$
– FKranhold
2 days ago
1
$begingroup$
In the universal property of the copower, as you state it in your comment, take $Y=Aodot X$. This gives $C(Aodot X,Aodot X)cong C(X,Aodot X)^A$. Combining this with the identity map you mention gives $1to C(X,Aodot X)^A$, which transposes to give the desired map $Ato C(X,Aodot X)$.
$endgroup$
– Peter LeFanu Lumsdaine
yesterday
$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
yesterday
add a comment |
$begingroup$
No. Consider the case where $(M,otimes,1)$ is $(mathbfSet,times,1)$, so the enrichment is vacuous, and $(C,oplus,0)$ is $(mathbfSet,+,0)$, with copowering $odot$ given by $times$.
Then the morphism you ask for would give a map
$$(A times X) + (B times Y) longrightarrow (A times B) times (X + Y) $$
which doesn’t exist in general: consider $A = X = Y = 1$, $B = 0$.
However, there is a natural map in the other direction. There are natural maps $A to C(X,A odot X)$ and $B to C(Y,B odot Y)$, the structure maps of the copowering. Also, the definition of enriched monoidal category includes the condition that $oplus$ is an enriched bifunctor, so there’s a general map $C(X,X') otimes C(Y,Y') to C(X oplus Y, X' oplus Y')$. Putting these together, we get a map
$$ A otimes B longrightarrow C(X, A odot X) otimes C(Y, B odot Y) longrightarrow C(X oplus Y, (A odot X) oplus (B odot Y)) $$
which corresponds under copowering to a map $(A otimes B) odot (X oplus Y) to (A odot X) oplus (B odot Y)$.
answered 2 days ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87113871
$endgroup$
No. Consider the case where $(M,otimes,1)$ is $(mathbfSet,times,1)$, so the enrichment is vacuous, and $(C,oplus,0)$ is $(mathbfSet,+,0)$, with copowering $odot$ given by $times$.
Then the morphism you ask for would give a map
$$(A times X) + (B times Y) longrightarrow (A times B) times (X + Y) $$
which doesn’t exist in general: consider $A = X = Y = 1$, $B = 0$.
However, there is a natural map in the other direction. There are natural maps $A to C(X,A odot X)$ and $B to C(Y,B odot Y)$, the structure maps of the copowering. Also, the definition of enriched monoidal category includes the condition that $oplus$ is an enriched bifunctor, so there’s a general map $C(X,X') otimes C(Y,Y') to C(X oplus Y, X' oplus Y')$. Putting these together, we get a map
$$ A otimes B longrightarrow C(X, A odot X) otimes C(Y, B odot Y) longrightarrow C(X oplus Y, (A odot X) oplus (B odot Y)) $$
which corresponds under copowering to a map $(A otimes B) odot (X oplus Y) to (A odot X) oplus (B odot Y)$.
answered 2 days ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87113871
edited 2 days ago
answered 2 days ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87113871
answered 2 days ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87113871
answered 2 days ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87113871
8,87113871
$begingroup$
Good counterexample! Then maybe there is a canonical morphism in the other direction? Otherwise, I have the above problem with the associativity axiom for algebras over operads, as long as we do not assume that the copowering is a monoidal functor …
$endgroup$
– FKranhold
2 days ago
$begingroup$
@FKranhold: Yes, there is a natural map in the converse direction — I’ll add the description of that in my answer.
$endgroup$
– Peter LeFanu Lumsdaine
2 days ago
$begingroup$
I don’t see the maps $Ato C(X,Aodot X)$ and $Bto C(Y,Bodot Y)$. It is clear that we have $1to C(X,X)$ and the only thing I know from the copower is that $C(Aodot X,Y)cong C(X,Y)^A$, right? How can I get $Ato C(X,Aodot X)$? The rest of your explanation is well understandable.
$endgroup$
– FKranhold
2 days ago
1
$begingroup$
In the universal property of the copower, as you state it in your comment, take $Y=Aodot X$. This gives $C(Aodot X,Aodot X)cong C(X,Aodot X)^A$. Combining this with the identity map you mention gives $1to C(X,Aodot X)^A$, which transposes to give the desired map $Ato C(X,Aodot X)$.
$endgroup$
– Peter LeFanu Lumsdaine
yesterday
$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
yesterday
add a comment |
$begingroup$
Good counterexample! Then maybe there is a canonical morphism in the other direction? Otherwise, I have the above problem with the associativity axiom for algebras over operads, as long as we do not assume that the copowering is a monoidal functor …
$endgroup$
– FKranhold
2 days ago
$begingroup$
@FKranhold: Yes, there is a natural map in the converse direction — I’ll add the description of that in my answer.
$endgroup$
– Peter LeFanu Lumsdaine
2 days ago
$begingroup$
I don’t see the maps $Ato C(X,Aodot X)$ and $Bto C(Y,Bodot Y)$. It is clear that we have $1to C(X,X)$ and the only thing I know from the copower is that $C(Aodot X,Y)cong C(X,Y)^A$, right? How can I get $Ato C(X,Aodot X)$? The rest of your explanation is well understandable.
$endgroup$
– FKranhold
2 days ago
1
$begingroup$
In the universal property of the copower, as you state it in your comment, take $Y=Aodot X$. This gives $C(Aodot X,Aodot X)cong C(X,Aodot X)^A$. Combining this with the identity map you mention gives $1to C(X,Aodot X)^A$, which transposes to give the desired map $Ato C(X,Aodot X)$.
$endgroup$
– Peter LeFanu Lumsdaine
yesterday
$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
yesterday
$begingroup$
Good counterexample! Then maybe there is a canonical morphism in the other direction? Otherwise, I have the above problem with the associativity axiom for algebras over operads, as long as we do not assume that the copowering is a monoidal functor …
$endgroup$
– FKranhold
2 days ago
$begingroup$
Good counterexample! Then maybe there is a canonical morphism in the other direction? Otherwise, I have the above problem with the associativity axiom for algebras over operads, as long as we do not assume that the copowering is a monoidal functor …
$endgroup$
– FKranhold
2 days ago
$begingroup$
@FKranhold: Yes, there is a natural map in the converse direction — I’ll add the description of that in my answer.
$endgroup$
– Peter LeFanu Lumsdaine
2 days ago
$begingroup$
@FKranhold: Yes, there is a natural map in the converse direction — I’ll add the description of that in my answer.
$endgroup$
– Peter LeFanu Lumsdaine
2 days ago
$begingroup$
I don’t see the maps $Ato C(X,Aodot X)$ and $Bto C(Y,Bodot Y)$. It is clear that we have $1to C(X,X)$ and the only thing I know from the copower is that $C(Aodot X,Y)cong C(X,Y)^A$, right? How can I get $Ato C(X,Aodot X)$? The rest of your explanation is well understandable.
$endgroup$
– FKranhold
2 days ago
$begingroup$
I don’t see the maps $Ato C(X,Aodot X)$ and $Bto C(Y,Bodot Y)$. It is clear that we have $1to C(X,X)$ and the only thing I know from the copower is that $C(Aodot X,Y)cong C(X,Y)^A$, right? How can I get $Ato C(X,Aodot X)$? The rest of your explanation is well understandable.
$endgroup$
– FKranhold
2 days ago
1
1
$begingroup$
In the universal property of the copower, as you state it in your comment, take $Y=Aodot X$. This gives $C(Aodot X,Aodot X)cong C(X,Aodot X)^A$. Combining this with the identity map you mention gives $1to C(X,Aodot X)^A$, which transposes to give the desired map $Ato C(X,Aodot X)$.
$endgroup$
– Peter LeFanu Lumsdaine
yesterday
$begingroup$
In the universal property of the copower, as you state it in your comment, take $Y=Aodot X$. This gives $C(Aodot X,Aodot X)cong C(X,Aodot X)^A$. Combining this with the identity map you mention gives $1to C(X,Aodot X)^A$, which transposes to give the desired map $Ato C(X,Aodot X)$.
$endgroup$
– Peter LeFanu Lumsdaine
yesterday
$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
yesterday
$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
yesterday
add a comment |
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Required, but never shown
Required, but never shown
1
$begingroup$
Okay, it seems to be equivalent to the formulation: “The copowering is a monoidal functor with respect to the component-wise monoidal structure.”
$endgroup$
– FKranhold
2 days ago