Two monoidal structures and copoweringDefinition 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
Definition 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 Mar 27 at 20:47
FKranholdFKranhold
3636
$endgroup$
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 Mar 27 at 20:47
FKranholdFKranhold
3636
$endgroup$
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
Mar 27 at 20:59
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 Mar 27 at 20:47
FKranholdFKranhold
3636
$endgroup$
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 Mar 27 at 20:47
FKranholdFKranhold
3636
asked Mar 27 at 20:47
FKranholdFKranhold
3636
asked Mar 27 at 20:47
FKranholdFKranhold
3636
asked Mar 27 at 20:47
FKranholdFKranhold
3636
asked Mar 27 at 20:47
FKranholdFKranhold
3636
3636
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
Mar 27 at 20:59
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
Mar 27 at 20:59
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
Mar 27 at 20:59
$begingroup$
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
Mar 27 at 20:59
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 Mar 27 at 20:59
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87613871
$endgroup$
$begingroup$
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
Mar 27 at 21:06
$begingroup$
@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
Mar 27 at 21:59
$begingroup$
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
Mar 27 at 22:30
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
Mar 28 at 8:26
$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
Mar 28 at 8:36
add a comment |
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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 Mar 27 at 20:59
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87613871
$endgroup$
$begingroup$
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
Mar 27 at 21:06
$begingroup$
@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
Mar 27 at 21:59
$begingroup$
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
Mar 27 at 22:30
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
Mar 28 at 8:26
$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
Mar 28 at 8:36
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 Mar 27 at 20:59
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87613871
$endgroup$
$begingroup$
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
Mar 27 at 21:06
$begingroup$
@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
Mar 27 at 21:59
$begingroup$
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
Mar 27 at 22:30
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
Mar 28 at 8:26
$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
Mar 28 at 8:36
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 Mar 27 at 20:59
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87613871
$endgroup$
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 Mar 27 at 20:59
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87613871
edited Mar 27 at 22:05
answered Mar 27 at 20:59
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87613871
answered Mar 27 at 20:59
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87613871
answered Mar 27 at 20:59
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,87613871
8,87613871
$begingroup$
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
Mar 27 at 21:06
$begingroup$
@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
Mar 27 at 21:59
$begingroup$
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
Mar 27 at 22:30
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
Mar 28 at 8:26
$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
Mar 28 at 8:36
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
Mar 27 at 21:06
$begingroup$
@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
Mar 27 at 21:59
$begingroup$
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
Mar 27 at 22:30
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
Mar 28 at 8:26
$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
Mar 28 at 8:36
$begingroup$
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
Mar 27 at 21:06
$begingroup$
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
Mar 27 at 21:06
$begingroup$
@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
Mar 27 at 21:59
$begingroup$
@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
Mar 27 at 21:59
$begingroup$
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
Mar 27 at 22:30
$begingroup$
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
Mar 27 at 22:30
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
Mar 28 at 8:26
$begingroup$
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
Mar 28 at 8:26
$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
Mar 28 at 8:36
$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
Mar 28 at 8:36
add a comment |
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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
Mar 27 at 20:59