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
Multi tool use
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 2 hours ago
FKranholdFKranhold
3046
$endgroup$
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 hours ago
FKranholdFKranhold
3046
$endgroup$
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 hours 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 hours ago
FKranholdFKranhold
3046
$endgroup$
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 hours ago
FKranholdFKranhold
3046
asked 2 hours ago
FKranholdFKranhold
3046
asked 2 hours ago
FKranholdFKranhold
3046
asked 2 hours ago
FKranholdFKranhold
3046
asked 2 hours ago
FKranholdFKranhold
3046
3046
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 hours 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 hours 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 hours 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 hours 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 hours ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,82613871
$endgroup$
$begingroup$
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 hours 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
1 hour 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
1 hour ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326520%2ftwo-monoidal-structures-and-copowering%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 hours ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,82613871
$endgroup$
$begingroup$
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 hours 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
1 hour 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
1 hour ago
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 hours ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,82613871
$endgroup$
$begingroup$
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 hours 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
1 hour 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
1 hour ago
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 hours ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,82613871
$endgroup$
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 hours ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,82613871
edited 1 hour ago
answered 2 hours ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,82613871
answered 2 hours ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,82613871
answered 2 hours ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
8,82613871
8,82613871
$begingroup$
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 hours 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
1 hour 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
1 hour ago
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 hours 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
1 hour 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
1 hour 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 hours 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 hours 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
add a comment |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326520%2ftwo-monoidal-structures-and-copowering%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
cRLJgkp982,weN8pWNVBjs,gzM 2Xo
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 hours ago