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










6












$begingroup$


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?










share|cite|improve this question









$endgroup$







  • 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
    yesterday















6












$begingroup$


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?










share|cite|improve this question









$endgroup$







  • 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
    yesterday













6












6








6





$begingroup$


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?










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









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
    yesterday












  • 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
    yesterday







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
yesterday




$begingroup$
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
yesterday










1 Answer
1






active

oldest

votes


















9












$begingroup$

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 $(A otimes B) odot (X oplus Y) to (A odot X) oplus (B odot Y)$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    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
    yesterday










  • $begingroup$
    @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
    yesterday










  • $begingroup$
    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
    yesterday







  • 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
    17 hours ago










  • $begingroup$
    Ah, of course! Thank you!
    $endgroup$
    – FKranhold
    16 hours ago











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
);



);













draft saved

draft discarded


















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









9












$begingroup$

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 $(A otimes B) odot (X oplus Y) to (A odot X) oplus (B odot Y)$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    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
    yesterday










  • $begingroup$
    @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
    yesterday










  • $begingroup$
    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
    yesterday







  • 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
    17 hours ago










  • $begingroup$
    Ah, of course! Thank you!
    $endgroup$
    – FKranhold
    16 hours ago















9












$begingroup$

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 $(A otimes B) odot (X oplus Y) to (A odot X) oplus (B odot Y)$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    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
    yesterday










  • $begingroup$
    @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
    yesterday










  • $begingroup$
    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
    yesterday







  • 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
    17 hours ago










  • $begingroup$
    Ah, of course! Thank you!
    $endgroup$
    – FKranhold
    16 hours ago













9












9








9





$begingroup$

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 $(A otimes B) odot (X oplus Y) to (A odot X) oplus (B odot Y)$.






share|cite|improve this answer











$endgroup$



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 $(A otimes B) odot (X oplus Y) to (A odot X) oplus (B odot Y)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









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
    yesterday










  • $begingroup$
    @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
    yesterday










  • $begingroup$
    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
    yesterday







  • 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
    17 hours ago










  • $begingroup$
    Ah, of course! Thank you!
    $endgroup$
    – FKranhold
    16 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
    yesterday










  • $begingroup$
    @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
    yesterday










  • $begingroup$
    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
    yesterday







  • 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
    17 hours ago










  • $begingroup$
    Ah, of course! Thank you!
    $endgroup$
    – FKranhold
    16 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
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
yesterday












$begingroup$
@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
yesterday




$begingroup$
@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
yesterday












$begingroup$
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
yesterday





$begingroup$
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
yesterday





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
17 hours ago




$begingroup$
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
17 hours ago












$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
16 hours ago




$begingroup$
Ah, of course! Thank you!
$endgroup$
– FKranhold
16 hours ago

















draft saved

draft discarded
















































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.




draft saved


draft discarded














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





















































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







Popular posts from this blog

រឿង រ៉ូមេអូ និង ហ្ស៊ុយលីយេ សង្ខេបរឿង តួអង្គ បញ្ជីណែនាំ

Crop image to path created in TikZ? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Crop an inserted image?TikZ pictures does not appear in posterImage behind and beyond crop marks?Tikz picture as large as possible on A4 PageTransparency vs image compression dilemmaHow to crop background from image automatically?Image does not cropTikzexternal capturing crop marks when externalizing pgfplots?How to include image path that contains a dollar signCrop image with left size given

Romeo and Juliet ContentsCharactersSynopsisSourcesDate and textThemes and motifsCriticism and interpretationLegacyScene by sceneSee alsoNotes and referencesSourcesExternal linksNavigation menu"Consumer Price Index (estimate) 1800–"10.2307/28710160037-3222287101610.1093/res/II.5.31910.2307/45967845967810.2307/2869925286992510.1525/jams.1982.35.3.03a00050"Dada Masilo: South African dancer who breaks the rules"10.1093/res/os-XV.57.1610.2307/28680942868094"Sweet Sorrow: Mann-Korman's Romeo and Juliet Closes Sept. 5 at MN's Ordway"the original10.2307/45957745957710.1017/CCOL0521570476.009"Ram Leela box office collections hit massive Rs 100 crore, pulverises prediction"Archived"Broadway Revival of Romeo and Juliet, Starring Orlando Bloom and Condola Rashad, Will Close Dec. 8"Archived10.1075/jhp.7.1.04hon"Wherefore art thou, Romeo? To make us laugh at Navy Pier"the original10.1093/gmo/9781561592630.article.O006772"Ram-leela Review Roundup: Critics Hail Film as Best Adaptation of Romeo and Juliet"Archived10.2307/31946310047-77293194631"Romeo and Juliet get Twitter treatment""Juliet's Nurse by Lois Leveen""Romeo and Juliet: Orlando Bloom's Broadway Debut Released in Theaters for Valentine's Day"Archived"Romeo and Juliet Has No Balcony"10.1093/gmo/9781561592630.article.O00778110.2307/2867423286742310.1076/enst.82.2.115.959510.1080/00138380601042675"A plague o' both your houses: error in GCSE exam paper forces apology""Juliet of the Five O'Clock Shadow, and Other Wonders"10.2307/33912430027-4321339124310.2307/28487440038-7134284874410.2307/29123140149-661129123144728341M"Weekender Guide: Shakespeare on The Drive""balcony"UK public library membership"romeo"UK public library membership10.1017/CCOL9780521844291"Post-Zionist Critique on Israel and the Palestinians Part III: Popular Culture"10.2307/25379071533-86140377-919X2537907"Capulets and Montagues: UK exam board admit mixing names up in Romeo and Juliet paper"Istoria Novellamente Ritrovata di Due Nobili Amanti2027/mdp.390150822329610820-750X"GCSE exam error: Board accidentally rewrites Shakespeare"10.2307/29176390149-66112917639"Exam board apologises after error in English GCSE paper which confused characters in Shakespeare's Romeo and Juliet""From Mariotto and Ganozza to Romeo and Guilietta: Metamorphoses of a Renaissance Tale"10.2307/37323537323510.2307/2867455286745510.2307/28678912867891"10 Questions for Taylor Swift"10.2307/28680922868092"Haymarket Theatre""The Zeffirelli Way: Revealing Talk by Florentine Director""Michael Smuin: 1938-2007 / Prolific dance director had showy career"The Life and Art of Edwin BoothRomeo and JulietRomeo and JulietRomeo and JulietRomeo and JulietEasy Read Romeo and JulietRomeo and Julieteeecb12003684p(data)4099369-3n8211610759dbe00d-a9e2-41a3-b2c1-977dd692899302814385X313670221313670221