Algebra operations as natural transformations The Next CEO of Stack OverflowMorphisms in the category of natural transformations?Understanding associators as natural transformationsWhat's the name of a morphism the morphism category of the category of categories?Natural transformations and the definition of Monoidal lax functorsWhat are the group objects in the category of finite sets and bijections, and its functor category?“Alternatives” to Natural TransformationsRepresenting natural transformations with diagramsBasic category theory: applying natural transformationsDefinition of category and natural transformationsCan natural transformations be viewed as functors between images of functors?

Why didn't Khan get resurrected in the Genesis Explosion?

Whats the best way to handle refactoring a big file?

Why didn't Theresa May consult with Parliament before negotiating a deal with the EU?

What do "high sea" and "carry" mean in this sentence?

Return the Closest Prime Number

How to count occurrences of text in a file?

When airplanes disconnect from a tanker during air to air refueling, why do they bank so sharply to the right?

Why does GHC infer a monomorphic type here, even with MonomorphismRestriction disabled?

Science fiction (dystopian) short story set after WWIII

Can a single photon have an energy density?

Visit to the USA with ESTA approved before trip to Iran

How do I construct this japanese bowl?

How should I support this large drywall patch?

How to get regions to plot as graphics

Should I tutor a student who I know has cheated on their homework?

How do spells that require an ability check vs. the caster's spell save DC work?

What makes a siege story/plot interesting?

Why is there a PLL in CPU?

Are there languages with no euphemisms?

Anatomically Correct Mesopelagic Aves

How to start emacs in "nothing" mode (`fundamental-mode`)

How to make a variable always equal to the result of some calculations?

Is it my responsibility to learn a new technology in my own time my employer wants to implement?

How do I solve this limit?



Algebra operations as natural transformations



The Next CEO of Stack OverflowMorphisms in the category of natural transformations?Understanding associators as natural transformationsWhat's the name of a morphism the morphism category of the category of categories?Natural transformations and the definition of Monoidal lax functorsWhat are the group objects in the category of finite sets and bijections, and its functor category?“Alternatives” to Natural TransformationsRepresenting natural transformations with diagramsBasic category theory: applying natural transformationsDefinition of category and natural transformationsCan natural transformations be viewed as functors between images of functors?










5












$begingroup$


Apologies in advance if the following makes little to no sense, but here goes ..



Denote $m_G : Gtimes Gto G$ the multiplication of a group $G$. Does it make sense to think of the map $m_G$ as some kind of morphism in a (at the moment unspecified) category?



Even more, could we think of the family $m_G$ (indexed by the class of groups) as a natural transformation of some or other functors?



Let's call our mystery categories $mathcal A$ and $mathcal B$ with mystery functors $X,Y :mathcal Atomathcal B$ such that the natural transformation condition holds:
$$ m_HX(f) = Y(f)m_G $$
where $f$ is a morphism in $mathcal A$ and $G,H$ are groups.



For this to make sense, we need a category with objects $Gtimes G$ and $G$. So, we extend (?) the category of groups by $mathcal B_0 := mboxGrp_0 cup Gtimes G mid GinmboxGrp_0$. Morphisms in 'separate components' remain as they are in $mboxGrp$ and $mboxGrptimesmboxGrp$ respectively. There would be no morphisms of the form $Gto Htimes H$ and
$$mathcal B(Gtimes G,H) := varphi m_G mid varphi in mboxHom(G,H) $$
where for every $x,yin G$ $varphi m_G (x,y) := varphi (xy) = varphi (x)varphi (y) =: m_H(varphi,varphi)(x,y)$. The identities are $1_G$ or $(1_G,1_G)$ depending on the component and composition of the morphisms would happen naturally.



  1. Is it guaranteed $(A,B) neq (A',B') implies mathcal B(A,B)capmathcal B(A',B') =emptyset, A,A',B,B'inmathcal B_0$?

  2. Taking $mathcal A = mboxGrp$ with $X :mathcal Atomathcal B$ given such that $X(G) = Gtimes G$ and for every morphism $f:Gto H$, $X(f) = (f,f)$. Put $Y:mathcal Atomathcal B$ as the embedding, then we would have $m_G$ as a natural transformation $XRightarrow Y$.

I omit the routine checks here, for they aren't important for this discussion. I am interested in whether this idea of regarding families of operations as natural transformations is, call it, well-founded



Questionnaire.



  1. Would such an approach be the only one? How else (if at all) would we regard the family $m_G$ as a natural transformation?

  2. Is this a more general thing in universal algebra? Given a class of algebras with certain operations of various arities, could we regard every family of operations as a natural transformation? (For instance, inverse operation or unit element operation of groups)









share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    @H.H. the components must also be morphisms themselves in the destination category so if it were to work, then that must involve some 'weird' categories. I'm concerned about the way I defined $mathcal B$ yet I can't find any rule that says a category can't contain objects of different 'type'.
    $endgroup$
    – Alvin Lepik
    Mar 18 at 11:27















5












$begingroup$


Apologies in advance if the following makes little to no sense, but here goes ..



Denote $m_G : Gtimes Gto G$ the multiplication of a group $G$. Does it make sense to think of the map $m_G$ as some kind of morphism in a (at the moment unspecified) category?



Even more, could we think of the family $m_G$ (indexed by the class of groups) as a natural transformation of some or other functors?



Let's call our mystery categories $mathcal A$ and $mathcal B$ with mystery functors $X,Y :mathcal Atomathcal B$ such that the natural transformation condition holds:
$$ m_HX(f) = Y(f)m_G $$
where $f$ is a morphism in $mathcal A$ and $G,H$ are groups.



For this to make sense, we need a category with objects $Gtimes G$ and $G$. So, we extend (?) the category of groups by $mathcal B_0 := mboxGrp_0 cup Gtimes G mid GinmboxGrp_0$. Morphisms in 'separate components' remain as they are in $mboxGrp$ and $mboxGrptimesmboxGrp$ respectively. There would be no morphisms of the form $Gto Htimes H$ and
$$mathcal B(Gtimes G,H) := varphi m_G mid varphi in mboxHom(G,H) $$
where for every $x,yin G$ $varphi m_G (x,y) := varphi (xy) = varphi (x)varphi (y) =: m_H(varphi,varphi)(x,y)$. The identities are $1_G$ or $(1_G,1_G)$ depending on the component and composition of the morphisms would happen naturally.



  1. Is it guaranteed $(A,B) neq (A',B') implies mathcal B(A,B)capmathcal B(A',B') =emptyset, A,A',B,B'inmathcal B_0$?

  2. Taking $mathcal A = mboxGrp$ with $X :mathcal Atomathcal B$ given such that $X(G) = Gtimes G$ and for every morphism $f:Gto H$, $X(f) = (f,f)$. Put $Y:mathcal Atomathcal B$ as the embedding, then we would have $m_G$ as a natural transformation $XRightarrow Y$.

I omit the routine checks here, for they aren't important for this discussion. I am interested in whether this idea of regarding families of operations as natural transformations is, call it, well-founded



Questionnaire.



  1. Would such an approach be the only one? How else (if at all) would we regard the family $m_G$ as a natural transformation?

  2. Is this a more general thing in universal algebra? Given a class of algebras with certain operations of various arities, could we regard every family of operations as a natural transformation? (For instance, inverse operation or unit element operation of groups)









share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    @H.H. the components must also be morphisms themselves in the destination category so if it were to work, then that must involve some 'weird' categories. I'm concerned about the way I defined $mathcal B$ yet I can't find any rule that says a category can't contain objects of different 'type'.
    $endgroup$
    – Alvin Lepik
    Mar 18 at 11:27













5












5








5


2



$begingroup$


Apologies in advance if the following makes little to no sense, but here goes ..



Denote $m_G : Gtimes Gto G$ the multiplication of a group $G$. Does it make sense to think of the map $m_G$ as some kind of morphism in a (at the moment unspecified) category?



Even more, could we think of the family $m_G$ (indexed by the class of groups) as a natural transformation of some or other functors?



Let's call our mystery categories $mathcal A$ and $mathcal B$ with mystery functors $X,Y :mathcal Atomathcal B$ such that the natural transformation condition holds:
$$ m_HX(f) = Y(f)m_G $$
where $f$ is a morphism in $mathcal A$ and $G,H$ are groups.



For this to make sense, we need a category with objects $Gtimes G$ and $G$. So, we extend (?) the category of groups by $mathcal B_0 := mboxGrp_0 cup Gtimes G mid GinmboxGrp_0$. Morphisms in 'separate components' remain as they are in $mboxGrp$ and $mboxGrptimesmboxGrp$ respectively. There would be no morphisms of the form $Gto Htimes H$ and
$$mathcal B(Gtimes G,H) := varphi m_G mid varphi in mboxHom(G,H) $$
where for every $x,yin G$ $varphi m_G (x,y) := varphi (xy) = varphi (x)varphi (y) =: m_H(varphi,varphi)(x,y)$. The identities are $1_G$ or $(1_G,1_G)$ depending on the component and composition of the morphisms would happen naturally.



  1. Is it guaranteed $(A,B) neq (A',B') implies mathcal B(A,B)capmathcal B(A',B') =emptyset, A,A',B,B'inmathcal B_0$?

  2. Taking $mathcal A = mboxGrp$ with $X :mathcal Atomathcal B$ given such that $X(G) = Gtimes G$ and for every morphism $f:Gto H$, $X(f) = (f,f)$. Put $Y:mathcal Atomathcal B$ as the embedding, then we would have $m_G$ as a natural transformation $XRightarrow Y$.

I omit the routine checks here, for they aren't important for this discussion. I am interested in whether this idea of regarding families of operations as natural transformations is, call it, well-founded



Questionnaire.



  1. Would such an approach be the only one? How else (if at all) would we regard the family $m_G$ as a natural transformation?

  2. Is this a more general thing in universal algebra? Given a class of algebras with certain operations of various arities, could we regard every family of operations as a natural transformation? (For instance, inverse operation or unit element operation of groups)









share|cite|improve this question











$endgroup$




Apologies in advance if the following makes little to no sense, but here goes ..



Denote $m_G : Gtimes Gto G$ the multiplication of a group $G$. Does it make sense to think of the map $m_G$ as some kind of morphism in a (at the moment unspecified) category?



Even more, could we think of the family $m_G$ (indexed by the class of groups) as a natural transformation of some or other functors?



Let's call our mystery categories $mathcal A$ and $mathcal B$ with mystery functors $X,Y :mathcal Atomathcal B$ such that the natural transformation condition holds:
$$ m_HX(f) = Y(f)m_G $$
where $f$ is a morphism in $mathcal A$ and $G,H$ are groups.



For this to make sense, we need a category with objects $Gtimes G$ and $G$. So, we extend (?) the category of groups by $mathcal B_0 := mboxGrp_0 cup Gtimes G mid GinmboxGrp_0$. Morphisms in 'separate components' remain as they are in $mboxGrp$ and $mboxGrptimesmboxGrp$ respectively. There would be no morphisms of the form $Gto Htimes H$ and
$$mathcal B(Gtimes G,H) := varphi m_G mid varphi in mboxHom(G,H) $$
where for every $x,yin G$ $varphi m_G (x,y) := varphi (xy) = varphi (x)varphi (y) =: m_H(varphi,varphi)(x,y)$. The identities are $1_G$ or $(1_G,1_G)$ depending on the component and composition of the morphisms would happen naturally.



  1. Is it guaranteed $(A,B) neq (A',B') implies mathcal B(A,B)capmathcal B(A',B') =emptyset, A,A',B,B'inmathcal B_0$?

  2. Taking $mathcal A = mboxGrp$ with $X :mathcal Atomathcal B$ given such that $X(G) = Gtimes G$ and for every morphism $f:Gto H$, $X(f) = (f,f)$. Put $Y:mathcal Atomathcal B$ as the embedding, then we would have $m_G$ as a natural transformation $XRightarrow Y$.

I omit the routine checks here, for they aren't important for this discussion. I am interested in whether this idea of regarding families of operations as natural transformations is, call it, well-founded



Questionnaire.



  1. Would such an approach be the only one? How else (if at all) would we regard the family $m_G$ as a natural transformation?

  2. Is this a more general thing in universal algebra? Given a class of algebras with certain operations of various arities, could we regard every family of operations as a natural transformation? (For instance, inverse operation or unit element operation of groups)






group-theory category-theory universal-algebra natural-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 18 at 11:13







Alvin Lepik

















asked Mar 18 at 10:54









Alvin LepikAlvin Lepik

2,7961924




2,7961924







  • 1




    $begingroup$
    @H.H. the components must also be morphisms themselves in the destination category so if it were to work, then that must involve some 'weird' categories. I'm concerned about the way I defined $mathcal B$ yet I can't find any rule that says a category can't contain objects of different 'type'.
    $endgroup$
    – Alvin Lepik
    Mar 18 at 11:27












  • 1




    $begingroup$
    @H.H. the components must also be morphisms themselves in the destination category so if it were to work, then that must involve some 'weird' categories. I'm concerned about the way I defined $mathcal B$ yet I can't find any rule that says a category can't contain objects of different 'type'.
    $endgroup$
    – Alvin Lepik
    Mar 18 at 11:27







1




1




$begingroup$
@H.H. the components must also be morphisms themselves in the destination category so if it were to work, then that must involve some 'weird' categories. I'm concerned about the way I defined $mathcal B$ yet I can't find any rule that says a category can't contain objects of different 'type'.
$endgroup$
– Alvin Lepik
Mar 18 at 11:27




$begingroup$
@H.H. the components must also be morphisms themselves in the destination category so if it were to work, then that must involve some 'weird' categories. I'm concerned about the way I defined $mathcal B$ yet I can't find any rule that says a category can't contain objects of different 'type'.
$endgroup$
– Alvin Lepik
Mar 18 at 11:27










1 Answer
1






active

oldest

votes


















4












$begingroup$

Let $DeclareMathOperatorSetSetSet$ denote the category of sets and $DeclareMathOperatorGrpGrpGrp$ denote the category of groups.
Let $Upsilon:GrptoSet$ denote the forgetful functor and $Delta:SettoSet$ be the diagonal functor $Delta(X)=Xtimes X$.



Then you are looking for a natural transformation $mu:DeltacircUpsilontoUpsilon$.
If $X$ is a group, then $Upsilon(X)$ is its underlying set.
For each group $X$ let $mu_X:(DeltacircUpsilon)(X)toUpsilon(X)$ be its composition law.
If $f:Xto Y$ is a group homomorphism, then we have a commutative diagram$requireAMScd$:
beginCD
Upsilon(X)timesUpsilon(X)@=(DeltacircUpsilon)(X)@>mu_X>>Upsilon(X)\
@VUpsilon(f)timesUpsilon(f)VV@V(DeltacircUpsilon)(f)VV@VVUpsilon(f)V\
Upsilon(Y)timesUpsilon(Y)@=(DeltacircUpsilon)(Y)@>>mu_Y >Upsilon(Y)
endCD

which proves naturalness of $mu_X$.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    This looks much neater and cleaner than my attempt. Thanks, I have a much better understanding of this idea now.
    $endgroup$
    – Alvin Lepik
    Mar 18 at 12:31










  • $begingroup$
    Also thanks for demonstrating the amscd package, looks much more convenient than tikz syntax-wise. Can't believe I didn't know about it!
    $endgroup$
    – Alvin Lepik
    Mar 18 at 12:44











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: "69"
;
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%2fmath.stackexchange.com%2fquestions%2f3152645%2falgebra-operations-as-natural-transformations%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









4












$begingroup$

Let $DeclareMathOperatorSetSetSet$ denote the category of sets and $DeclareMathOperatorGrpGrpGrp$ denote the category of groups.
Let $Upsilon:GrptoSet$ denote the forgetful functor and $Delta:SettoSet$ be the diagonal functor $Delta(X)=Xtimes X$.



Then you are looking for a natural transformation $mu:DeltacircUpsilontoUpsilon$.
If $X$ is a group, then $Upsilon(X)$ is its underlying set.
For each group $X$ let $mu_X:(DeltacircUpsilon)(X)toUpsilon(X)$ be its composition law.
If $f:Xto Y$ is a group homomorphism, then we have a commutative diagram$requireAMScd$:
beginCD
Upsilon(X)timesUpsilon(X)@=(DeltacircUpsilon)(X)@>mu_X>>Upsilon(X)\
@VUpsilon(f)timesUpsilon(f)VV@V(DeltacircUpsilon)(f)VV@VVUpsilon(f)V\
Upsilon(Y)timesUpsilon(Y)@=(DeltacircUpsilon)(Y)@>>mu_Y >Upsilon(Y)
endCD

which proves naturalness of $mu_X$.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    This looks much neater and cleaner than my attempt. Thanks, I have a much better understanding of this idea now.
    $endgroup$
    – Alvin Lepik
    Mar 18 at 12:31










  • $begingroup$
    Also thanks for demonstrating the amscd package, looks much more convenient than tikz syntax-wise. Can't believe I didn't know about it!
    $endgroup$
    – Alvin Lepik
    Mar 18 at 12:44















4












$begingroup$

Let $DeclareMathOperatorSetSetSet$ denote the category of sets and $DeclareMathOperatorGrpGrpGrp$ denote the category of groups.
Let $Upsilon:GrptoSet$ denote the forgetful functor and $Delta:SettoSet$ be the diagonal functor $Delta(X)=Xtimes X$.



Then you are looking for a natural transformation $mu:DeltacircUpsilontoUpsilon$.
If $X$ is a group, then $Upsilon(X)$ is its underlying set.
For each group $X$ let $mu_X:(DeltacircUpsilon)(X)toUpsilon(X)$ be its composition law.
If $f:Xto Y$ is a group homomorphism, then we have a commutative diagram$requireAMScd$:
beginCD
Upsilon(X)timesUpsilon(X)@=(DeltacircUpsilon)(X)@>mu_X>>Upsilon(X)\
@VUpsilon(f)timesUpsilon(f)VV@V(DeltacircUpsilon)(f)VV@VVUpsilon(f)V\
Upsilon(Y)timesUpsilon(Y)@=(DeltacircUpsilon)(Y)@>>mu_Y >Upsilon(Y)
endCD

which proves naturalness of $mu_X$.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    This looks much neater and cleaner than my attempt. Thanks, I have a much better understanding of this idea now.
    $endgroup$
    – Alvin Lepik
    Mar 18 at 12:31










  • $begingroup$
    Also thanks for demonstrating the amscd package, looks much more convenient than tikz syntax-wise. Can't believe I didn't know about it!
    $endgroup$
    – Alvin Lepik
    Mar 18 at 12:44













4












4








4





$begingroup$

Let $DeclareMathOperatorSetSetSet$ denote the category of sets and $DeclareMathOperatorGrpGrpGrp$ denote the category of groups.
Let $Upsilon:GrptoSet$ denote the forgetful functor and $Delta:SettoSet$ be the diagonal functor $Delta(X)=Xtimes X$.



Then you are looking for a natural transformation $mu:DeltacircUpsilontoUpsilon$.
If $X$ is a group, then $Upsilon(X)$ is its underlying set.
For each group $X$ let $mu_X:(DeltacircUpsilon)(X)toUpsilon(X)$ be its composition law.
If $f:Xto Y$ is a group homomorphism, then we have a commutative diagram$requireAMScd$:
beginCD
Upsilon(X)timesUpsilon(X)@=(DeltacircUpsilon)(X)@>mu_X>>Upsilon(X)\
@VUpsilon(f)timesUpsilon(f)VV@V(DeltacircUpsilon)(f)VV@VVUpsilon(f)V\
Upsilon(Y)timesUpsilon(Y)@=(DeltacircUpsilon)(Y)@>>mu_Y >Upsilon(Y)
endCD

which proves naturalness of $mu_X$.






share|cite|improve this answer











$endgroup$



Let $DeclareMathOperatorSetSetSet$ denote the category of sets and $DeclareMathOperatorGrpGrpGrp$ denote the category of groups.
Let $Upsilon:GrptoSet$ denote the forgetful functor and $Delta:SettoSet$ be the diagonal functor $Delta(X)=Xtimes X$.



Then you are looking for a natural transformation $mu:DeltacircUpsilontoUpsilon$.
If $X$ is a group, then $Upsilon(X)$ is its underlying set.
For each group $X$ let $mu_X:(DeltacircUpsilon)(X)toUpsilon(X)$ be its composition law.
If $f:Xto Y$ is a group homomorphism, then we have a commutative diagram$requireAMScd$:
beginCD
Upsilon(X)timesUpsilon(X)@=(DeltacircUpsilon)(X)@>mu_X>>Upsilon(X)\
@VUpsilon(f)timesUpsilon(f)VV@V(DeltacircUpsilon)(f)VV@VVUpsilon(f)V\
Upsilon(Y)timesUpsilon(Y)@=(DeltacircUpsilon)(Y)@>>mu_Y >Upsilon(Y)
endCD

which proves naturalness of $mu_X$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 19 at 16:23









Litho

3,4671716




3,4671716










answered Mar 18 at 12:18









Fabio LucchiniFabio Lucchini

9,56611426




9,56611426







  • 1




    $begingroup$
    This looks much neater and cleaner than my attempt. Thanks, I have a much better understanding of this idea now.
    $endgroup$
    – Alvin Lepik
    Mar 18 at 12:31










  • $begingroup$
    Also thanks for demonstrating the amscd package, looks much more convenient than tikz syntax-wise. Can't believe I didn't know about it!
    $endgroup$
    – Alvin Lepik
    Mar 18 at 12:44












  • 1




    $begingroup$
    This looks much neater and cleaner than my attempt. Thanks, I have a much better understanding of this idea now.
    $endgroup$
    – Alvin Lepik
    Mar 18 at 12:31










  • $begingroup$
    Also thanks for demonstrating the amscd package, looks much more convenient than tikz syntax-wise. Can't believe I didn't know about it!
    $endgroup$
    – Alvin Lepik
    Mar 18 at 12:44







1




1




$begingroup$
This looks much neater and cleaner than my attempt. Thanks, I have a much better understanding of this idea now.
$endgroup$
– Alvin Lepik
Mar 18 at 12:31




$begingroup$
This looks much neater and cleaner than my attempt. Thanks, I have a much better understanding of this idea now.
$endgroup$
– Alvin Lepik
Mar 18 at 12:31












$begingroup$
Also thanks for demonstrating the amscd package, looks much more convenient than tikz syntax-wise. Can't believe I didn't know about it!
$endgroup$
– Alvin Lepik
Mar 18 at 12:44




$begingroup$
Also thanks for demonstrating the amscd package, looks much more convenient than tikz syntax-wise. Can't believe I didn't know about it!
$endgroup$
– Alvin Lepik
Mar 18 at 12:44

















draft saved

draft discarded
















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • 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%2fmath.stackexchange.com%2fquestions%2f3152645%2falgebra-operations-as-natural-transformations%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

How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer

random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye