Is it meaningful to distinguish a one-object category and its opposite?












2












$begingroup$


As in the title, for a category with only one object $mathcal C$, $mathcal C^{op}$ seems the same with the original one but I still believe there is a difference between them, can someone give an example where $mathcal C$ and $mathcal C^{op}$ play different roles?



Thank you in advance.






category-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
    $endgroup$
    – Musa Al-hassy
    Jan 31 at 0:31
















2












$begingroup$


As in the title, for a category with only one object $mathcal C$, $mathcal C^{op}$ seems the same with the original one but I still believe there is a difference between them, can someone give an example where $mathcal C$ and $mathcal C^{op}$ play different roles?



Thank you in advance.






category-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
    $endgroup$
    – Musa Al-hassy
    Jan 31 at 0:31














2












2








2





$begingroup$


As in the title, for a category with only one object $mathcal C$, $mathcal C^{op}$ seems the same with the original one but I still believe there is a difference between them, can someone give an example where $mathcal C$ and $mathcal C^{op}$ play different roles?



Thank you in advance.






category-theory







share|cite|improve this question









$endgroup$




As in the title, for a category with only one object $mathcal C$, $mathcal C^{op}$ seems the same with the original one but I still believe there is a difference between them, can someone give an example where $mathcal C$ and $mathcal C^{op}$ play different roles?



Thank you in advance.






category-theory




category-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 28 at 20:01









Display NameDisplay Name

444212




444212












  • $begingroup$
    Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
    $endgroup$
    – Musa Al-hassy
    Jan 31 at 0:31


















  • $begingroup$
    Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
    $endgroup$
    – Musa Al-hassy
    Jan 31 at 0:31
















$begingroup$
Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
$endgroup$
– Musa Al-hassy
Jan 31 at 0:31




$begingroup$
Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
$endgroup$
– Musa Al-hassy
Jan 31 at 0:31










2 Answers
2






active

oldest

votes


















4












$begingroup$

$mathcal{C}$ corresponds to a monoid $M$ which is the set of morphisms $hom(cdot,cdot)$. Let us write the multiplication by $circ$. Then $mathcal{C}^{op}$ corresponds to $M^{op}$ which is the same set of $M$ but with multiplication $xcirc' y=ycirc x$.



This is useful sometimes. For example, if $R$ is a ring, then $R^{op}$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^{op}$-module.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BMto Set$ can be identified with a left $M$-set, while a functor $BM^{mathrm{op}}to Set$ is similarly identified with a right $M$-set.



    The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is
    $$M=langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=brangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^{mathrm{op}}$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Perhaps providing an example where $M$ and $M^{op}$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
      $endgroup$
      – Max
      Jan 28 at 20:32








    • 1




      $begingroup$
      @Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
      $endgroup$
      – Kevin Carlson
      Jan 28 at 21:56










    • $begingroup$
      Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
      $endgroup$
      – Max
      Jan 28 at 21:58






    • 1




      $begingroup$
      Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
      $endgroup$
      – Kevin Carlson
      Jan 28 at 22:17











    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%2f3091320%2fis-it-meaningful-to-distinguish-a-one-object-category-and-its-opposite%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    $mathcal{C}$ corresponds to a monoid $M$ which is the set of morphisms $hom(cdot,cdot)$. Let us write the multiplication by $circ$. Then $mathcal{C}^{op}$ corresponds to $M^{op}$ which is the same set of $M$ but with multiplication $xcirc' y=ycirc x$.



    This is useful sometimes. For example, if $R$ is a ring, then $R^{op}$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^{op}$-module.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      $mathcal{C}$ corresponds to a monoid $M$ which is the set of morphisms $hom(cdot,cdot)$. Let us write the multiplication by $circ$. Then $mathcal{C}^{op}$ corresponds to $M^{op}$ which is the same set of $M$ but with multiplication $xcirc' y=ycirc x$.



      This is useful sometimes. For example, if $R$ is a ring, then $R^{op}$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^{op}$-module.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        $mathcal{C}$ corresponds to a monoid $M$ which is the set of morphisms $hom(cdot,cdot)$. Let us write the multiplication by $circ$. Then $mathcal{C}^{op}$ corresponds to $M^{op}$ which is the same set of $M$ but with multiplication $xcirc' y=ycirc x$.



        This is useful sometimes. For example, if $R$ is a ring, then $R^{op}$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^{op}$-module.






        share|cite|improve this answer









        $endgroup$



        $mathcal{C}$ corresponds to a monoid $M$ which is the set of morphisms $hom(cdot,cdot)$. Let us write the multiplication by $circ$. Then $mathcal{C}^{op}$ corresponds to $M^{op}$ which is the same set of $M$ but with multiplication $xcirc' y=ycirc x$.



        This is useful sometimes. For example, if $R$ is a ring, then $R^{op}$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^{op}$-module.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 28 at 20:24









        Eclipse SunEclipse Sun

        7,5021437




        7,5021437























            3












            $begingroup$

            Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BMto Set$ can be identified with a left $M$-set, while a functor $BM^{mathrm{op}}to Set$ is similarly identified with a right $M$-set.



            The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is
            $$M=langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=brangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^{mathrm{op}}$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Perhaps providing an example where $M$ and $M^{op}$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
              $endgroup$
              – Max
              Jan 28 at 20:32








            • 1




              $begingroup$
              @Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
              $endgroup$
              – Kevin Carlson
              Jan 28 at 21:56










            • $begingroup$
              Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
              $endgroup$
              – Max
              Jan 28 at 21:58






            • 1




              $begingroup$
              Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
              $endgroup$
              – Kevin Carlson
              Jan 28 at 22:17
















            3












            $begingroup$

            Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BMto Set$ can be identified with a left $M$-set, while a functor $BM^{mathrm{op}}to Set$ is similarly identified with a right $M$-set.



            The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is
            $$M=langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=brangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^{mathrm{op}}$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Perhaps providing an example where $M$ and $M^{op}$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
              $endgroup$
              – Max
              Jan 28 at 20:32








            • 1




              $begingroup$
              @Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
              $endgroup$
              – Kevin Carlson
              Jan 28 at 21:56










            • $begingroup$
              Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
              $endgroup$
              – Max
              Jan 28 at 21:58






            • 1




              $begingroup$
              Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
              $endgroup$
              – Kevin Carlson
              Jan 28 at 22:17














            3












            3








            3





            $begingroup$

            Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BMto Set$ can be identified with a left $M$-set, while a functor $BM^{mathrm{op}}to Set$ is similarly identified with a right $M$-set.



            The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is
            $$M=langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=brangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^{mathrm{op}}$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to






            share|cite|improve this answer











            $endgroup$



            Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BMto Set$ can be identified with a left $M$-set, while a functor $BM^{mathrm{op}}to Set$ is similarly identified with a right $M$-set.



            The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is
            $$M=langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=brangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^{mathrm{op}}$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 28 at 22:16

























            answered Jan 28 at 20:21









            Kevin CarlsonKevin Carlson

            32.9k23372




            32.9k23372












            • $begingroup$
              Perhaps providing an example where $M$ and $M^{op}$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
              $endgroup$
              – Max
              Jan 28 at 20:32








            • 1




              $begingroup$
              @Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
              $endgroup$
              – Kevin Carlson
              Jan 28 at 21:56










            • $begingroup$
              Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
              $endgroup$
              – Max
              Jan 28 at 21:58






            • 1




              $begingroup$
              Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
              $endgroup$
              – Kevin Carlson
              Jan 28 at 22:17


















            • $begingroup$
              Perhaps providing an example where $M$ and $M^{op}$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
              $endgroup$
              – Max
              Jan 28 at 20:32








            • 1




              $begingroup$
              @Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
              $endgroup$
              – Kevin Carlson
              Jan 28 at 21:56










            • $begingroup$
              Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
              $endgroup$
              – Max
              Jan 28 at 21:58






            • 1




              $begingroup$
              Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
              $endgroup$
              – Kevin Carlson
              Jan 28 at 22:17
















            $begingroup$
            Perhaps providing an example where $M$ and $M^{op}$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
            $endgroup$
            – Max
            Jan 28 at 20:32






            $begingroup$
            Perhaps providing an example where $M$ and $M^{op}$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
            $endgroup$
            – Max
            Jan 28 at 20:32






            1




            1




            $begingroup$
            @Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
            $endgroup$
            – Kevin Carlson
            Jan 28 at 21:56




            $begingroup$
            @Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
            $endgroup$
            – Kevin Carlson
            Jan 28 at 21:56












            $begingroup$
            Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
            $endgroup$
            – Max
            Jan 28 at 21:58




            $begingroup$
            Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
            $endgroup$
            – Max
            Jan 28 at 21:58




            1




            1




            $begingroup$
            Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
            $endgroup$
            – Kevin Carlson
            Jan 28 at 22:17




            $begingroup$
            Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
            $endgroup$
            – Kevin Carlson
            Jan 28 at 22:17


















            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%2f3091320%2fis-it-meaningful-to-distinguish-a-one-object-category-and-its-opposite%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

            mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

            Image
            0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
            Read more

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more