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









$endgroup$

















    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









    $endgroup$















      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 2 hours ago









      Display NameDisplay Name

      442212




      442212






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          $mathcal{C}$ 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
              2 hours ago








            • 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
              51 mins ago










            • $begingroup$
              Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
              $endgroup$
              – Max
              50 mins ago






            • 1




              $begingroup$
              Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
              $endgroup$
              – Kevin Carlson
              31 mins 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: "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









            3












            $begingroup$

            $mathcal{C}$ 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












              $begingroup$

              $mathcal{C}$ 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












                3








                3





                $begingroup$

                $mathcal{C}$ 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 2 hours ago









                Eclipse SunEclipse Sun

                7,2141437




                7,2141437























                    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
                      2 hours ago








                    • 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
                      51 mins ago










                    • $begingroup$
                      Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
                      $endgroup$
                      – Max
                      50 mins ago






                    • 1




                      $begingroup$
                      Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
                      $endgroup$
                      – Kevin Carlson
                      31 mins ago
















                    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
                      2 hours ago








                    • 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
                      51 mins ago










                    • $begingroup$
                      Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
                      $endgroup$
                      – Max
                      50 mins ago






                    • 1




                      $begingroup$
                      Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
                      $endgroup$
                      – Kevin Carlson
                      31 mins ago














                    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 32 mins ago

























                    answered 2 hours ago









                    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
                      2 hours ago








                    • 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
                      51 mins ago










                    • $begingroup$
                      Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
                      $endgroup$
                      – Max
                      50 mins ago






                    • 1




                      $begingroup$
                      Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
                      $endgroup$
                      – Kevin Carlson
                      31 mins ago


















                    • $begingroup$
                      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
                      2 hours ago








                    • 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
                      51 mins ago










                    • $begingroup$
                      Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
                      $endgroup$
                      – Max
                      50 mins ago






                    • 1




                      $begingroup$
                      Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
                      $endgroup$
                      – Kevin Carlson
                      31 mins ago
















                    $begingroup$
                    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
                    2 hours ago






                    $begingroup$
                    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
                    2 hours ago






                    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
                    51 mins ago




                    $begingroup$
                    @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
                    51 mins ago












                    $begingroup$
                    Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
                    $endgroup$
                    – Max
                    50 mins ago




                    $begingroup$
                    Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
                    $endgroup$
                    – Max
                    50 mins ago




                    1




                    1




                    $begingroup$
                    Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
                    $endgroup$
                    – Kevin Carlson
                    31 mins ago




                    $begingroup$
                    Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
                    $endgroup$
                    – Kevin Carlson
                    31 mins ago


















                    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

                    Olav Thon

                    Image
                    Olav Thon Fødd 29. juni 1923 Ål Yrke forretningsdrivande Olav Thon på Commons Olav Thon (fødd 29. juni 1923) er ein norsk, stort sett sjølvlært forretningsmann, fødd i Ål i Hallingdal, som har gjort store pengar på særleg kjøp, bygging og utvikling av fast eigedom, mellom anna kjøpesentre, hotell og restaurantar, i både Noreg og somme andre land. Han vart Ridder av 1. klasse av St. Olavs Orden i 2003, og er som ivrig friluftsmann æresmedlem i Den Norske Turistforening etter å ha ytt store økonomiske bidrag til foreininga sine hytter og ruter i fjellet. Han er kjend for alltid å gå med ei karakteristisk strikkelue. Olav Thon i samtale med Trond Helleland på sentralbanksjefens årstale i februar 2016 I ung alder reiste Olav Thon til Newcastle for å lære seg engelsk, og han var seinare ei tid i Sveits, der han gjorde forretningar i import og eksport. Deretter tok han til å kjøpe faste eigedomar i Noreg med stor økonomisk suksess, verksemda voks snøgt og v...
                    Read more

                    Waikiki

                    Image
                    Denne artikkelen kan ha godt av ein språkvask , som reinskar opp målføringa og/eller innfører same språkstilen overalt. Denne artikkelen kan ha godt av ei opprydding for å nå ein høgare standard og/eller for å verta i tråd med standardoppsettet. Sjå korleis du redigerer ei side og stilmanualen for hjelp. Waikiki Land   USA Stat Hawaii Stad Honolulu Wikimedia Commons: Waikiki Utsyn over Waikiki Waikīkī er ein bydel i Honolulu, sør på øya Oahu, som er ei av Hawaii-øyane. Waikīkīstranda er ein av dei meste berømte strendene i verda. Waikīkī strekkjer seg langs stillehavskysten og har ei rekkje velrenommerte hotell. Bydelen er Hawaii-turismens høyborg. På Waikīkī Beach står ei statue av den lokale sportsstjerna Duka Kahanamoku. Kjelder | Denne artikkelen bygger på «Waikiki» frå Wikipedia på bokmål, den 4. november 2018. Denne artikkelen manglar kjelder eller referansar . Hjelp Wikipedia med å finna truverdige kjelder...
                    Read more

                    Tårekanal

                    Image
                    Tåreapparatet ( a ) Tårekjertel ( b ) Øvre tårepunkt ( c ) Øvre tårekanalikkel ( d ) Tåresekk ( e ) Nedre tårepunkt ( f ) Nedre tårekanalikkel ( g ) Tårekanal Tårekanalen eller nase-tårekanalen (av latin ductus nasolacrimalis ) er ein ductus som frakter overflødige tårer frå tåresekken til nasehola. Opninga inn mot nedre nasegang er delvis tildekt av ei ufullstendig slimhinnefold kalla Hasners ventil. Overflødige tårer strøymer gjennom tårekanalen og trekker inn i nedre nasegang. Dette er grunnen til at ein kan få rennande nase når ein gret, og at ein somme gonger kan kjenne saltsmaken av tårer i munnen. Både tårekanalen og tåresekken er tekt med fleirlaga sylinderepitel som inneheld slimproduserande begerceller, og er begge omgjevne av bindevev. Dakryostenose (tårekanalstenose) er ei innsnevring av tårekanalen som fører til epifora, ei overflod av tårer. Medfødd dakryostenose kan føra til cystisk utviding av tårekanalen ved at det dannast ein dakryocystocele. Personar m...
                    Read more