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







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    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






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          $begingroup$

          $mathcal{C}$ 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









          $endgroup$





















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            $begingroup$

            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











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            2 Answers
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            $begingroup$

            $mathcal{C}$ 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


















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