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











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    2 Answers
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    2 Answers
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    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


















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