How many morphisms from 1 to 1+1 can there be? Planned maintenance scheduled April 23, 2019 at...



How many morphisms from 1 to 1+1 can there be?



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$begingroup$


Here is an interesting question raised by Alice Rhyl.



Let $C$ be a category with a terminal object $1$ and finite coproducts. How many different morphisms $f : 1 to 1 + 1$ can there be?



There are always two obvious morphisms $f : 1 to 1 + 1$, coming from the definition of coproduct. But if $C$ is the category with one object and one morphism, $1 + 1 = 1$ so these two obvious morphisms are equal and there's really just one.



Can there be three different morphisms $f : 1 to 1 + 1$? I don't know.



There can be four. Take $C = mathrm{Set}^2$; then the terminal object in $C$ is $(1,1)$ (where $1$ is your favorite one-element set), and there are four different morphisms $f: (1,1) to (1,1) + (1,1)$.



Indeed, any power of two is possible; just take $C = mathrm{Set}^n$.



What other numbers are possible? (I find finite cardinals more interesting here.)



So far $C$ has just been any category with a terminal object and finite coproducts. In a paper I'm writing with Christian Williams, I'm more interested in the case where $C$ is a cartesian closed category with finite coproducts. How many morphisms $f : 1 to 1 + 1$ can there be in this case?



(All the examples I've given above are categories of this sort.)










share|cite|improve this question











$endgroup$

















    6












    $begingroup$


    Here is an interesting question raised by Alice Rhyl.



    Let $C$ be a category with a terminal object $1$ and finite coproducts. How many different morphisms $f : 1 to 1 + 1$ can there be?



    There are always two obvious morphisms $f : 1 to 1 + 1$, coming from the definition of coproduct. But if $C$ is the category with one object and one morphism, $1 + 1 = 1$ so these two obvious morphisms are equal and there's really just one.



    Can there be three different morphisms $f : 1 to 1 + 1$? I don't know.



    There can be four. Take $C = mathrm{Set}^2$; then the terminal object in $C$ is $(1,1)$ (where $1$ is your favorite one-element set), and there are four different morphisms $f: (1,1) to (1,1) + (1,1)$.



    Indeed, any power of two is possible; just take $C = mathrm{Set}^n$.



    What other numbers are possible? (I find finite cardinals more interesting here.)



    So far $C$ has just been any category with a terminal object and finite coproducts. In a paper I'm writing with Christian Williams, I'm more interested in the case where $C$ is a cartesian closed category with finite coproducts. How many morphisms $f : 1 to 1 + 1$ can there be in this case?



    (All the examples I've given above are categories of this sort.)










    share|cite|improve this question











    $endgroup$















      6












      6








      6


      1



      $begingroup$


      Here is an interesting question raised by Alice Rhyl.



      Let $C$ be a category with a terminal object $1$ and finite coproducts. How many different morphisms $f : 1 to 1 + 1$ can there be?



      There are always two obvious morphisms $f : 1 to 1 + 1$, coming from the definition of coproduct. But if $C$ is the category with one object and one morphism, $1 + 1 = 1$ so these two obvious morphisms are equal and there's really just one.



      Can there be three different morphisms $f : 1 to 1 + 1$? I don't know.



      There can be four. Take $C = mathrm{Set}^2$; then the terminal object in $C$ is $(1,1)$ (where $1$ is your favorite one-element set), and there are four different morphisms $f: (1,1) to (1,1) + (1,1)$.



      Indeed, any power of two is possible; just take $C = mathrm{Set}^n$.



      What other numbers are possible? (I find finite cardinals more interesting here.)



      So far $C$ has just been any category with a terminal object and finite coproducts. In a paper I'm writing with Christian Williams, I'm more interested in the case where $C$ is a cartesian closed category with finite coproducts. How many morphisms $f : 1 to 1 + 1$ can there be in this case?



      (All the examples I've given above are categories of this sort.)










      share|cite|improve this question











      $endgroup$




      Here is an interesting question raised by Alice Rhyl.



      Let $C$ be a category with a terminal object $1$ and finite coproducts. How many different morphisms $f : 1 to 1 + 1$ can there be?



      There are always two obvious morphisms $f : 1 to 1 + 1$, coming from the definition of coproduct. But if $C$ is the category with one object and one morphism, $1 + 1 = 1$ so these two obvious morphisms are equal and there's really just one.



      Can there be three different morphisms $f : 1 to 1 + 1$? I don't know.



      There can be four. Take $C = mathrm{Set}^2$; then the terminal object in $C$ is $(1,1)$ (where $1$ is your favorite one-element set), and there are four different morphisms $f: (1,1) to (1,1) + (1,1)$.



      Indeed, any power of two is possible; just take $C = mathrm{Set}^n$.



      What other numbers are possible? (I find finite cardinals more interesting here.)



      So far $C$ has just been any category with a terminal object and finite coproducts. In a paper I'm writing with Christian Williams, I'm more interested in the case where $C$ is a cartesian closed category with finite coproducts. How many morphisms $f : 1 to 1 + 1$ can there be in this case?



      (All the examples I've given above are categories of this sort.)







      ct.category-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago







      John Baez

















      asked 2 hours ago









      John BaezJohn Baez

      8,7704699




      8,7704699






















          1 Answer
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          active

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          2












          $begingroup$

          Let $V$ be any variety of idempotent operations, such as the varieties of idempotent groupoids (aka magmas), or idempotent semigroups, or semilattices, or lattices. Then $V$ is complete and cocomplete (if we include in $V$ an empty algebra); $1$ is just the $1$-element algebra, and morphisms $1to A$ are in 1–1 correspondence with set-theoretical elements of $A$. The object $1$ is also the free algebra on $1$ generator, hence $1+1$ is the free algebra on $2$ generators. So, all in all, there are as many morphisms $1to1+1$ as is the cardinality of the free $V$-algebra on two generators. For example:




          • If $V$ is the variety of semilattices, the number is $3$.


          • If $V$ is the variety of idempotent semigroups, the number is $6$.


          • If $V$ is the variety of idempotent groupoids, the number is $aleph_0$. More generally, if $V$ is the variety of all algebras with $kappagealeph_0$ idempotent binary operations, then the number is $kappa$.







          share|cite|improve this answer











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            1 Answer
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            $begingroup$

            Let $V$ be any variety of idempotent operations, such as the varieties of idempotent groupoids (aka magmas), or idempotent semigroups, or semilattices, or lattices. Then $V$ is complete and cocomplete (if we include in $V$ an empty algebra); $1$ is just the $1$-element algebra, and morphisms $1to A$ are in 1–1 correspondence with set-theoretical elements of $A$. The object $1$ is also the free algebra on $1$ generator, hence $1+1$ is the free algebra on $2$ generators. So, all in all, there are as many morphisms $1to1+1$ as is the cardinality of the free $V$-algebra on two generators. For example:




            • If $V$ is the variety of semilattices, the number is $3$.


            • If $V$ is the variety of idempotent semigroups, the number is $6$.


            • If $V$ is the variety of idempotent groupoids, the number is $aleph_0$. More generally, if $V$ is the variety of all algebras with $kappagealeph_0$ idempotent binary operations, then the number is $kappa$.







            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Let $V$ be any variety of idempotent operations, such as the varieties of idempotent groupoids (aka magmas), or idempotent semigroups, or semilattices, or lattices. Then $V$ is complete and cocomplete (if we include in $V$ an empty algebra); $1$ is just the $1$-element algebra, and morphisms $1to A$ are in 1–1 correspondence with set-theoretical elements of $A$. The object $1$ is also the free algebra on $1$ generator, hence $1+1$ is the free algebra on $2$ generators. So, all in all, there are as many morphisms $1to1+1$ as is the cardinality of the free $V$-algebra on two generators. For example:




              • If $V$ is the variety of semilattices, the number is $3$.


              • If $V$ is the variety of idempotent semigroups, the number is $6$.


              • If $V$ is the variety of idempotent groupoids, the number is $aleph_0$. More generally, if $V$ is the variety of all algebras with $kappagealeph_0$ idempotent binary operations, then the number is $kappa$.







              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Let $V$ be any variety of idempotent operations, such as the varieties of idempotent groupoids (aka magmas), or idempotent semigroups, or semilattices, or lattices. Then $V$ is complete and cocomplete (if we include in $V$ an empty algebra); $1$ is just the $1$-element algebra, and morphisms $1to A$ are in 1–1 correspondence with set-theoretical elements of $A$. The object $1$ is also the free algebra on $1$ generator, hence $1+1$ is the free algebra on $2$ generators. So, all in all, there are as many morphisms $1to1+1$ as is the cardinality of the free $V$-algebra on two generators. For example:




                • If $V$ is the variety of semilattices, the number is $3$.


                • If $V$ is the variety of idempotent semigroups, the number is $6$.


                • If $V$ is the variety of idempotent groupoids, the number is $aleph_0$. More generally, if $V$ is the variety of all algebras with $kappagealeph_0$ idempotent binary operations, then the number is $kappa$.







                share|cite|improve this answer











                $endgroup$



                Let $V$ be any variety of idempotent operations, such as the varieties of idempotent groupoids (aka magmas), or idempotent semigroups, or semilattices, or lattices. Then $V$ is complete and cocomplete (if we include in $V$ an empty algebra); $1$ is just the $1$-element algebra, and morphisms $1to A$ are in 1–1 correspondence with set-theoretical elements of $A$. The object $1$ is also the free algebra on $1$ generator, hence $1+1$ is the free algebra on $2$ generators. So, all in all, there are as many morphisms $1to1+1$ as is the cardinality of the free $V$-algebra on two generators. For example:




                • If $V$ is the variety of semilattices, the number is $3$.


                • If $V$ is the variety of idempotent semigroups, the number is $6$.


                • If $V$ is the variety of idempotent groupoids, the number is $aleph_0$. More generally, if $V$ is the variety of all algebras with $kappagealeph_0$ idempotent binary operations, then the number is $kappa$.








                share|cite|improve this answer














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                share|cite|improve this answer








                edited 41 secs ago

























                answered 29 mins ago









                Emil JeřábekEmil Jeřábek

                30.4k389143




                30.4k389143






























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