Why are epimorphisms in the category of Monoids not necessarily surjections?












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The book "basic category theory" states that in the category Mon, epimorphisms are not necessarily surjections, but doesn't explain why. Why is this the case?






category-theory







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    3












    $begingroup$


    The book "basic category theory" states that in the category Mon, epimorphisms are not necessarily surjections, but doesn't explain why. Why is this the case?






    category-theory







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      The book "basic category theory" states that in the category Mon, epimorphisms are not necessarily surjections, but doesn't explain why. Why is this the case?






      category-theory







      share|cite|improve this question









      $endgroup$




      The book "basic category theory" states that in the category Mon, epimorphisms are not necessarily surjections, but doesn't explain why. Why is this the case?






      category-theory




      category-theory






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      asked Feb 1 at 17:40









      user56834user56834

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          3 Answers
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          $begingroup$

          Let $Bbb Z$ be the additive group of integers, and $Bbb N_0={0,1,2,ldots}$.
          Then the inclusion $Bbb N_0toBbb Z$ is an epimorphism in Mon.






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            Quoting Wikipedia, we see:



            In the category of monoids, Mon, the inclusion map $mathbb{N}tomathbb{Z}$ is a non-surjective epimorphism. To see this, suppose that $g_1$ and $g_2$ are two distinct maps from $mathbb{Z}$ to some monoid $M$. Then for some $ninmathbb{Z}$, $g_1(n)neq g_2(n)$, so $g_1(-n)neq g_2(-n)$. Either $n$ or $-n$ is in $mathbb{N}$,, so the restrictions of $g_1$ and $g_2$ to $mathbb{N}$ are unequal.






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            • $begingroup$
              I don't understand the argument? specifically, why does $g_1(-n)neq g_2(-n)$ need to hold? That would hold if we were in the category of groups, I see that. But I don't see why it would hold in the category of monoids. also, I don't get the last sentence.
              $endgroup$
              – user56834
              Feb 1 at 18:17








            • 1




              $begingroup$
              $0=g_1(0)=g_1(n+(-n))=g_1(n)+g_1(-n)$
              $endgroup$
              – vadim123
              Feb 1 at 20:53





















            1












            $begingroup$

            Consider the inclusion $i: mathbb{N}hookrightarrowmathbb{Z}$. This is clearly not surjective. However, for any monoid $M$ and any two maps $f,g:mathbb{Z} to M$ such that $f circ i = g circ i$, we must have $f = g$, by a simple argument.






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              3 Answers
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              3 Answers
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              4












              $begingroup$

              Let $Bbb Z$ be the additive group of integers, and $Bbb N_0={0,1,2,ldots}$.
              Then the inclusion $Bbb N_0toBbb Z$ is an epimorphism in Mon.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                Let $Bbb Z$ be the additive group of integers, and $Bbb N_0={0,1,2,ldots}$.
                Then the inclusion $Bbb N_0toBbb Z$ is an epimorphism in Mon.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  Let $Bbb Z$ be the additive group of integers, and $Bbb N_0={0,1,2,ldots}$.
                  Then the inclusion $Bbb N_0toBbb Z$ is an epimorphism in Mon.






                  share|cite|improve this answer









                  $endgroup$



                  Let $Bbb Z$ be the additive group of integers, and $Bbb N_0={0,1,2,ldots}$.
                  Then the inclusion $Bbb N_0toBbb Z$ is an epimorphism in Mon.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 1 at 17:45









                  Lord Shark the UnknownLord Shark the Unknown

                  104k1160132




                  104k1160132























                      3












                      $begingroup$

                      Quoting Wikipedia, we see:



                      In the category of monoids, Mon, the inclusion map $mathbb{N}tomathbb{Z}$ is a non-surjective epimorphism. To see this, suppose that $g_1$ and $g_2$ are two distinct maps from $mathbb{Z}$ to some monoid $M$. Then for some $ninmathbb{Z}$, $g_1(n)neq g_2(n)$, so $g_1(-n)neq g_2(-n)$. Either $n$ or $-n$ is in $mathbb{N}$,, so the restrictions of $g_1$ and $g_2$ to $mathbb{N}$ are unequal.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I don't understand the argument? specifically, why does $g_1(-n)neq g_2(-n)$ need to hold? That would hold if we were in the category of groups, I see that. But I don't see why it would hold in the category of monoids. also, I don't get the last sentence.
                        $endgroup$
                        – user56834
                        Feb 1 at 18:17








                      • 1




                        $begingroup$
                        $0=g_1(0)=g_1(n+(-n))=g_1(n)+g_1(-n)$
                        $endgroup$
                        – vadim123
                        Feb 1 at 20:53


















                      3












                      $begingroup$

                      Quoting Wikipedia, we see:



                      In the category of monoids, Mon, the inclusion map $mathbb{N}tomathbb{Z}$ is a non-surjective epimorphism. To see this, suppose that $g_1$ and $g_2$ are two distinct maps from $mathbb{Z}$ to some monoid $M$. Then for some $ninmathbb{Z}$, $g_1(n)neq g_2(n)$, so $g_1(-n)neq g_2(-n)$. Either $n$ or $-n$ is in $mathbb{N}$,, so the restrictions of $g_1$ and $g_2$ to $mathbb{N}$ are unequal.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I don't understand the argument? specifically, why does $g_1(-n)neq g_2(-n)$ need to hold? That would hold if we were in the category of groups, I see that. But I don't see why it would hold in the category of monoids. also, I don't get the last sentence.
                        $endgroup$
                        – user56834
                        Feb 1 at 18:17








                      • 1




                        $begingroup$
                        $0=g_1(0)=g_1(n+(-n))=g_1(n)+g_1(-n)$
                        $endgroup$
                        – vadim123
                        Feb 1 at 20:53
















                      3












                      3








                      3





                      $begingroup$

                      Quoting Wikipedia, we see:



                      In the category of monoids, Mon, the inclusion map $mathbb{N}tomathbb{Z}$ is a non-surjective epimorphism. To see this, suppose that $g_1$ and $g_2$ are two distinct maps from $mathbb{Z}$ to some monoid $M$. Then for some $ninmathbb{Z}$, $g_1(n)neq g_2(n)$, so $g_1(-n)neq g_2(-n)$. Either $n$ or $-n$ is in $mathbb{N}$,, so the restrictions of $g_1$ and $g_2$ to $mathbb{N}$ are unequal.






                      share|cite|improve this answer









                      $endgroup$



                      Quoting Wikipedia, we see:



                      In the category of monoids, Mon, the inclusion map $mathbb{N}tomathbb{Z}$ is a non-surjective epimorphism. To see this, suppose that $g_1$ and $g_2$ are two distinct maps from $mathbb{Z}$ to some monoid $M$. Then for some $ninmathbb{Z}$, $g_1(n)neq g_2(n)$, so $g_1(-n)neq g_2(-n)$. Either $n$ or $-n$ is in $mathbb{N}$,, so the restrictions of $g_1$ and $g_2$ to $mathbb{N}$ are unequal.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Feb 1 at 17:46









                      vadim123vadim123

                      76k897189




                      76k897189












                      • $begingroup$
                        I don't understand the argument? specifically, why does $g_1(-n)neq g_2(-n)$ need to hold? That would hold if we were in the category of groups, I see that. But I don't see why it would hold in the category of monoids. also, I don't get the last sentence.
                        $endgroup$
                        – user56834
                        Feb 1 at 18:17








                      • 1




                        $begingroup$
                        $0=g_1(0)=g_1(n+(-n))=g_1(n)+g_1(-n)$
                        $endgroup$
                        – vadim123
                        Feb 1 at 20:53




















                      • $begingroup$
                        I don't understand the argument? specifically, why does $g_1(-n)neq g_2(-n)$ need to hold? That would hold if we were in the category of groups, I see that. But I don't see why it would hold in the category of monoids. also, I don't get the last sentence.
                        $endgroup$
                        – user56834
                        Feb 1 at 18:17








                      • 1




                        $begingroup$
                        $0=g_1(0)=g_1(n+(-n))=g_1(n)+g_1(-n)$
                        $endgroup$
                        – vadim123
                        Feb 1 at 20:53


















                      $begingroup$
                      I don't understand the argument? specifically, why does $g_1(-n)neq g_2(-n)$ need to hold? That would hold if we were in the category of groups, I see that. But I don't see why it would hold in the category of monoids. also, I don't get the last sentence.
                      $endgroup$
                      – user56834
                      Feb 1 at 18:17






                      $begingroup$
                      I don't understand the argument? specifically, why does $g_1(-n)neq g_2(-n)$ need to hold? That would hold if we were in the category of groups, I see that. But I don't see why it would hold in the category of monoids. also, I don't get the last sentence.
                      $endgroup$
                      – user56834
                      Feb 1 at 18:17






                      1




                      1




                      $begingroup$
                      $0=g_1(0)=g_1(n+(-n))=g_1(n)+g_1(-n)$
                      $endgroup$
                      – vadim123
                      Feb 1 at 20:53






                      $begingroup$
                      $0=g_1(0)=g_1(n+(-n))=g_1(n)+g_1(-n)$
                      $endgroup$
                      – vadim123
                      Feb 1 at 20:53













                      1












                      $begingroup$

                      Consider the inclusion $i: mathbb{N}hookrightarrowmathbb{Z}$. This is clearly not surjective. However, for any monoid $M$ and any two maps $f,g:mathbb{Z} to M$ such that $f circ i = g circ i$, we must have $f = g$, by a simple argument.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Consider the inclusion $i: mathbb{N}hookrightarrowmathbb{Z}$. This is clearly not surjective. However, for any monoid $M$ and any two maps $f,g:mathbb{Z} to M$ such that $f circ i = g circ i$, we must have $f = g$, by a simple argument.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Consider the inclusion $i: mathbb{N}hookrightarrowmathbb{Z}$. This is clearly not surjective. However, for any monoid $M$ and any two maps $f,g:mathbb{Z} to M$ such that $f circ i = g circ i$, we must have $f = g$, by a simple argument.






                          share|cite|improve this answer









                          $endgroup$



                          Consider the inclusion $i: mathbb{N}hookrightarrowmathbb{Z}$. This is clearly not surjective. However, for any monoid $M$ and any two maps $f,g:mathbb{Z} to M$ such that $f circ i = g circ i$, we must have $f = g$, by a simple argument.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Feb 1 at 17:47









                          user3482749user3482749

                          4,266919




                          4,266919






























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