Is a categorical coproduct of epimorphisms (monomorphisms) always an epimorphism (a monomorphism)?











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Let $mathbf{C}$ be a category (that does not necessary have a coproduct for every collection of objects). Suppose that we have two families of objects $(A_i)_{iin I}$ and $(B_i)_{iin I}$ in $mathbf{C}$ indexed by the same index set $I$. Assume further that there exist coproducts $A$ and $B$ of $(A_i)_{iin I}$ and of $(B_i)_{iin I}$, respectively. If $f_i:A_ito B_i$ are morphisms for all $iin I$, then there exists a unique morphism $f:Ato B$, which is the coproduct of $(f_i)_{iin I}$.



I know the following results. The morphism $f$ need not be monic, even if all $f_i$ are monic. If $mathbf{C}$ is an abelian category and if all $f_i$ are epic, then $f$ is epic. Therefore, I have two related questions. All references are very welcome.




Question I. In general (where $mathbf{C}$ need not be an abelian category), does $f$ have to be epic, provided that all $f_i$ are epic? If so, could you please provide a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?



Question II. If $mathbf{C}$ is an abelian category and if all $f_i$ are monic, then does it follow that $f$ is monic? If so, could you please provide
a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?











share|cite|improve this question




























    up vote
    7
    down vote

    favorite












    Let $mathbf{C}$ be a category (that does not necessary have a coproduct for every collection of objects). Suppose that we have two families of objects $(A_i)_{iin I}$ and $(B_i)_{iin I}$ in $mathbf{C}$ indexed by the same index set $I$. Assume further that there exist coproducts $A$ and $B$ of $(A_i)_{iin I}$ and of $(B_i)_{iin I}$, respectively. If $f_i:A_ito B_i$ are morphisms for all $iin I$, then there exists a unique morphism $f:Ato B$, which is the coproduct of $(f_i)_{iin I}$.



    I know the following results. The morphism $f$ need not be monic, even if all $f_i$ are monic. If $mathbf{C}$ is an abelian category and if all $f_i$ are epic, then $f$ is epic. Therefore, I have two related questions. All references are very welcome.




    Question I. In general (where $mathbf{C}$ need not be an abelian category), does $f$ have to be epic, provided that all $f_i$ are epic? If so, could you please provide a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?



    Question II. If $mathbf{C}$ is an abelian category and if all $f_i$ are monic, then does it follow that $f$ is monic? If so, could you please provide
    a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?











    share|cite|improve this question


























      up vote
      7
      down vote

      favorite









      up vote
      7
      down vote

      favorite











      Let $mathbf{C}$ be a category (that does not necessary have a coproduct for every collection of objects). Suppose that we have two families of objects $(A_i)_{iin I}$ and $(B_i)_{iin I}$ in $mathbf{C}$ indexed by the same index set $I$. Assume further that there exist coproducts $A$ and $B$ of $(A_i)_{iin I}$ and of $(B_i)_{iin I}$, respectively. If $f_i:A_ito B_i$ are morphisms for all $iin I$, then there exists a unique morphism $f:Ato B$, which is the coproduct of $(f_i)_{iin I}$.



      I know the following results. The morphism $f$ need not be monic, even if all $f_i$ are monic. If $mathbf{C}$ is an abelian category and if all $f_i$ are epic, then $f$ is epic. Therefore, I have two related questions. All references are very welcome.




      Question I. In general (where $mathbf{C}$ need not be an abelian category), does $f$ have to be epic, provided that all $f_i$ are epic? If so, could you please provide a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?



      Question II. If $mathbf{C}$ is an abelian category and if all $f_i$ are monic, then does it follow that $f$ is monic? If so, could you please provide
      a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?











      share|cite|improve this question















      Let $mathbf{C}$ be a category (that does not necessary have a coproduct for every collection of objects). Suppose that we have two families of objects $(A_i)_{iin I}$ and $(B_i)_{iin I}$ in $mathbf{C}$ indexed by the same index set $I$. Assume further that there exist coproducts $A$ and $B$ of $(A_i)_{iin I}$ and of $(B_i)_{iin I}$, respectively. If $f_i:A_ito B_i$ are morphisms for all $iin I$, then there exists a unique morphism $f:Ato B$, which is the coproduct of $(f_i)_{iin I}$.



      I know the following results. The morphism $f$ need not be monic, even if all $f_i$ are monic. If $mathbf{C}$ is an abelian category and if all $f_i$ are epic, then $f$ is epic. Therefore, I have two related questions. All references are very welcome.




      Question I. In general (where $mathbf{C}$ need not be an abelian category), does $f$ have to be epic, provided that all $f_i$ are epic? If so, could you please provide a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?



      Question II. If $mathbf{C}$ is an abelian category and if all $f_i$ are monic, then does it follow that $f$ is monic? If so, could you please provide
      a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?








      reference-request ct.category-theory abelian-categories products






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      edited Dec 10 at 0:55

























      asked Dec 9 at 23:44









      Batominovski

      22815




      22815






















          1 Answer
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          Question 1: Yes. The $I$-coproduct-functor $bigsqcup_Icolonprod_{iin I}mathbf{C}tomathbf{C}$ is left-adjoint (its right adjoint is the diagonal functor $Delta_{mathbf{C}}^Icolon mathbf{C}toprod_{iin I}mathbf{C}$), hence always preserves epimorphisms.



          Question 2: No, in general (even if $mathbf{C}$ is $I$-coproduct-complete). Cocomplete abelian categories with such property are called satisfying axiom AB4. See nlab article on Grothendieck axioms. Some counterexamples were discussed on SE; see, for example Zhen Lin's answer on MSE. The original source for this axiom is the Tôhoku paper (A.Grothendieck, "Sur quelques points d'algèbre homologique", 1957).



          Regarding the changes in your question after edits:



          In the situation when $mathbf{C}$ is not $I$-coproduct-complete, the previous proof becomes not entirely correct (probably it may be improved by regarding relative adjoint functors, but I don't think it's relevant to this question). However, there is a straightforward proof of this fact:
          $$
          (gcirc f=hcirc f)Rightarrow(gcirc fcirc s_i=hcirc fcirc s_i)Rightarrow(gcirc s_icirc f_i=hcirc s_icirc f_i)Rightarrow(gcirc s_i=hcirc s_i)Rightarrow(gcirc h),
          $$

          where $g$ and $h$ are arbitrary morphisms of $mathbf{C}$ with domain $B$ and $s_i$, $iin I$, are canonical injections of coproducts.



          As for the finite $I$ in your second question: there is no such counterexamples. An abelian category is always finitely complete, finitely cocomplete and finite coproducts coincide with finite products, hence if $I$ is finite, then the $I$-coproduct functor of an abelian (or even additive) category is both left- and right-adjoint, hence exact. So the axiom AB4 requires only that the infinite coproducts of monomorphisms should be monomorphisms.






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            1 Answer
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            active

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            active

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            active

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            up vote
            11
            down vote



            accepted










            Question 1: Yes. The $I$-coproduct-functor $bigsqcup_Icolonprod_{iin I}mathbf{C}tomathbf{C}$ is left-adjoint (its right adjoint is the diagonal functor $Delta_{mathbf{C}}^Icolon mathbf{C}toprod_{iin I}mathbf{C}$), hence always preserves epimorphisms.



            Question 2: No, in general (even if $mathbf{C}$ is $I$-coproduct-complete). Cocomplete abelian categories with such property are called satisfying axiom AB4. See nlab article on Grothendieck axioms. Some counterexamples were discussed on SE; see, for example Zhen Lin's answer on MSE. The original source for this axiom is the Tôhoku paper (A.Grothendieck, "Sur quelques points d'algèbre homologique", 1957).



            Regarding the changes in your question after edits:



            In the situation when $mathbf{C}$ is not $I$-coproduct-complete, the previous proof becomes not entirely correct (probably it may be improved by regarding relative adjoint functors, but I don't think it's relevant to this question). However, there is a straightforward proof of this fact:
            $$
            (gcirc f=hcirc f)Rightarrow(gcirc fcirc s_i=hcirc fcirc s_i)Rightarrow(gcirc s_icirc f_i=hcirc s_icirc f_i)Rightarrow(gcirc s_i=hcirc s_i)Rightarrow(gcirc h),
            $$

            where $g$ and $h$ are arbitrary morphisms of $mathbf{C}$ with domain $B$ and $s_i$, $iin I$, are canonical injections of coproducts.



            As for the finite $I$ in your second question: there is no such counterexamples. An abelian category is always finitely complete, finitely cocomplete and finite coproducts coincide with finite products, hence if $I$ is finite, then the $I$-coproduct functor of an abelian (or even additive) category is both left- and right-adjoint, hence exact. So the axiom AB4 requires only that the infinite coproducts of monomorphisms should be monomorphisms.






            share|cite|improve this answer



























              up vote
              11
              down vote



              accepted










              Question 1: Yes. The $I$-coproduct-functor $bigsqcup_Icolonprod_{iin I}mathbf{C}tomathbf{C}$ is left-adjoint (its right adjoint is the diagonal functor $Delta_{mathbf{C}}^Icolon mathbf{C}toprod_{iin I}mathbf{C}$), hence always preserves epimorphisms.



              Question 2: No, in general (even if $mathbf{C}$ is $I$-coproduct-complete). Cocomplete abelian categories with such property are called satisfying axiom AB4. See nlab article on Grothendieck axioms. Some counterexamples were discussed on SE; see, for example Zhen Lin's answer on MSE. The original source for this axiom is the Tôhoku paper (A.Grothendieck, "Sur quelques points d'algèbre homologique", 1957).



              Regarding the changes in your question after edits:



              In the situation when $mathbf{C}$ is not $I$-coproduct-complete, the previous proof becomes not entirely correct (probably it may be improved by regarding relative adjoint functors, but I don't think it's relevant to this question). However, there is a straightforward proof of this fact:
              $$
              (gcirc f=hcirc f)Rightarrow(gcirc fcirc s_i=hcirc fcirc s_i)Rightarrow(gcirc s_icirc f_i=hcirc s_icirc f_i)Rightarrow(gcirc s_i=hcirc s_i)Rightarrow(gcirc h),
              $$

              where $g$ and $h$ are arbitrary morphisms of $mathbf{C}$ with domain $B$ and $s_i$, $iin I$, are canonical injections of coproducts.



              As for the finite $I$ in your second question: there is no such counterexamples. An abelian category is always finitely complete, finitely cocomplete and finite coproducts coincide with finite products, hence if $I$ is finite, then the $I$-coproduct functor of an abelian (or even additive) category is both left- and right-adjoint, hence exact. So the axiom AB4 requires only that the infinite coproducts of monomorphisms should be monomorphisms.






              share|cite|improve this answer

























                up vote
                11
                down vote



                accepted







                up vote
                11
                down vote



                accepted






                Question 1: Yes. The $I$-coproduct-functor $bigsqcup_Icolonprod_{iin I}mathbf{C}tomathbf{C}$ is left-adjoint (its right adjoint is the diagonal functor $Delta_{mathbf{C}}^Icolon mathbf{C}toprod_{iin I}mathbf{C}$), hence always preserves epimorphisms.



                Question 2: No, in general (even if $mathbf{C}$ is $I$-coproduct-complete). Cocomplete abelian categories with such property are called satisfying axiom AB4. See nlab article on Grothendieck axioms. Some counterexamples were discussed on SE; see, for example Zhen Lin's answer on MSE. The original source for this axiom is the Tôhoku paper (A.Grothendieck, "Sur quelques points d'algèbre homologique", 1957).



                Regarding the changes in your question after edits:



                In the situation when $mathbf{C}$ is not $I$-coproduct-complete, the previous proof becomes not entirely correct (probably it may be improved by regarding relative adjoint functors, but I don't think it's relevant to this question). However, there is a straightforward proof of this fact:
                $$
                (gcirc f=hcirc f)Rightarrow(gcirc fcirc s_i=hcirc fcirc s_i)Rightarrow(gcirc s_icirc f_i=hcirc s_icirc f_i)Rightarrow(gcirc s_i=hcirc s_i)Rightarrow(gcirc h),
                $$

                where $g$ and $h$ are arbitrary morphisms of $mathbf{C}$ with domain $B$ and $s_i$, $iin I$, are canonical injections of coproducts.



                As for the finite $I$ in your second question: there is no such counterexamples. An abelian category is always finitely complete, finitely cocomplete and finite coproducts coincide with finite products, hence if $I$ is finite, then the $I$-coproduct functor of an abelian (or even additive) category is both left- and right-adjoint, hence exact. So the axiom AB4 requires only that the infinite coproducts of monomorphisms should be monomorphisms.






                share|cite|improve this answer














                Question 1: Yes. The $I$-coproduct-functor $bigsqcup_Icolonprod_{iin I}mathbf{C}tomathbf{C}$ is left-adjoint (its right adjoint is the diagonal functor $Delta_{mathbf{C}}^Icolon mathbf{C}toprod_{iin I}mathbf{C}$), hence always preserves epimorphisms.



                Question 2: No, in general (even if $mathbf{C}$ is $I$-coproduct-complete). Cocomplete abelian categories with such property are called satisfying axiom AB4. See nlab article on Grothendieck axioms. Some counterexamples were discussed on SE; see, for example Zhen Lin's answer on MSE. The original source for this axiom is the Tôhoku paper (A.Grothendieck, "Sur quelques points d'algèbre homologique", 1957).



                Regarding the changes in your question after edits:



                In the situation when $mathbf{C}$ is not $I$-coproduct-complete, the previous proof becomes not entirely correct (probably it may be improved by regarding relative adjoint functors, but I don't think it's relevant to this question). However, there is a straightforward proof of this fact:
                $$
                (gcirc f=hcirc f)Rightarrow(gcirc fcirc s_i=hcirc fcirc s_i)Rightarrow(gcirc s_icirc f_i=hcirc s_icirc f_i)Rightarrow(gcirc s_i=hcirc s_i)Rightarrow(gcirc h),
                $$

                where $g$ and $h$ are arbitrary morphisms of $mathbf{C}$ with domain $B$ and $s_i$, $iin I$, are canonical injections of coproducts.



                As for the finite $I$ in your second question: there is no such counterexamples. An abelian category is always finitely complete, finitely cocomplete and finite coproducts coincide with finite products, hence if $I$ is finite, then the $I$-coproduct functor of an abelian (or even additive) category is both left- and right-adjoint, hence exact. So the axiom AB4 requires only that the infinite coproducts of monomorphisms should be monomorphisms.







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                edited Dec 10 at 3:48

























                answered Dec 10 at 0:28









                Oskar

                481159




                481159






























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