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











up vote
2
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
    2
    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
      2
      down vote

      favorite









      up vote
      2
      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 4 hours ago

























      asked 5 hours ago









      Batominovski

      20114




      20114






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          4
          down vote













          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























            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "504"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f317280%2fis-a-categorical-coproduct-of-epimorphisms-monomorphisms-always-an-epimorphism%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            4
            down vote













            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
              4
              down vote













              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
                4
                down vote










                up vote
                4
                down vote









                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.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 1 hour ago

























                answered 4 hours ago









                Oskar

                396138




                396138






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to MathOverflow!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f317280%2fis-a-categorical-coproduct-of-epimorphisms-monomorphisms-always-an-epimorphism%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    What visual should I use to simply compare current year value vs last year in Power BI desktop

                    How to ignore python UserWarning in pytest?

                    Alexandru Averescu