Regarding the construction of quotient category The 2019 Stack Overflow Developer Survey Results Are InEquivalent conditions for a preabelian category to be abelianThe definition of the quotient category in abelian category.Properties of quotient categories.Why does the pushout preserve monic in an abelian category?Find limit of sequence in category of cones.Are poset-shaped limits of finite groups profinite?A technical question on the category of metric spacesderived category of quotient categoryTriangulated category associated to a subcategory of an abelian categoryComposition of morphisms in Quotient category

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Regarding the construction of quotient category



The 2019 Stack Overflow Developer Survey Results Are InEquivalent conditions for a preabelian category to be abelianThe definition of the quotient category in abelian category.Properties of quotient categories.Why does the pushout preserve monic in an abelian category?Find limit of sequence in category of cones.Are poset-shaped limits of finite groups profinite?A technical question on the category of metric spacesderived category of quotient categoryTriangulated category associated to a subcategory of an abelian categoryComposition of morphisms in Quotient category










0












$begingroup$


Let $mathcalA$ be an abelian category and $mathcalT$ be a thick subcategory (i.e., closed under taking subquotient and extensions) of $mathcalA$. Then we construct the quotient category $mathcalA/T$ with objects same as category $mathcalA$, and as for morphisms $Mor_mathcalA/T(A,B)$ we consider the set $$mathcalI=(A',B'): A' subseteq A, B' subseteq B,; ; A/A' in mathcalT,; B' in mathcal
T$$

w.r.t to the order $$(A',B')leq (A'',B'') iff A'supseteq A'',; ; B'subseteq B''.$$
Indeed, $; mathcalI; $ is a directed set w.r.t. pre-order $; leq; $. So we define a diagram $$D:mathcalIrightarrow Mor_mathcalA\
(A',B')rightarrow Mor_mathcalA(A',B/B'),$$

and then we go on to define$$Mor_mathcalA/T(A,B)= Colim_(A',B')in mathcalI Mor_mathcalA(A',B/B').$$



My first question is, how is the diagram $D$ well defined? Since the object $B/B'$ depends on the choice of the monic map $B'rightarrow B$ and we choose the fixed pair $(A',B')$ for all the monic maps $A' rightarrow A$ and $B'rightarrow B$.



Indeed, if we take $(A'rightarrow A, B'rightarrow B)$ pairs for fixed monic maps instead of $(A',B')$, the problem goes away and we have unique map $Mor_mathcalA (A',B/B')rightarrow Mor_mathcalA(A'',B/B''). $



Secondly, why is it enough to have $B''supseteq B'$, i.e., a monic map $B' rightarrow B''$? Don't we need the condition $B' rightarrow B'' rightarrow B= B' rightarrow B$ to have a map $B/B'rightarrow B/B''$?










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    Let $mathcalA$ be an abelian category and $mathcalT$ be a thick subcategory (i.e., closed under taking subquotient and extensions) of $mathcalA$. Then we construct the quotient category $mathcalA/T$ with objects same as category $mathcalA$, and as for morphisms $Mor_mathcalA/T(A,B)$ we consider the set $$mathcalI=(A',B'): A' subseteq A, B' subseteq B,; ; A/A' in mathcalT,; B' in mathcal
    T$$

    w.r.t to the order $$(A',B')leq (A'',B'') iff A'supseteq A'',; ; B'subseteq B''.$$
    Indeed, $; mathcalI; $ is a directed set w.r.t. pre-order $; leq; $. So we define a diagram $$D:mathcalIrightarrow Mor_mathcalA\
    (A',B')rightarrow Mor_mathcalA(A',B/B'),$$

    and then we go on to define$$Mor_mathcalA/T(A,B)= Colim_(A',B')in mathcalI Mor_mathcalA(A',B/B').$$



    My first question is, how is the diagram $D$ well defined? Since the object $B/B'$ depends on the choice of the monic map $B'rightarrow B$ and we choose the fixed pair $(A',B')$ for all the monic maps $A' rightarrow A$ and $B'rightarrow B$.



    Indeed, if we take $(A'rightarrow A, B'rightarrow B)$ pairs for fixed monic maps instead of $(A',B')$, the problem goes away and we have unique map $Mor_mathcalA (A',B/B')rightarrow Mor_mathcalA(A'',B/B''). $



    Secondly, why is it enough to have $B''supseteq B'$, i.e., a monic map $B' rightarrow B''$? Don't we need the condition $B' rightarrow B'' rightarrow B= B' rightarrow B$ to have a map $B/B'rightarrow B/B''$?










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      Let $mathcalA$ be an abelian category and $mathcalT$ be a thick subcategory (i.e., closed under taking subquotient and extensions) of $mathcalA$. Then we construct the quotient category $mathcalA/T$ with objects same as category $mathcalA$, and as for morphisms $Mor_mathcalA/T(A,B)$ we consider the set $$mathcalI=(A',B'): A' subseteq A, B' subseteq B,; ; A/A' in mathcalT,; B' in mathcal
      T$$

      w.r.t to the order $$(A',B')leq (A'',B'') iff A'supseteq A'',; ; B'subseteq B''.$$
      Indeed, $; mathcalI; $ is a directed set w.r.t. pre-order $; leq; $. So we define a diagram $$D:mathcalIrightarrow Mor_mathcalA\
      (A',B')rightarrow Mor_mathcalA(A',B/B'),$$

      and then we go on to define$$Mor_mathcalA/T(A,B)= Colim_(A',B')in mathcalI Mor_mathcalA(A',B/B').$$



      My first question is, how is the diagram $D$ well defined? Since the object $B/B'$ depends on the choice of the monic map $B'rightarrow B$ and we choose the fixed pair $(A',B')$ for all the monic maps $A' rightarrow A$ and $B'rightarrow B$.



      Indeed, if we take $(A'rightarrow A, B'rightarrow B)$ pairs for fixed monic maps instead of $(A',B')$, the problem goes away and we have unique map $Mor_mathcalA (A',B/B')rightarrow Mor_mathcalA(A'',B/B''). $



      Secondly, why is it enough to have $B''supseteq B'$, i.e., a monic map $B' rightarrow B''$? Don't we need the condition $B' rightarrow B'' rightarrow B= B' rightarrow B$ to have a map $B/B'rightarrow B/B''$?










      share|cite|improve this question











      $endgroup$




      Let $mathcalA$ be an abelian category and $mathcalT$ be a thick subcategory (i.e., closed under taking subquotient and extensions) of $mathcalA$. Then we construct the quotient category $mathcalA/T$ with objects same as category $mathcalA$, and as for morphisms $Mor_mathcalA/T(A,B)$ we consider the set $$mathcalI=(A',B'): A' subseteq A, B' subseteq B,; ; A/A' in mathcalT,; B' in mathcal
      T$$

      w.r.t to the order $$(A',B')leq (A'',B'') iff A'supseteq A'',; ; B'subseteq B''.$$
      Indeed, $; mathcalI; $ is a directed set w.r.t. pre-order $; leq; $. So we define a diagram $$D:mathcalIrightarrow Mor_mathcalA\
      (A',B')rightarrow Mor_mathcalA(A',B/B'),$$

      and then we go on to define$$Mor_mathcalA/T(A,B)= Colim_(A',B')in mathcalI Mor_mathcalA(A',B/B').$$



      My first question is, how is the diagram $D$ well defined? Since the object $B/B'$ depends on the choice of the monic map $B'rightarrow B$ and we choose the fixed pair $(A',B')$ for all the monic maps $A' rightarrow A$ and $B'rightarrow B$.



      Indeed, if we take $(A'rightarrow A, B'rightarrow B)$ pairs for fixed monic maps instead of $(A',B')$, the problem goes away and we have unique map $Mor_mathcalA (A',B/B')rightarrow Mor_mathcalA(A'',B/B''). $



      Secondly, why is it enough to have $B''supseteq B'$, i.e., a monic map $B' rightarrow B''$? Don't we need the condition $B' rightarrow B'' rightarrow B= B' rightarrow B$ to have a map $B/B'rightarrow B/B''$?







      category-theory homological-algebra abelian-categories






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Mar 24 at 9:12









      Later

      634




      634










      asked Mar 24 at 6:54









      solgaleosolgaleo

      8712




      8712




















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          $begingroup$

          1) $A' subseteq A, ; B' subseteq B$ are treated as subobjects, so I would say that your reformulation (fixing the whole monomorphisms rather than just domains) seems to be the intended meaning (and the one that makes sense).



          2) Similarly, the comparison of objects ``$A'' subseteq A'$ '' seems to be meant as comparison in the category of subobjects of $A$, i.e. the monomorphism should be compatible with those to $A$.






          share|cite|improve this answer









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            0












            $begingroup$

            1) $A' subseteq A, ; B' subseteq B$ are treated as subobjects, so I would say that your reformulation (fixing the whole monomorphisms rather than just domains) seems to be the intended meaning (and the one that makes sense).



            2) Similarly, the comparison of objects ``$A'' subseteq A'$ '' seems to be meant as comparison in the category of subobjects of $A$, i.e. the monomorphism should be compatible with those to $A$.






            share|cite|improve this answer









            $endgroup$

















              0












              $begingroup$

              1) $A' subseteq A, ; B' subseteq B$ are treated as subobjects, so I would say that your reformulation (fixing the whole monomorphisms rather than just domains) seems to be the intended meaning (and the one that makes sense).



              2) Similarly, the comparison of objects ``$A'' subseteq A'$ '' seems to be meant as comparison in the category of subobjects of $A$, i.e. the monomorphism should be compatible with those to $A$.






              share|cite|improve this answer









              $endgroup$















                0












                0








                0





                $begingroup$

                1) $A' subseteq A, ; B' subseteq B$ are treated as subobjects, so I would say that your reformulation (fixing the whole monomorphisms rather than just domains) seems to be the intended meaning (and the one that makes sense).



                2) Similarly, the comparison of objects ``$A'' subseteq A'$ '' seems to be meant as comparison in the category of subobjects of $A$, i.e. the monomorphism should be compatible with those to $A$.






                share|cite|improve this answer









                $endgroup$



                1) $A' subseteq A, ; B' subseteq B$ are treated as subobjects, so I would say that your reformulation (fixing the whole monomorphisms rather than just domains) seems to be the intended meaning (and the one that makes sense).



                2) Similarly, the comparison of objects ``$A'' subseteq A'$ '' seems to be meant as comparison in the category of subobjects of $A$, i.e. the monomorphism should be compatible with those to $A$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 26 at 5:03









                Pavel ČoupekPavel Čoupek

                4,57611126




                4,57611126



























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