Dual statement category theoryDo opposite categories always exist?Why learn Category Theory in order to study Group Theory?Definition of a Functor of Abelian CategoriesCommuting with kernels implies left exactness in Abelian categoryIntuition for AB5 and Grothendieck categoriesWhy is Grp not an Abelian Category?Dual statements involving functorsQuestion on dual category.Category theory on Kahler geometryfirst homomorphism theorem in category theoryApplying the duality principle to adjunction

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Dual statement category theory


Do opposite categories always exist?Why learn Category Theory in order to study Group Theory?Definition of a Functor of Abelian CategoriesCommuting with kernels implies left exactness in Abelian categoryIntuition for AB5 and Grothendieck categoriesWhy is Grp not an Abelian Category?Dual statements involving functorsQuestion on dual category.Category theory on Kahler geometryfirst homomorphism theorem in category theoryApplying the duality principle to adjunction






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2












$begingroup$


I think I don't understand the categorical duality principle. For instance, if you prove that a certain category has kernels, by duality principle it has cokernels. But why this doesn't apply with Grothendieck axioms for abelian categories? For example, if a category has arbitrary direct sums, why the dual statement doesn't hold in general?










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  • 1




    $begingroup$
    May be helpful for you: math.stackexchange.com/questions/2092872/…
    $endgroup$
    – Sunny Rathore
    9 hours ago










  • $begingroup$
    Thanks, it has been very useful!
    $endgroup$
    – Smm
    8 hours ago


















2












$begingroup$


I think I don't understand the categorical duality principle. For instance, if you prove that a certain category has kernels, by duality principle it has cokernels. But why this doesn't apply with Grothendieck axioms for abelian categories? For example, if a category has arbitrary direct sums, why the dual statement doesn't hold in general?










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    May be helpful for you: math.stackexchange.com/questions/2092872/…
    $endgroup$
    – Sunny Rathore
    9 hours ago










  • $begingroup$
    Thanks, it has been very useful!
    $endgroup$
    – Smm
    8 hours ago














2












2








2





$begingroup$


I think I don't understand the categorical duality principle. For instance, if you prove that a certain category has kernels, by duality principle it has cokernels. But why this doesn't apply with Grothendieck axioms for abelian categories? For example, if a category has arbitrary direct sums, why the dual statement doesn't hold in general?










share|cite|improve this question









$endgroup$




I think I don't understand the categorical duality principle. For instance, if you prove that a certain category has kernels, by duality principle it has cokernels. But why this doesn't apply with Grothendieck axioms for abelian categories? For example, if a category has arbitrary direct sums, why the dual statement doesn't hold in general?







category-theory duality-theorems






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 9 hours ago









SmmSmm

381 silver badge7 bronze badges




381 silver badge7 bronze badges







  • 1




    $begingroup$
    May be helpful for you: math.stackexchange.com/questions/2092872/…
    $endgroup$
    – Sunny Rathore
    9 hours ago










  • $begingroup$
    Thanks, it has been very useful!
    $endgroup$
    – Smm
    8 hours ago













  • 1




    $begingroup$
    May be helpful for you: math.stackexchange.com/questions/2092872/…
    $endgroup$
    – Sunny Rathore
    9 hours ago










  • $begingroup$
    Thanks, it has been very useful!
    $endgroup$
    – Smm
    8 hours ago








1




1




$begingroup$
May be helpful for you: math.stackexchange.com/questions/2092872/…
$endgroup$
– Sunny Rathore
9 hours ago




$begingroup$
May be helpful for you: math.stackexchange.com/questions/2092872/…
$endgroup$
– Sunny Rathore
9 hours ago












$begingroup$
Thanks, it has been very useful!
$endgroup$
– Smm
8 hours ago





$begingroup$
Thanks, it has been very useful!
$endgroup$
– Smm
8 hours ago











2 Answers
2






active

oldest

votes


















4












$begingroup$


For instance, if you prove that a certain category has kernels, by duality principle it has cokernels.




You have to be careful what certain means: If for a property $mathsf P$ of categories you prove $mathsf P(mathscr C)Rightarrow mathscr Ctext has kernels$ for any $mathscr C$, then this also gives you $mathsf P(mathscr C^textop)Rightarrow left(mathscr C^textoptext has kernelsLeftrightarrow mathscr Ctext has cokernelsright)$. However, $mathsf P((-)^textop)$ is a different property than $mathsf P$ in general, and only if you know that $mathsf P(mathscr C^textop)Leftrightarrow mathsf P(mathscr C)$ you get the dual statement for categories satisfying $mathsf P$ for free.



Examples:



  • If $mathsf P(mathscr C):=mathscrCtext is abelian$, then $mathsf P(mathscr C^textop)Leftrightarrow mathsf P(mathscr C)$ because being abelian is a self-dual property. Therefore, with any statement about abelian categories you get another, 'dual', statement about abelian categories for free.


  • If $mathsf P(mathscr C):=mathscrCtext is Grothendieck$, then $mathsf P((-)^textop)notLeftrightarrow mathsf P$: In fact, the only Grothendieck abelian categories with Grothendieck dual are the trivial categories (those where every object is a zero object). Hence, statements about Grothendieck abelian categories do not come in dual pairs! Instead, with every statement about a Grothendieck abelian category, you get a dual statement about co-Grothendieck abelian categories, but as these aren't very popular (the only concrete example I can think of is $textabelian groups^textopcong textcompact abelian topological groups$), that's usually not very useful. It's important to notice this striking asymmetry in the focus on Grothendieck abelian categories in homological algebra.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    This is ok, but for example: suppose that I want to prove that a cateogory is abelian, let's say the cateogry of sheaves on a topological space X: $mathbfSh(X)$. Why I only have to verify the half of the axioms? Why if I have kernels in $mathbfSh(X)$, then I have cokernels in $mathbfSh(X)$? Sorry if this question is too obvious, considering the explanations you have given me.
    $endgroup$
    – Smm
    8 hours ago











  • $begingroup$
    Oh, What I just said is false, isn't it? Sorry!
    $endgroup$
    – Smm
    7 hours ago






  • 1




    $begingroup$
    @Smm I think that's false - you do indeed need to verify the full set of axioms. Consider for example the category of finitely generated $R$-modules. This category always has cokernels, but it has kernels (and is abelian) if and only if $R$ is Noetherian.
    $endgroup$
    – Hanno
    7 hours ago










  • $begingroup$
    Yes, you are right. Thanks everyone!
    $endgroup$
    – Smm
    7 hours ago


















3












$begingroup$

If you have some dual notions like kernels and cokernels or more generally limits and colimits then duality states that if $mathcalC$ has kernels, the opposite category $mathcalC^textop$ has cokernels. That is because a colimit is the same as a limit when all involved arrows are reversed. In general you do not get the dual notions in the same category. You can construct an easy example for that by considering a poset as a category.






share|cite|improve this answer











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    2 Answers
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    active

    oldest

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    4












    $begingroup$


    For instance, if you prove that a certain category has kernels, by duality principle it has cokernels.




    You have to be careful what certain means: If for a property $mathsf P$ of categories you prove $mathsf P(mathscr C)Rightarrow mathscr Ctext has kernels$ for any $mathscr C$, then this also gives you $mathsf P(mathscr C^textop)Rightarrow left(mathscr C^textoptext has kernelsLeftrightarrow mathscr Ctext has cokernelsright)$. However, $mathsf P((-)^textop)$ is a different property than $mathsf P$ in general, and only if you know that $mathsf P(mathscr C^textop)Leftrightarrow mathsf P(mathscr C)$ you get the dual statement for categories satisfying $mathsf P$ for free.



    Examples:



    • If $mathsf P(mathscr C):=mathscrCtext is abelian$, then $mathsf P(mathscr C^textop)Leftrightarrow mathsf P(mathscr C)$ because being abelian is a self-dual property. Therefore, with any statement about abelian categories you get another, 'dual', statement about abelian categories for free.


    • If $mathsf P(mathscr C):=mathscrCtext is Grothendieck$, then $mathsf P((-)^textop)notLeftrightarrow mathsf P$: In fact, the only Grothendieck abelian categories with Grothendieck dual are the trivial categories (those where every object is a zero object). Hence, statements about Grothendieck abelian categories do not come in dual pairs! Instead, with every statement about a Grothendieck abelian category, you get a dual statement about co-Grothendieck abelian categories, but as these aren't very popular (the only concrete example I can think of is $textabelian groups^textopcong textcompact abelian topological groups$), that's usually not very useful. It's important to notice this striking asymmetry in the focus on Grothendieck abelian categories in homological algebra.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      This is ok, but for example: suppose that I want to prove that a cateogory is abelian, let's say the cateogry of sheaves on a topological space X: $mathbfSh(X)$. Why I only have to verify the half of the axioms? Why if I have kernels in $mathbfSh(X)$, then I have cokernels in $mathbfSh(X)$? Sorry if this question is too obvious, considering the explanations you have given me.
      $endgroup$
      – Smm
      8 hours ago











    • $begingroup$
      Oh, What I just said is false, isn't it? Sorry!
      $endgroup$
      – Smm
      7 hours ago






    • 1




      $begingroup$
      @Smm I think that's false - you do indeed need to verify the full set of axioms. Consider for example the category of finitely generated $R$-modules. This category always has cokernels, but it has kernels (and is abelian) if and only if $R$ is Noetherian.
      $endgroup$
      – Hanno
      7 hours ago










    • $begingroup$
      Yes, you are right. Thanks everyone!
      $endgroup$
      – Smm
      7 hours ago















    4












    $begingroup$


    For instance, if you prove that a certain category has kernels, by duality principle it has cokernels.




    You have to be careful what certain means: If for a property $mathsf P$ of categories you prove $mathsf P(mathscr C)Rightarrow mathscr Ctext has kernels$ for any $mathscr C$, then this also gives you $mathsf P(mathscr C^textop)Rightarrow left(mathscr C^textoptext has kernelsLeftrightarrow mathscr Ctext has cokernelsright)$. However, $mathsf P((-)^textop)$ is a different property than $mathsf P$ in general, and only if you know that $mathsf P(mathscr C^textop)Leftrightarrow mathsf P(mathscr C)$ you get the dual statement for categories satisfying $mathsf P$ for free.



    Examples:



    • If $mathsf P(mathscr C):=mathscrCtext is abelian$, then $mathsf P(mathscr C^textop)Leftrightarrow mathsf P(mathscr C)$ because being abelian is a self-dual property. Therefore, with any statement about abelian categories you get another, 'dual', statement about abelian categories for free.


    • If $mathsf P(mathscr C):=mathscrCtext is Grothendieck$, then $mathsf P((-)^textop)notLeftrightarrow mathsf P$: In fact, the only Grothendieck abelian categories with Grothendieck dual are the trivial categories (those where every object is a zero object). Hence, statements about Grothendieck abelian categories do not come in dual pairs! Instead, with every statement about a Grothendieck abelian category, you get a dual statement about co-Grothendieck abelian categories, but as these aren't very popular (the only concrete example I can think of is $textabelian groups^textopcong textcompact abelian topological groups$), that's usually not very useful. It's important to notice this striking asymmetry in the focus on Grothendieck abelian categories in homological algebra.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      This is ok, but for example: suppose that I want to prove that a cateogory is abelian, let's say the cateogry of sheaves on a topological space X: $mathbfSh(X)$. Why I only have to verify the half of the axioms? Why if I have kernels in $mathbfSh(X)$, then I have cokernels in $mathbfSh(X)$? Sorry if this question is too obvious, considering the explanations you have given me.
      $endgroup$
      – Smm
      8 hours ago











    • $begingroup$
      Oh, What I just said is false, isn't it? Sorry!
      $endgroup$
      – Smm
      7 hours ago






    • 1




      $begingroup$
      @Smm I think that's false - you do indeed need to verify the full set of axioms. Consider for example the category of finitely generated $R$-modules. This category always has cokernels, but it has kernels (and is abelian) if and only if $R$ is Noetherian.
      $endgroup$
      – Hanno
      7 hours ago










    • $begingroup$
      Yes, you are right. Thanks everyone!
      $endgroup$
      – Smm
      7 hours ago













    4












    4








    4





    $begingroup$


    For instance, if you prove that a certain category has kernels, by duality principle it has cokernels.




    You have to be careful what certain means: If for a property $mathsf P$ of categories you prove $mathsf P(mathscr C)Rightarrow mathscr Ctext has kernels$ for any $mathscr C$, then this also gives you $mathsf P(mathscr C^textop)Rightarrow left(mathscr C^textoptext has kernelsLeftrightarrow mathscr Ctext has cokernelsright)$. However, $mathsf P((-)^textop)$ is a different property than $mathsf P$ in general, and only if you know that $mathsf P(mathscr C^textop)Leftrightarrow mathsf P(mathscr C)$ you get the dual statement for categories satisfying $mathsf P$ for free.



    Examples:



    • If $mathsf P(mathscr C):=mathscrCtext is abelian$, then $mathsf P(mathscr C^textop)Leftrightarrow mathsf P(mathscr C)$ because being abelian is a self-dual property. Therefore, with any statement about abelian categories you get another, 'dual', statement about abelian categories for free.


    • If $mathsf P(mathscr C):=mathscrCtext is Grothendieck$, then $mathsf P((-)^textop)notLeftrightarrow mathsf P$: In fact, the only Grothendieck abelian categories with Grothendieck dual are the trivial categories (those where every object is a zero object). Hence, statements about Grothendieck abelian categories do not come in dual pairs! Instead, with every statement about a Grothendieck abelian category, you get a dual statement about co-Grothendieck abelian categories, but as these aren't very popular (the only concrete example I can think of is $textabelian groups^textopcong textcompact abelian topological groups$), that's usually not very useful. It's important to notice this striking asymmetry in the focus on Grothendieck abelian categories in homological algebra.






    share|cite|improve this answer









    $endgroup$




    For instance, if you prove that a certain category has kernels, by duality principle it has cokernels.




    You have to be careful what certain means: If for a property $mathsf P$ of categories you prove $mathsf P(mathscr C)Rightarrow mathscr Ctext has kernels$ for any $mathscr C$, then this also gives you $mathsf P(mathscr C^textop)Rightarrow left(mathscr C^textoptext has kernelsLeftrightarrow mathscr Ctext has cokernelsright)$. However, $mathsf P((-)^textop)$ is a different property than $mathsf P$ in general, and only if you know that $mathsf P(mathscr C^textop)Leftrightarrow mathsf P(mathscr C)$ you get the dual statement for categories satisfying $mathsf P$ for free.



    Examples:



    • If $mathsf P(mathscr C):=mathscrCtext is abelian$, then $mathsf P(mathscr C^textop)Leftrightarrow mathsf P(mathscr C)$ because being abelian is a self-dual property. Therefore, with any statement about abelian categories you get another, 'dual', statement about abelian categories for free.


    • If $mathsf P(mathscr C):=mathscrCtext is Grothendieck$, then $mathsf P((-)^textop)notLeftrightarrow mathsf P$: In fact, the only Grothendieck abelian categories with Grothendieck dual are the trivial categories (those where every object is a zero object). Hence, statements about Grothendieck abelian categories do not come in dual pairs! Instead, with every statement about a Grothendieck abelian category, you get a dual statement about co-Grothendieck abelian categories, but as these aren't very popular (the only concrete example I can think of is $textabelian groups^textopcong textcompact abelian topological groups$), that's usually not very useful. It's important to notice this striking asymmetry in the focus on Grothendieck abelian categories in homological algebra.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 9 hours ago









    HannoHanno

    15k2 gold badges18 silver badges43 bronze badges




    15k2 gold badges18 silver badges43 bronze badges











    • $begingroup$
      This is ok, but for example: suppose that I want to prove that a cateogory is abelian, let's say the cateogry of sheaves on a topological space X: $mathbfSh(X)$. Why I only have to verify the half of the axioms? Why if I have kernels in $mathbfSh(X)$, then I have cokernels in $mathbfSh(X)$? Sorry if this question is too obvious, considering the explanations you have given me.
      $endgroup$
      – Smm
      8 hours ago











    • $begingroup$
      Oh, What I just said is false, isn't it? Sorry!
      $endgroup$
      – Smm
      7 hours ago






    • 1




      $begingroup$
      @Smm I think that's false - you do indeed need to verify the full set of axioms. Consider for example the category of finitely generated $R$-modules. This category always has cokernels, but it has kernels (and is abelian) if and only if $R$ is Noetherian.
      $endgroup$
      – Hanno
      7 hours ago










    • $begingroup$
      Yes, you are right. Thanks everyone!
      $endgroup$
      – Smm
      7 hours ago
















    • $begingroup$
      This is ok, but for example: suppose that I want to prove that a cateogory is abelian, let's say the cateogry of sheaves on a topological space X: $mathbfSh(X)$. Why I only have to verify the half of the axioms? Why if I have kernels in $mathbfSh(X)$, then I have cokernels in $mathbfSh(X)$? Sorry if this question is too obvious, considering the explanations you have given me.
      $endgroup$
      – Smm
      8 hours ago











    • $begingroup$
      Oh, What I just said is false, isn't it? Sorry!
      $endgroup$
      – Smm
      7 hours ago






    • 1




      $begingroup$
      @Smm I think that's false - you do indeed need to verify the full set of axioms. Consider for example the category of finitely generated $R$-modules. This category always has cokernels, but it has kernels (and is abelian) if and only if $R$ is Noetherian.
      $endgroup$
      – Hanno
      7 hours ago










    • $begingroup$
      Yes, you are right. Thanks everyone!
      $endgroup$
      – Smm
      7 hours ago















    $begingroup$
    This is ok, but for example: suppose that I want to prove that a cateogory is abelian, let's say the cateogry of sheaves on a topological space X: $mathbfSh(X)$. Why I only have to verify the half of the axioms? Why if I have kernels in $mathbfSh(X)$, then I have cokernels in $mathbfSh(X)$? Sorry if this question is too obvious, considering the explanations you have given me.
    $endgroup$
    – Smm
    8 hours ago





    $begingroup$
    This is ok, but for example: suppose that I want to prove that a cateogory is abelian, let's say the cateogry of sheaves on a topological space X: $mathbfSh(X)$. Why I only have to verify the half of the axioms? Why if I have kernels in $mathbfSh(X)$, then I have cokernels in $mathbfSh(X)$? Sorry if this question is too obvious, considering the explanations you have given me.
    $endgroup$
    – Smm
    8 hours ago













    $begingroup$
    Oh, What I just said is false, isn't it? Sorry!
    $endgroup$
    – Smm
    7 hours ago




    $begingroup$
    Oh, What I just said is false, isn't it? Sorry!
    $endgroup$
    – Smm
    7 hours ago




    1




    1




    $begingroup$
    @Smm I think that's false - you do indeed need to verify the full set of axioms. Consider for example the category of finitely generated $R$-modules. This category always has cokernels, but it has kernels (and is abelian) if and only if $R$ is Noetherian.
    $endgroup$
    – Hanno
    7 hours ago




    $begingroup$
    @Smm I think that's false - you do indeed need to verify the full set of axioms. Consider for example the category of finitely generated $R$-modules. This category always has cokernels, but it has kernels (and is abelian) if and only if $R$ is Noetherian.
    $endgroup$
    – Hanno
    7 hours ago












    $begingroup$
    Yes, you are right. Thanks everyone!
    $endgroup$
    – Smm
    7 hours ago




    $begingroup$
    Yes, you are right. Thanks everyone!
    $endgroup$
    – Smm
    7 hours ago













    3












    $begingroup$

    If you have some dual notions like kernels and cokernels or more generally limits and colimits then duality states that if $mathcalC$ has kernels, the opposite category $mathcalC^textop$ has cokernels. That is because a colimit is the same as a limit when all involved arrows are reversed. In general you do not get the dual notions in the same category. You can construct an easy example for that by considering a poset as a category.






    share|cite|improve this answer











    $endgroup$

















      3












      $begingroup$

      If you have some dual notions like kernels and cokernels or more generally limits and colimits then duality states that if $mathcalC$ has kernels, the opposite category $mathcalC^textop$ has cokernels. That is because a colimit is the same as a limit when all involved arrows are reversed. In general you do not get the dual notions in the same category. You can construct an easy example for that by considering a poset as a category.






      share|cite|improve this answer











      $endgroup$















        3












        3








        3





        $begingroup$

        If you have some dual notions like kernels and cokernels or more generally limits and colimits then duality states that if $mathcalC$ has kernels, the opposite category $mathcalC^textop$ has cokernels. That is because a colimit is the same as a limit when all involved arrows are reversed. In general you do not get the dual notions in the same category. You can construct an easy example for that by considering a poset as a category.






        share|cite|improve this answer











        $endgroup$



        If you have some dual notions like kernels and cokernels or more generally limits and colimits then duality states that if $mathcalC$ has kernels, the opposite category $mathcalC^textop$ has cokernels. That is because a colimit is the same as a limit when all involved arrows are reversed. In general you do not get the dual notions in the same category. You can construct an easy example for that by considering a poset as a category.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 9 hours ago

























        answered 9 hours ago









        ThorWittichThorWittich

        6929 bronze badges




        6929 bronze badges



























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