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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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2 Answers 2

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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 ${\mathsf P}({\mathscr C})\Rightarrow {\mathscr C}\text{ has kernels}$ for any ${\mathscr C}$, then this also gives you ${\mathsf P}({\mathscr C}^\text{op})\Rightarrow \left({\mathscr C}^\text{op}\text{ has kernels}\Leftrightarrow {\mathscr C}\text{ has cokernels}\right)$. However, ${\mathsf P}((-)^\text{op})$ is a different property than ${\mathsf P}$ in general, and only if you know that ${\mathsf P}({\mathscr C}^\text{op})\Leftrightarrow {\mathsf P}({\mathscr C})$ you get the dual statement for categories satisfying ${\mathsf P}$ for free.

Examples:

  • If ${\mathsf P}({\mathscr C}):=\mathscr{C}\text{ is abelian}$, then ${\mathsf P}({\mathscr C}^\text{op})\Leftrightarrow {\mathsf P}({\mathscr C})$ 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 ${\mathsf P}({\mathscr C}):=\mathscr{C}\text{ is Grothendieck}$, then ${\mathsf P}((-)^\text{op})\not\Leftrightarrow {\mathsf P}$: 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 $\{\text{abelian groups}\}^{\text{op}}\cong \{\text{compact abelian topological groups}\}$), that's usually not very useful. It's important to notice this striking asymmetry in the focus on Grothendieck abelian categories in homological algebra.

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  • $\begingroup$ 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: $\mathbf{Sh}(X)$. Why I only have to verify the half of the axioms? Why if I have kernels in $\mathbf{Sh}(X)$, then I have cokernels in $\mathbf{Sh}(X)$? Sorry if this question is too obvious, considering the explanations you have given me. $\endgroup$
    – user_12345
    Jun 23, 2019 at 15:48
  • $\begingroup$ Oh, What I just said is false, isn't it? Sorry! $\endgroup$
    – user_12345
    Jun 23, 2019 at 16:02
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    $\begingroup$ @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
    Jun 23, 2019 at 16:46
  • $\begingroup$ Yes, you are right. Thanks everyone! $\endgroup$
    – user_12345
    Jun 23, 2019 at 16:50
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If you have some dual notions like kernels and cokernels or more generally limits and colimits then duality states that if $\mathcal{C}$ has kernels, the opposite category $\mathcal{C}^{\text{op}}$ 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.

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