# Dual statement category theory

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?

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.

• 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. Jun 23, 2019 at 15:48
• Oh, What I just said is false, isn't it? Sorry! Jun 23, 2019 at 16:02
• @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. Jun 23, 2019 at 16:46
• Yes, you are right. Thanks everyone! Jun 23, 2019 at 16:50

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.