Why didn't dualization reverse all arrows in this case?

Question

Here are two quotes from Leinster's book:

According to the first quote a dual is obtained by reversing all the arrows.

Accrording to the second quote, the second pair of functors is the dual of the first.

But if so, why aren't the arrows (in the category of categories and functors) in the second pair reversed (they still go from left to right)? And why aren't the arrows in $\mathbf{Set}$ reversed (i.e., why isn't $\mathbf{Set}$ replaced with $\mathbf{Set}^{op}$)? As far as I can see, only the arrows in $\mathscr A$ are reversed. Why so?

If it's easier to answer what exactly "dual" means (in the framework of Leinster up to p.90), that would be great too.

Answer 1

The way the duality principle is phrased is meant to give some initial intuition, but is not suitable (and, considering the qualification as 'informal', not intended to be used) for formal application. One reason you've already mentioned: There are categorical constructions where the 'dual' is not obtained by reversing 'all' arrows. Secondly, while the introductory paragraph can be read to suggest this, dual constructions or proofs need not be carried out 'again', but they are in fact formal consequences of the original construction.

An attempt for a formal statement of the duality principle would be:

For every generic statement $\forall{\mathscr A}: {\mathsf P}({\mathscr A})$ about categories, formulated in some fixed context (perhaps involving other categories) categories, there is a 'dual' generic statement $\forall{\mathscr A}: {\mathsf P}({\mathscr A}^{\text{op}})$, which is equivalent to the original one.

Note that the context of the statement stays the same - dualization applies to the single category the statement is parametric over. In your example of the Yoneda Lemma, the parameter is ${\mathscr A}$, and the statement is that a concrete canonical assignment describes a functor ${\mathscr A}\to [{\mathscr A}^{\text{op}},\textsf{Set}]$. The dual statement hence says that for any ${\mathscr A}$ there is a concrete canonical assignment describing a functor ${\mathscr A}^{\text{op}}\to [{\mathscr A}^\text{op op}(\equiv {\mathscr A}),\textsf{Set}]$.

Answer 2

Duality doesn't work the way it seems you think it does. The point is that if $\mathcal{A}$ is a category, so is $\mathcal{A}^{op}$, and if something is true of every category, then it is also true of the dual of every category. The reason we don't reverse everything is because the theorem above is a theorem about an arbitrary category $\mathcal{A}$, not about an arbitrary diagram $\mathcal{A}\to \mathrm{whatever}$.

Here, the statement is that for any category $\mathcal{A}$, there exists a functor $H^*:\mathcal{A}^{op}\to [\mathcal{A},\mathbf{Set}]$. Since it's true for every category, it's also true for $\mathcal{A}^{op}$; that is, there's a functor $\mathcal{A}=(\mathcal{A}^{op})^{op}\to [\mathcal{A}^{op},\mathbf{Set}]$ taking $a\in \mathcal{A}$ to $$\mathcal{A}^{op}(a,-):\mathcal{A}^{op}\to\mathbf{Set}$$ which, by the definition of $\mathcal{A}^{op}$, is exactly the same as the functor $\mathcal{A}(-,a)=H_a$.

Answer 3

• Every categorical definition, theorem, and proof has a dual, obtained by reversing all the arrows.

• In this case, what specifically are we dualizing—a definition, theorem, proof?

• Given a category $\mathscr{A}$, we have a recipe for building a functor $\hom^\bullet : \mathscr{A}^{op} \rightarrow [\mathscr{A}, \mathrm{Set}]$.

• If we choose to dualize by reversing the arrows within $\mathscr{A}$— that is, exchanging the labels of domain and codomain within $\mathscr{A}$—and making no other changes, we get a different functor recipe. This is because we have dualized all references to the recipe's argument (the category $\mathscr{A}$).

• The resulting functor is $\hom_\bullet : \mathscr{A}\rightarrow [\mathscr{A}^{op}, \mathrm{Set}]$. You can confirm that we've made exactly the changes that result from reversing the arrows within $\mathscr{A}$.

• Or in other words, we ask what our functor recipe would be if we exchanged every mention of "the domain in $\mathscr{A}$" and "the codomain within $\mathscr{A}$".

• We ask this dualization question and not a different dualization question because it seemed we had a systematic recipe for building a functor out of a category $\mathscr{A}$— and dualization ought to give us another, possibly interesting, functor recipe for the category $\mathscr{A}$.

• The original functor recipe allows you to take any category and build a functor $\hom^\bullet$ out of it. By duality, there must be another functor recipe (obtained by dualizing the references to its "ingredient", the category $\mathscr{A}$) which allows you to take any category and build a similar hom functor out of it.