# Applying the duality principle to adjunction

In Tom Leinster's book "Basic Category theory", Construction 1.1.9 gives a brief introduction of the concept "dual":

Every category $\cal A$ has an opposite or dual category $\cal A^{\sf op}$, deﬁned by reversing the arrows.

Also it says:

The principle of duality is fundamental to category theory. Informally, it states that every categorical deﬁnition, theorem and proof has a dual, obtained by reversing all the arrows.

Later in Chapter 2, it talks about adjunctions. And it seems that the principle of duality is applied here to get some theorems. For instance, we have:

For each $A∈ \cal A$, we have a map $(A \overset{η_A}\rightarrow GF(A))=\overline{F(A) \overset{1} → F(A)}$. Dually, for each $B∈\cal B$, we have a map $(FG(B)\overset{ε_B} →B)=\overline{G(B)\overset{1} →G(B)}$.

Or something like this. I don't quite understand why the two equations are dual of each other. I cannot see where we "reverse all the arrows". My understand is taking the dual of a statement is passing from a category to it opposite category. But I cannot figure out in detail what is going on here, why it is obtained from passing to an opposite category.

• What is the bar over a morphism? And, are you sure that the "dually" that baffles you really appeals the duality principle? – Fosco Feb 22 '18 at 16:04
• @FoscoLoregian Adding the bar is applying the natural isomorphism $Hom(F(A), F(A))\to Hom(A,GF(A))$ for the adjoint $F\dashv G$. And I think a mathematics textbook should not be careless: If it is not obtained by applying the "dual" according to the definition, it would not be called the "dual". – PropositionX Feb 22 '18 at 20:25

I think that you should take Loregian's advice in not taking that "dually" to be the one he speaks of when he defines the duality principle.

If you insist in your search for answers and refuse to settle for a no the point is in looking at $Cat$ as a 2-category. Here you have that the two functors $$F:\mathcal{A}\to\mathcal{B}$$ $$G:\mathcal{B}\to\mathcal{A}$$ are just two morphisms between two object in this 2-category. The two natural transformations are what you call 2-morphisms or 2-cell (with a geometric flavour): $$\eta:Id_{\mathcal{A}}\to GF$$ $$\epsilon:FG\to Id_{\mathcal{B}}$$

Small remark is that the definition of adjunction makes sense in any 2-category. Be it $Cat$ or anything else.

The idea becomes to apply the duality principle to the 2-category $Cat$. As you already noted you can invert the morphisms, or the 2-morphisms, or both. The notation is $Cat^{op}$ if I invert the functors, $Cat^{co}$ if I invert the natural transformations and $Cat^{coop}$ if I invert both.

So by considering the 4 different categories $Cat$, $Cat^{co}$, $Cat^{op}$, $Cat^{coop}$ we get that the adjunction $F\dashv G$ becomes $G\dashv F$, $G\dashv F$ when we invert either the functors or the natural transformations, and $F\dashv G$ when we invert both. The point here is that when we invert both we get that the unit becomes the counit and the counit becomes the unit.

As you wanted the unit and the counit are "dual" to each other in the sense of inverting arrows.

P.S. the small remark in between is needed because the other three 2-categories are no longer $Cat$ and we cannot think of the objects as categories. So we need to point out that the construction is not a privilege of $Cat$.

Here you have $F : \mathcal{A}\to \mathcal{B}, G: \mathcal{B}\to \mathcal{A}$, with $F \dashv G$ . On the one hand you have $\eta_A: A\to GFA$.

But now if you look at $F^{op} : \mathcal{A}^{op} \to \mathcal{B}^{op}, G^{op} :\mathcal{B}^{op} \to \mathcal{A}^{op}$, then you have an adjunction in the other direction, that is $G^{op} \dashv F^{op}$; and you thus get a dual to $\eta_A$ which is $\epsilon_B : B\to F^{op}G^{op}B$. However, this arrow lives in $\mathcal{B}^{op}$ so returning to $\mathcal{B}$ yields $\epsilon_B : FGB \to B$.

Hence when you reverse all the arrows (go from $\mathcal{A}$ to $\mathcal{A}^{op}$, $\mathcal{B}$ to $\mathcal{B}^{op}$) you do get that $\epsilon$ is dual to $\eta$