# Constructing functors between categories and their duals

Perhaps this question has a trivial answer, but it eludes me. I am thinking about functors between categories and their duals, specifically for the category $\textbf{Rel}\$ and its dual $\textbf{Rel*}\$, but if possible I'd like to approach this as a general question concerning the notion of "turning around the arrows" to form a dual.

Here's my problem: Suppose we want to find a functor $F$ from a category $C$ to its dual $C*$ such that the functor takes each morphism and object in $C$ to a exactly one morphism or object in $C*$. For me, the problem arises because "turn the arrow around" is not a valid operation for a functor.

Of course for a single morphism $f \colon A \to B$ we can effectively turn that particular arrow around by letting $F(A)=B$ and $F(B) = A$. But, thinking geometrically, whenever we simply switch $A$ and $B$ , we no longer have other arrows pointing correctly from $A$ or $B$ to all the other objects, $C, D, E,...$, etc.

Using the strict definition of a functor as two functions (one function for objects to objects, and one function for morphisms to morphisms), how do we know there is a functor that connects a category to its dual in the bijective fashion as described above?

[P.S. My specific problem arise from Awodey p.25, part (c) of the first exercise for $\textbf{Rel}\$ and its dual but, if possible, I'd like to concentrate on the most general case.]

• Could you please remind those of us without Awodey's book to hand, what the category Rel is? – Lord Shark the Unknown Aug 11 '17 at 4:53
• @LordSharktheUnknown $\mathbf{Rel}$ is the category whose objects are sets and arrows are relations, with the usual composition of relations (i.e. $S\circ R=\{(x,z)|\exists y \ s.t.\ (xRy)\wedge (ySz)\}$). – Arnaud D. Aug 11 '17 at 8:16

For a general category $C$, there is in general no way to define a funtor $C\to C^{op}$. For this problem, you really need to use the fact that your arrows are relations between sets to construct such a functor; it will exist because every relation $R:X\to Y$ has an "opposite relation" $R^{o}:Y\to X$ defined by $yR^{o}x\Leftrightarrow xRy$. You can check that this defines a functor $\mathbf{Rel}\to \mathbf{Rel}^{op}$ (which is the identity on objects). You can also check that it is bijective; in fact, it is its own inverse (or rather, its dual is). This is makes it an example of dagger category.
• Thanks for that answer, but I have one question. Where is that definition for $R^0$ coming from? Although it makes sense, I can't seem to derive it from other definitions or theorems. – MPitts Aug 11 '17 at 22:51
• @MPitts $X\times Y$ is naturally isomorphic to $Y\times X$; since a relation $R:X\to Y$ is a subset of $X\times Y$, it's easy to transform it into a subset of $Y\times X$, i.e. a relation $R^{o}:Y\to X$. In some sense, with relations the domain and codomain are more symmetric than with functions, so it's easy to turn them around. – Arnaud D. Aug 12 '17 at 13:37
• Let's take an example. Suppose the relation in question is ' < ', and we look at $f \colon \mathbb { N } \to \mathbb{N}$. So the morphism is the set $\lbrace(1,2),(1,3),(1,4),...(2,3),(2,4),...(3,4),(3,5),...\rbrace$. Now consider f* (the asterisk indicates the morphism f in the context of the dual). There seem be be two possibilities here. (continued...) – MPitts Aug 12 '17 at 20:17
• (a) the morphism f* becomes ' > ' in the dual and so pairs like (1,2) become (2,1) to reflect the change in the morphism, or alternatively, (b) the morphism f stays the same, except now pairs such as $(1,2) \in A \times B$ are simply expressed differently as $(2,1) \in B \times A$. I assume (b) is the right way to look at things, right? – MPitts Aug 12 '17 at 20:18