# Definition of functors between dual categories

I am attempting to learn some cat theory on my own, but so far I am confused over the simple definitions of duals and functors (so pls bear with me). The following 3-object example demonstrates my confusion:

Suppose we have a simple category $\mathcal{D}$ with three objects (or alternatively, these could be three typical objects in some larger category). Let $f : A \to B$, $g : B \to C$, and $g \circ f : A \to C$. We also have the dual category $\mathcal{D}^{op}$ with $f^* : B^* \to A^*$, $g^* : C^* \to B^*$, and $(g \circ f)^* : C^* \to A^*$. (Here the asterisks are used only to indicate whether we are referring to objects and arrows used in the original category, or the dual.)

Now my question concerns how to create a functor $F$ from $\mathcal{D}$ to $\mathcal{D}^{op}$. I think we know there must be such a functor between these two categories (or between subsets of larger categories), but I am unable to define such a functor that conforms to the definitions as I understand (or more likely, misunderstand) them.

This question is probably very straightforward, but a simple answer would help a great deal in clarifying my confusion. Thanks.

Either use a constant functor, like the $G$ in Tsemo Aristide's answer, or use $F$ defined by $F(A)=C^*$, $F(B)=B^*$, $F(C)=A^*$, $F(f)=g^*$, and $F(g)=f^*$.
• @MPitts Yes, Tsemo's answer has been deleted. His $G$ sent all three objects to $A$ and all morphisms to the identity morphism of $A$. Of course you could do the same with either of the other two objects instead of $A$. There are also some other functors like sending both $A$ and $B$ to $C$ and sending $C$ to $B$ (and sending morphisms to the only ones that have the right domain and codomain). – Andreas Blass Jul 8 '17 at 19:22
• The problem in your last comment is "and their morphisms"; a pair of objects might have more than one morphism between them. Consider a category that is like your $\mathcal D$ except that it has two morphisms $f$ and $f'$ from $A$ to $B$ (and they both have the same composite with $g:B\to C$, so the new category just has one more morphism than $\mathcal D$). Then this new category is not isomorphic to its dual. I think that, once you think about this example a bit, you'll see why $\mathcal D$ is not a typical picture of three objects and their morphisms in other categories. – Andreas Blass Jul 9 '17 at 22:56