Does rigidity imply a unique dualizing functor? Let $\mathcal{C}$ be a rigid, monoidal category. Can I talk about $\mathcal{C}$ as having a unique, well-defined, dualizing functor (i.e. one that maps objects and morphisms onto their respective duals)?
What is clear to me is that dual objects are unique up to unique isomorphism. However, all that seems to tell me is that dualizing functors are also unique up to unique isomorphism. In particular, the question of whether or not any dualizing functors exist never seems to come up. Is it not possible to imagine a rigid category which doesn't admit a functorial mapping of its objects onto their duals?
 A: For rigid symmetric monoidal categories, there is in fact always a duality functor, unique up to isomorphism (without symmetry you would have to specify what you mean by "dual").
Here is a possible proof of existence:
Consider the following category $\tilde C$, its objects are quadruples $(x,y,\eta,\epsilon)$ where $x,y$ are objects of $C$ and $\eta: 1_C\to x\otimes y$ is a morphism in $C$, and $\epsilon : y\otimes x\to 1_C$ is another morphism, together exhibiting $y$ as a dual of $x$.
A morphism $(x_0,y_0,\eta_0,\epsilon_0)\to (x_1,y_1,\eta_1,\epsilon_1)$ is a pair of morphisms, $f :x_0\to x_1, g : y_1\to y_0$ such that $g$ "is dual to $f$" in the following sense: the composite $$y_1\cong y_1\otimes 1_C \to y_1\otimes x_0\otimes y_0\to y_1\otimes x_1\otimes y_0 \to 1_C\otimes y_0\cong y_0$$
is equal to $g$, where I've used the unitors of $C$, the map $\eta_0$, then $f$, and $\epsilon_1$ .
Composition is the obvious one, where you have to check that the "dual" of $f\circ f'$ is the dual of $f'\circ$ the dual of $f$. This check is essentially going to come up in any proof, and is the essential part of the proof - you should try to do that check.
I then claim the following two things: the forgetful functor $\tilde C\to C, (x,y,\eta,\epsilon)\mapsto x$ and $(f,g)\mapsto f$ is an equivalence of categories.
The fact that it is essentially surjective comes from the fact that $ C$ is rigid, and fully faithfulness comes from the fact that for any $f$ there is a unique $g$ which works (namely, the composite that I described !).
Then you get the following zigzag $C\overset\sim\leftarrow \tilde C\to C^{op}$ where the second arrow is projection to $y$ (and $g$)
Now, equivalences have quasi-inverses (if you have the axiom of choice, or a canonical choice of a dual and duality data for any $x$ - which you don't need to be functorial a priori), so that gives you a well defined functor $D: C\to C^{op}$ which is obviously an equivalence.
Unicity (up to natural isomorphism) is not hard to prove, you should try that !
It gets a little bit more complicated for higher categories. A functor $D$ is always definable, in a similar way as above (although the definition of $\tilde C$ needs to be changed a bit, because you need to specify a $2$-morphism witnessing one of the triangle identities), but saying what it does on objects and morphisms is no longer enough to fully determine it as a functor - the obstruction is exactly the same obstruction as the unicity of a functor which is the identity on objects and on $1$-morphisms.
