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?

  • $\begingroup$ I'm not sure I understand the question: if $C$ is rigid, i.e. if you have a chosen left and right dual for each object, then the assignment to any object of its left dual, say, and each morphism to its left transpose is already a functor on the nose (and its inverse is given by taking right duals) so there's nothing to do. Am I missing something ? $\endgroup$
    – Adrien
    Jul 17, 2021 at 15:19
  • $\begingroup$ @Adrien Your comment goes to the heart of my question, in particular your use of the world 'chosen'. In all the examples of rigid categories I have seen rigidity is proved by providing a dualizing functor. However, this is structure (as we could have chosen a different functor) yet being rigid is supposed to be a property. $\endgroup$
    – Arthur
    Jul 17, 2021 at 15:41
  • $\begingroup$ Not quite, I might be splitting hair but usually rigidity is proved by providing a dual for each object (ie often there one or several preferred way to pick a dual for each object) and then it just so happens that it's always a functor, ie the compatibility with morphisms is automatic. But of course it's also the case if you just pick a dual for each object "randomly". In other words, the meaning of "choosing" or "picking" is exaclty the same as when you pick an inverse to an equivalence, or an adjoint to a functor. $\endgroup$
    – Adrien
    Jul 17, 2021 at 17:59
  • $\begingroup$ In particular the choice I'm talking about is literally the same choice as the one in Maxime's answer when you choose an inverse to the functor $\tilde C \rightarrow C$. SO his answer is definitely a nice way to formulate it but at the end of the day we're saying pretty much the same thing. Indeed, as he also says, the key step is that at some point you need to choose a dual for each object in a way which is not à priori a functor but then it turns out it canonically and automatically becomes one. $\endgroup$
    – Adrien
    Jul 17, 2021 at 18:01
  • $\begingroup$ I agree with everything you've said and rereading your initial comment it's true that it's what you've been saying from the start. I still feel that something like Maxime's argument or at least its conclusion should be emphasized in expositions of the material, as the fact that you do automatically get a functor 'on the nose' isn't a trivial point (at least for someone who is relatively new to CT like me). $\endgroup$
    – Arthur
    Jul 17, 2021 at 18:34

1 Answer 1


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.

  • $\begingroup$ This seems like a very nice answer, however I do have a few follow up questions: 1) where do we use symmetry? Dual here simply means an object together with the correct structure maps. 2) if I don't want to use the action of choice, I don't see how a choice of non functorial duals get me there anyway. $\endgroup$
    – Arthur
    Jul 17, 2021 at 15:45
  • $\begingroup$ I'm just not sure what the correct notion of dual is in a non symmetric setting but the argument can probably be adapted to that setting, yes. For 2), the point is that if you have a fully faithful essentially surjective functor, to define a quasi-inverse you only need an object-wise choice of an antecedent. With no additional information, that will require the axiom of choice; but if you have a choice of object-wise (non functorial) duals, that is enough to get you that quasi-inverse $\endgroup$ Jul 17, 2021 at 16:17
  • $\begingroup$ Ah yes, that makes sense. Thanks a lot, that you've provided exactly the kind of answer I was hoping for. $\endgroup$
    – Arthur
    Jul 17, 2021 at 16:26
  • $\begingroup$ I think I know what was confusing me with respect to symmetry. If my original question were stated more carefully I would have decided whether or not I was talking about left or right duals. Your argument seems insensitive to either choice, so nothing to worry about. $\endgroup$
    – Arthur
    Jul 17, 2021 at 16:41

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