# Functors on rigid tensor categories.

This is a question about the proof of proposition 1.13 in Deligne and Milne, Tannakian Categories. Let $C,C'$ be two rigid tensor categories and $F,G : C \rightarrow C'$ be two tensor functors. Let $u : F \rightarrow G$ be a morphism of functors. Define the morphism $v : G \rightarrow F$ by $$v(X) : G(X) \simeq G(X^\vee)^\vee \xrightarrow{{}^t u(X^\vee)} F(X^\vee)^\vee \simeq F(X).$$

Why is $v$ the inverse of $u$ ?

• Isn't a well-placed question, is $v$ a funtor or a transformation between funtors $F$ and $G$? Commented Dec 11, 2012 at 18:39
• This is a job for... String Diagram Man! :-) Commented Dec 11, 2012 at 21:04
• Nitpick: instead of saying "morphism of functors", you should say "morphism of tensor functors" (there's a distinction). Commented Dec 12, 2012 at 0:29

Okay, here is a link to my web at the nLab which provides a diagrammatic proof (for one of the two equations that must be verified; the other equation is established similarly).

Of course, the gigantic diagram which you will find did not spring from my head like Pallas Athena. It was assembled by first studying a simple string diagram proof, which unfortunately I can't draw for you in a convenient way. The big commutative diagram which results only looks intimidating.

• I have added a condensed version of the diagram which might be easier to comprehend, together with a similar diagram for the other equation $v(X) \circ u(X) = 1$. Also added is a generalization of this result to 2-categories. Commented Dec 13, 2012 at 16:13
• Why are the diagrams on nlab so often badly rendered @ToddTrimble? It seems like it would be much clearer if you used tikzcd/quiver-format. Also, why do you not use notation consistent with the article? It would make it much easier to understand your post. For example, you use $\lambda$ for one of the unit constraints, which Tannakian Categories use for the natural transformation $\lambda:F \Rightarrow G$ we want to show is an isomorphism. Commented Jul 2 at 1:44
• @Ben123 Hmm, "no good deed goes unpunished". Don't ask me about the technical details of the nLab software, but tikzcd became integrated with it long after I wrote that diagram up. Meanwhile, due to further technical reasons which I won't explain, I am currently unable to edit my articles there. Finally, I use notation that I am accustomed to and that I find satisfying. In particular, I'm not going to bend over backwards to try to suit everyone's taste. Perhaps your time is better spent grasping the essence of the shorter, condensed version that I thoughtfully supplied? Commented Jul 3 at 1:07
• No, you just have to keep in mind how you construct transposes in terms of the unit and counit. $^t(u(X^\vee))$ is by definition the composition $$G(X) \overset{\eta \otimes 1}{\to} F(X) \otimes F(X^\vee) \otimes G(X) \overset{1 \otimes u(X^\vee) \otimes 1}{\to} F(X) \otimes G(X^\vee) \otimes G(X) \overset{1 \otimes \varepsilon} \to F(X)$$ where $\eta: I \to F(X) \otimes F(X^\vee)$ is the "unit" for a monoidal dual pair, and $\varepsilon: G(X^\vee) \otimes G(X) \to I$ is the counit for another. This looks much simpler if you use string diagrams. Commented Jul 21 at 12:28
• It's only a difference in convention. See for instance ncatlab.org/nlab/show/dualizable+object#in_a_monoidal_category, especially Remark 2.2. The mathematics is the same; you modulate between these two definitions of dual by reversing the order of tensor factors. It's hard to enforce universal agreement with these sorts of conventions, just like the fact that some people compose functions in diagrammatic order and others in the Leibnizian order. Commented Jul 23 at 20:11

Here's an answer with all the relevant diagrams following Deligne's article "Catégories Tannakiennes" (look at 2.7 and 2.4). It gives the slightly stronger result that any natural transformation is a natural isomorphism when restricted to dualizable objects.

1. The dual of an object $$X$$ is given by an object $$X^\vee$$ together with morphisms $$\epsilon\colon X\otimes X^\vee \to 1$$ and $$\eta\colon 1\to X\otimes X^\vee$$ such that the compositions: $$X \xrightarrow{1_X\otimes \eta} X\otimes X^\vee \otimes X \xrightarrow{\epsilon\otimes 1_X} X$$ $$X^\vee \xrightarrow{\eta\otimes 1_{X^\vee}} X^\vee \otimes X \otimes X^\vee \xrightarrow{1_{X^\vee}\otimes \epsilon} X^\vee$$ are the identities. It easily follows that a symmetric monoidal functor $$F\colon C\to C'$$ takes duals to duals: the dual of $$F(X)$$ is $$F(X^\vee)$$ together with the morphisms $$F(\epsilon)$$ and $$F(\eta)$$ composed with the canonical isomorphisms relating tensor products and units in $$C$$ and $$C'$$.
2. If $$X$$ and $$Y$$ are dualizable objects, then there is a bijection between morphisms $$X\to Y$$ and morphisms $$Y^\vee\to X^\vee$$. This bijection takes a morphism $$f\colon X\to Y$$ to its transpose $$f^t\colon Y^\vee\to X^\vee$$ given by the composition $$Y^\vee\xrightarrow{1\otimes \eta_X} Y^\vee\otimes X\otimes X^\vee \xrightarrow{1\otimes f \otimes 1} Y^\vee\otimes Y\otimes X^\vee \xrightarrow{\epsilon_Y\otimes 1} X^\vee$$ with a similar formula for the other direction of the bijection.
3. Let $$u$$ be a natural transformation of tensor functors $$F,G\colon C\to C'$$. Then for any dualizable object $$X$$ we have morphisms $$u:=u_X\colon F(X)\to G(X), \qquad u^\vee:=u_{X^\vee}\colon F(X^\vee)\to G(X^\vee)$$ The map $$u^\vee$$ looks like $$u^t$$ but goes in the opposite direction. We will see that it is in fact the inverse of $$u^t$$. The naturality of $$u$$ for $$\epsilon$$ (after applying canonical isomorphisms) gives that $$\epsilon_{F(X)} = \epsilon_{G(X)}\circ (u\otimes u^\vee)\colon F(X)\otimes F(X)^\vee \to G(X)\otimes G(X)^\vee\to 1\tag{1}$$ and the naturality for $$\eta$$ gives that $$\eta_{G(X)} = (u\otimes u^\vee)\circ \eta_{F(X)}\colon 1\to F(X)\otimes F(X)^\vee \to G(X)\otimes G(X)^\vee\tag{2}$$
4. It now remains to prove that if $$A$$ and $$B$$ are dualizable objects in a tensor category and $$u\colon A\to B$$ and $$u^\vee\colon A^\vee\to B^\vee$$ are maps such that (1) $$\epsilon_A = \epsilon_B\circ (u\otimes u^\vee)$$ and (2) $$\eta_B = (u\otimes u^\vee)\circ \eta_A$$, then $$u^\vee$$ is an inverse to $$u^t$$. That $$u^t\circ u^\vee = 1$$ follows from the commutative diagram $$\require{AMScd} \begin{CD} A^\vee @>{1\otimes \eta_A}>> A^\vee\otimes A\otimes A^\vee @>{\epsilon_A\otimes 1}>> A^\vee \\ @V{u^\vee}VV @V{u^\vee\otimes 1\otimes 1}VV @A{\epsilon_B\otimes 1}AA\\ B^\vee @>{1\otimes \eta_A}>> B^\vee\otimes A\otimes A^\vee @>{1\otimes u\otimes 1}>> B^\vee\otimes B\otimes A^\vee \end{CD}$$ where the top row is the identity and the composition along the bottom row and the right-hand map is $$u^t$$ by the formula for the transpose. The commutativity of the left hand square is obvious and the commutativity of the right hand square is (1). A similar diagram which is commutative by (2) shows that $$u^\vee\circ u^t = 1$$.

The condensed diagrams in Todd's answer ("Reader's digest") are the diagrams in 4 except that Todd does $$u\circ (u^\vee)^t=1$$ and $$(u^\vee)^t\circ u=1$$.

• All this is also known to category theorists under the catchphrase "doctrinal adjunction". See for example the second corollary here. It also generalizes well from monoidal categories to bicategories. Commented Jul 23 at 16:23