Functors on rigid tensor categories.

Question

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 $u$ ?

Answer 1

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.

Answer 2

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 $(u^\vee)^t\circ u=1$.