# 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 $u$ ?

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

## 1 Answer

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.

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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. – Todd Trimble Dec 13 '12 at 16:13