MathOverflow User Auguste Hoang Duc asked after a proof of proposition 1.13 in Tannakian Categories by Deligne and Milne, to the effect that given two rigid tensor categories $C$, $D$, the category $[C, D]$ of tensor functors and morphisms between them is a groupoid.
Thus, let $F, G: C \to D$ be tensor functors and let $u: F \to G$ be a morphism between tensor functors. For convenience, we assume associativity contraints are strict (i.e., identities). We will denote monoidal units by $I$ (without specifying the monoidal category to which it belongs, which will be clear from context). The unit constraints are denoted $\lambda_a: I \otimes a \to a$ and $\rho_a: a \otimes I \to a$ (usually suppressing the subscript).
Recall that rigidity means that to each object $X$ there is a dual object $X^\vee$ together with
An evaluation morphism $\varepsilon: X^\vee \otimes X \to I$,
A co-evaluation morphism $\eta: I \to X \otimes X^\vee$
satisfying familiar triangular equations.
Recall that tensor functors $G$ come equipped with structural constraints (that are natural isomorphisms in the lower-case subscripts that appear):
$\alpha_{G; a, b}: G(a) \otimes G(b) \to G(a \otimes b)$,
$i_G: I \to G(I)$
subject to some coherence conditions. A morphism $u: F \to G$ of tensor functors is a natural transformation that is moreover compatible with the structural constraints of $F$ and $G$.
Given all this data, Deligne-Milne write down the inverse of $u X$ (for an object $X$ of $C$ as
and the OP wishes to verify this is indeed inverse to $u(X)$.
The proof that $u(X) \circ v(X) = 1_{G X}$ is displayed in the following large diagram. The proof that $v(X) \circ u(X) = 1_{F X}$ is similar and will be left to the reader. Starting from the occurrence of $G X$ in the top row and following the string of arrows in the counterclockwise direction along the perimeter gives the composite $u(X) \circ v(X)$. Each of the subdiagrams is readily verified to commute, using either naturality, functoriality, a coherence condition on structural constraints, a triangular equation for $X$, or the compatibility of $u$ with structural constraints. (Note: subscripts $X$, $X^\vee$, etc. of components of the structural constraints $\alpha_F$, $\alpha_G$ are suppressed for the sake of legibility. The symbol $1$ is used to indicate identity arrows on unnamed objects; which objects these are can be inferred from context.)
Here is the condensed, “Reader’s Digest” version of the diagram above (suppressing instances of the monoidal unit):
Here we are using the fact that a tensor functor preserves takes dual objects to dual objects, so that the bottom horizontal composite is an identity. Most of the details that are suppressed go into showing the commutativity of the triangle on the left, using compatibility of $u$ with the structural constraints of tensor functors.
There is a similar diagram for the other equation $v(X) \circ u(X) = 1$:
Tensor categories may be regarded as 2-categories (i.e., bicategories) with one object, and there is an often-used 2-categorical generalization of the result above.
The analogy works as follows:
Tensor functors generalize to 2-functors (also called “homomorphisms” [Bénabou] or “strong functors” or “pseudofunctors”),
Objects $X$ of tensor categories correspond to 1-cells $f$ of bicategories.
The role of dual object $X^\vee$ in a tensor category will be played by a right adjoint $f^\ast$ of $f$.
Morphisms of tensor functors correspond to oplax natural transformations between 2-functors (according to Bénabou’s convention on the direction of 2-cells; I myself would prefer to follow the convention in Johnstone’s Elephant and call them lax natural transformations).
Here is the result:
Let $B, C$ be 2-categories, let $F, G: B \to C$ be 2-functors, and let $\theta: F \to G$ be an oplax natural transformation. If $f: b \to c$ has a right adjoint $f^\ast$ in $B$, then the 2-cell $\theta \cdot f$ is invertible.
Given an adjunction $f \dashv f^\ast$ and 2-functors $F$, $G$, we have $F f \dashv F f^\ast$ and $Gf \dashv G f^\ast$. It is easily verified that the diagram
pastes to give the inverse $(\theta \cdot f)^{-1}$, where the left triangle 2-cell is the unit of the adjunctions $F f \dashv F f^*$, and the right triangle 2-cell is the counit of $G f \dashv G f^*$.