Why is $\mathcal{C}$ equivalent to $\mathcal{C}^{\text{op}}$ when $\mathcal{C}$ is a compact category?

Question

I came across the statement that a compact closed category $\mathcal{C}$ is equivalent to its dual category $\mathcal{C}^{\text{op}}$ (see e.g. this StackExchange post).

This fact seems to be regarded as obvious as it would follow basically from the definition of a compact category.

The definitions I'm working with are the following:

Definition 1: A compact closed category, or simply a compact category, is a symmetric monoidal category in which every object is dualizable (i.e. it has left and right duals).

Definition 2.1: A right duality between objects $A, A^* \in (\mathcal{C}, \otimes, I)$ consists of a morphism $\varepsilon_A : A^* \otimes A \to I$ called the counit and a morphism $\eta_A : I \to A \oplus A^*$ called the unit, such that the following two diagrams commute:

$$ \array{ A^\ast \otimes (A \otimes A^\ast) &\overset{\text{id}_{A^\ast} \otimes \eta_A}{\longleftarrow}& A^\ast \otimes 1 \\ {}^{{\alpha^{-1}_{A^\ast,A, A^\ast}}}_{{\simeq}}\downarrow && \downarrow^{{\lambda^{-1}_{A^\ast} \circ \rho_{A^\ast}}}_{{\simeq}} \\ (A^\ast \otimes A) \otimes A^\ast &\underset{\varepsilon_A \otimes \text{id}_{A^\ast}}{\longrightarrow}& 1 \otimes A^\ast } $$ $$ \array{ (A \otimes A^\ast) \otimes A &\overset{\eta_A \otimes \text{id}_A}{\longleftarrow}& 1 \otimes A \\ {}^{{\alpha_{A,A^\ast, A}}}_{{\simeq}}\downarrow && \downarrow^{{\rho_A^{-1}\circ \lambda_A}}_{{\simeq}} \\ A \otimes (A^\ast \otimes A) &\underset{\text{id}_A \otimes \varepsilon_A}{\longrightarrow}& A \otimes 1 } $$ where $\lambda$ and $\rho$ are the left and right unitor of the monoidal category.

Left duality is defined similarly.

Definition 2.1': An equivalence between two categories $\mathcal{C}$ and $\mathcal{D}$ is an equivalence in the 2-category Cat of all categories, hence a pair of inverse functors, hence it is

1. a pair of functors

$$ \mathcal{C} \xrightarrow{F} \mathcal{D} \xrightarrow{G} \mathcal{C}, $$

2. natural isomorphisms

$$ F \circ G \cong \text{Id}_{\mathcal{D}} $$

and

$$ G \circ F \cong \text{Id}_{\mathcal{C}} \,. $$

This is called an adjoint equivalence if the natural isomorphisms above satisfy the triangle identities, thus exhibiting $F$ and $G$ as a pair of adjoint functors.

Two categories are called equivalent if there exists an equivalence between them.

I guess that one can somehow construct an adjoint equivalence between $\mathcal{C}$ and $\mathcal{C}^{\text{op}}$, where the components of the unit and counit of the adjunction are given by the $\eta_A$ and $\varepsilon_A$ of the duality. I cannot figure out how this would work in detail though and I have the suspicion that I am overcomplicating this probably trivial problem.

Can someone please help me understand why $\mathcal{C}$ would obviously be equivalent to $\mathcal{C}^{\text{op}}$?

Answer 1

In a compact closed category, we have for any object $A$ adjunctions $-\otimes A\dashv -\otimes A^* \dashv -\otimes A$ (see Propositions 2.14 and 2.15 on this nLab page). Therefore, for any $A,B\in\mathcal{C}$, we have $\mathcal{C}(A,B)\cong \mathcal{C}(A\otimes B^*,I)\cong\mathcal{C}(B^*\otimes A,I)\cong\mathcal{C}(B^*,A^*)$, where $I$ is the monoidal unit. This isomorphism is natural in $A$ and $B$. Let $F\colon\mathcal{C}\to\mathcal{C}^\mathrm{op}$ be given by $FA=A^*$ and declare $F$ to act on morphisms according to the isomorphism $\mathcal{C}(A,B)\cong\mathcal{C}(B^*,A^*)$. Using naturality of this isomorphism, one may check that $F$ is indeed a functor, i.e. respects composition and identities. Now, $F$ is essentially surjective because $F(A^{*})=A^{**}\cong A$, and is fully faithful by construction. Therefore, $F$ is an equivalence $\mathcal{C}\to\mathcal{C}^\mathrm{op}$.