Show that duality functor is anti-monoidal

Question

Let $\mathcal{C}$ be a right rigid (not strict) monoidal category with associativity constraint $\Phi$. Let $J_{U,V}: U^*\otimes V^*\to (V\otimes U)^*$ be the canonical isomorphism for every objects $U,V\in\mathcal{C}$ . I would like to show that the pair $((-)^*, J)$ is an anti-monoidal functor, i.e. for any three objects $U,V,W\in\mathcal{C}$ $$J_{U,(W\otimes V)} \circ (U^*\otimes J_{V,W}) \circ \Phi_{U^*,V^*,W^*}=\Phi_{W,V,U}^* \circ J_{(V\otimes U), W}\circ (J_{U,V}\otimes W^*)$$ It should be an easy exercise of diagram chasing, but... I am stuck.

Answer 1

First of all, by Mac Lane Coherence Theorem we may assume that $\mathcal{C}$ is strict. Therefore we may omit associativity and unit constraints and we are left to check that $$J_{U,W \otimes V} \circ \left(U^* \otimes J_{V,W}\right) = J_{V \otimes U,W} \circ \left(J_{U,V} \otimes W^*\right).$$ Recall that right rigidity means that for $U$ in $\mathcal{C}$ there exists $U^*$ in $\mathcal{C}$ and $\mathsf{ev}_U : U \otimes U^* \to 1$ and $\mathsf{db}_U:1\to U^* \otimes U$ such that $$\left(\mathsf{ev}_U \otimes U\right)\circ \left(U \otimes \mathsf{db}_U\right) = \mathsf{id}_U \quad \text{and} \quad \left(U^* \otimes \mathsf{ev}_U\right)\circ \left(\mathsf{db}_U \otimes U^*\right) = \mathsf{id}_{U^*}.\tag{$\star$}$$ In particular, this implies that for every object $U$ in $\mathcal{C}$, the endofunctor $U \otimes (-)$ has a right adjoint $U^* \otimes (-)$. Now, recall the definition of $J_{U,V} : U^* \otimes V^* \to (V \otimes U)^*$: by adjunction, it is the unique morphism such that $$\mathsf{ev}_{V \otimes U} \circ \left(V \otimes U \otimes J_{U,V}\right) = \mathsf{ev}_V \circ \left(V \otimes \mathsf{ev}_U \otimes V^*\right). \tag{$*$}$$ On the one hand, we may compute $$\mathsf{ev}_{W \otimes V \otimes U} \circ \left(W \otimes V \otimes U\otimes J_{V \otimes U,W}\right) \circ \left(W \otimes V \otimes U\otimes J_{U,V} \otimes W^*\right) = \\ = \mathsf{ev}_W \circ \left(W \otimes \mathsf{ev}_{V \otimes U} \otimes W^*\right) \circ \left(W \otimes V \otimes U\otimes J_{U,V} \otimes W^*\right) \\ = \mathsf{ev}_W \circ \left(W \otimes \mathsf{ev}_{V } \otimes W^*\right) \circ \left(W \otimes V \otimes \mathsf{ev}_U \otimes V^* \otimes W^*\right) . $$ On the other hand, $$\mathsf{ev}_{W \otimes V \otimes U} \circ \left(W \otimes V \otimes U\otimes J_{U,W \otimes V}\right) \circ \left(W \otimes V \otimes U\otimes U^* \otimes J_{V,W}\right) = \\ = \mathsf{ev}_{W \otimes V} \circ \left(W \otimes V \otimes \mathsf{ev}_{U} \otimes (W\otimes V)^*\right) \circ \left(W \otimes V \otimes U\otimes U^* \otimes J_{V,W}\right) \\ = \mathsf{ev}_{W \otimes V} \circ \left(W \otimes V \otimes J_{V,W}\right) \circ \left(W \otimes V \otimes \mathsf{ev}_{U} \otimes V^* \otimes W^*\right) \\ = \mathsf{ev}_W \circ \left(W \otimes \mathsf{ev}_{V } \otimes W^*\right) \circ \left(W \otimes V \otimes \mathsf{ev}_U \otimes V^* \otimes W^*\right) . $$ As a consequence, by bijectivity of the adjunction isomorphism $$\mathcal{C}\left(W \otimes V \otimes U \otimes U^* \otimes V^* \otimes W^*,1\right) \cong \mathcal{C}\left(U^* \otimes V^* \otimes W^*,(W \otimes V \otimes U)^*\right),$$ the desired equality holds.

If now you would like to rewrite the argument by explicitly reporting the constraints, it is enough to add them to $(\star)$ and $(*)$ and check that the pentagon and triangle axioms do their job.