# Show that duality functor is anti-monoidal

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.

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.