In [A duality formalism in the spirit of Grothendieck and Verdier](https://arxiv.org/abs/1108.6020), Boyarchenko and Drinfeld consider a monoidal category $(\mathcal{M}, \otimes, \mathbf{1})$ together with an object $K \in \mathcal{M}$ such that there is an equivalence $D \colon \mathcal{M}^\mathrm{op} \to \mathcal{M}$ where for all $Y \in \mathcal{M}$ the functor $\operatorname{Hom}((-) \otimes Y, K)$ is representable by $DY$.


In this setup they define a new product via $X \odot Y := D^{-1}(DY \otimes DX)$ and claim that this defines a monoidal structure on $\mathcal{M}$. I fail to see what the associator morphism $$D^{-1}(DZ \otimes D(D^{-1}(DY \otimes DX))) \to D^{-1}( D(D^{-1}(DZ \otimes DY)) \otimes DX)$$ should be. How can one construct such a morphism?


**Edit:** I was missing a $D$ in the domain and codomain. With the correct domain and codomain, one can get an evident associator using that $D^{-1}D \cong \mathrm{Id}$ (cf. the answer of HeinrichD).



Associator of the "dual" monoidal structure of a Grothendieck--Verdier Category
Let $F : \mathcal{C} \to \mathcal{D}$ be an equivalence of categories and $(\mathcal{D},\otimes,1)$ be a monoidal structure on $\mathcal{D}$. Then we have an induced monoidal structure on $\mathcal{C}\,$: Choose a quasi-inverse $F^{-1} : \mathcal{D} \to \mathcal{C}$ and isomorphisms $\mathrm{id}\_{\mathcal{C}} \cong F^{-1} \circ F$, $\mathrm{id}\_{\mathcal{D}} \cong F \circ F^{-1}$ satisfying the triangle identities. For $x,y \in \mathcal{C}$ we define $x \otimes y := F^{-1}(F(x) \otimes F(y))$. Hence, there is an isomorphism $F(x \otimes y) \cong F(x) \otimes F(y)$. We also let $1 := F^{-1}(1)$. The associator is defined by
$$x \otimes (y \otimes z) = F^{-1}(F(x) \otimes F(y \otimes z)) \cong F^{-1}(F(x) \otimes (F(y) \otimes F(z)))$$
$$ \cong F^{-1}((F(x) \otimes F(y)) \otimes F(z)) \cong F^{-1}(F(x \otimes y) \otimes F(z)) = (x \otimes y) \otimes z.$$


