How does the duality functor with respect to $K$ behave on morphisms?

Question

In A duality formalism in the spirit of Grothendieck and Verdier Boyarchenko and Drinfeld give the following definition of the terms dualizing object and duality functor:

An object $K$ in a monoidal category $M$ is said to be dualizing if for every $Y \in M$ the functor $X \mapsto Hom(X \otimes Y, K)$ is representable by some object $DY \in M$ and the contravariant functor $D : M \rightarrow M$ is an antiequivalence. $D$ is called the duality functor with respect to $K$.

After chosing a representing object $DY$ of the functor $Hom(- \otimes Y, K)$ for every object $Y \in Obj(M)$ we obtain an assignment $Y \mapsto DY$ on objects of $M$. However, what morphism $DZ \rightarrow DY$ do we assign to a morphism $f: Y \rightarrow Z$ in $M$? That is, how does $D$ become a functor?

According to page 33 in this paper, one can use the Yoneda lemma to show that "the assignment $X \mapsto DX$ extends to a functor." I do not see how. Any hints?

Answer 1

Your morphism $f:Y\to Z$ produces (because $\otimes$ is functorial) a morphism $X\otimes f:X\otimes Y\to X\otimes Z$, which in turn induces by composition a function $Hom(X\otimes Z,K)\to Hom(X\otimes Y,K)$. By the definition of $D$, this amounts to a morphism $Hom(X,DZ)\to Hom(X,DY)$.

Furthermore, all of the preceding works for all objects $X$ and the constructions are natural with respect to $X$. That is, we have a natural transformation $Hom(-,DZ)\to Hom(-,DY)$. By Yoneda, this corresponds to a morphism $DZ\to DY$. That morphism is $Df$.

Answer 2

A morphism $X \rightarrow Y$ induces a natural transformation $Hom(-\otimes Y, Z) \rightarrow Hom(-\otimes X, Z)$ and by the Yoneda Lemma, since $DX$ and $DY$ represent these functor this natural transformation corresponds to a unique morphism $DY \rightarrow DX$ (the Yoneda embedding is fully faithful). Functoriality follows from the naturality of the Yoneda Lemma.