Remark 4.43.7. Lemma 4.43.6 says in particular that $Z \mapsto Z \otimes Y$ is the right adjoint of $Z' \mapsto Z' \otimes X$. In particular, uniqueness of adjoint functors guarantees that a left dual of $X$, if it exists, is unique up to unique isomorphism. Conversely, assume the functor $Z \mapsto Z \otimes Y$ is a right adjoint of the functor $Z' \mapsto Z' \otimes X$, i.e., we're given a bijection

functorial in both $Z$ and $Z'$. The unit of the adjunction produces maps

functorial in $Z$ and the counit of the adjoint produces maps

functorial in $Z'$. In particular, we find $\eta = \eta _\mathbf {1} : \mathbf{1} \to X \otimes Y$ and $\epsilon = \epsilon _\mathbf {1} : Y \otimes X \to \mathbf{1}$. As an exercise in the relationship between units, counits, and the adjunction isomorphism, the reader can show that we have

However, this isn't enough to show that $(\epsilon \otimes \text{id}_ Y) \circ (\text{id}_ Y \otimes \eta ) = \text{id}_ Y$ and $(\text{id}_ X \otimes \epsilon ) \circ (\eta \otimes \text{id}_ X) = \text{id}_ X$, because we don't know in general that $\eta _ Y = \text{id}_ Y \otimes \eta $ and we don't know that $\epsilon _ X = \epsilon \otimes \text{id}_ X$. For this it would suffice to know that our adjunction isomorphism has the following property: for every $W, Z, Z'$ the diagram

If this holds, we will say *the adjunction is compatible with the given tensor structure*. Thus the requirement that $Z \mapsto Z \otimes Y$ be the right adjoint of $Z' \mapsto Z' \otimes X$ compatible with the given tensor structure is an equivalent formulation of the property of being a left dual.

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