$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proposition 6.1.6.15. Let $\operatorname{\mathcal{C}}$ be a monoidal category and let $\operatorname{ev}: X \otimes Y \rightarrow {\bf 1}$ be a morphism of $\operatorname{\mathcal{C}}$. Then:
- $(1)$
If the morphism $\operatorname{ev}$ exhibits $Y$ as a right dual of $X$ (Definition 6.1.6.6), then it exhibits $Y$ as a weak right dual of $X$ (Definition 6.1.6.13). The converse holds if $X$ is right dualizable.
- $(2)$
If the morphism $\operatorname{ev}$ exhibits $X$ as a left dual of $Y$, then it exhibits $X$ as a weak left dual of $Y$. The converse holds if $Y$ is left dualizable.
Proof.
We will prove $(1)$; the proof of $(2)$ is similar. If $\operatorname{ev}: X \otimes Y \rightarrow {\bf 1}$ is a duality datum, then it exhibits $Y$ as a weak right dual of $X$ by virtue of Variant 6.1.3.2 (applied to the $2$-category $B\operatorname{\mathcal{C}}$). Conversely, suppose that $\operatorname{ev}$ exhibits $Y$ as a weak right dual of $X$. If there exists another object $Y' \in \operatorname{\mathcal{C}}$ and a duality datum $\operatorname{ev}': X \otimes Y' \rightarrow {\bf 1}$, then the universal property of $Y$ guarantees that there is a unique morphism $u: Y' \rightarrow Y$ for which $\operatorname{ev}'$ is equal to the composite map $X \otimes Y' \xrightarrow {\operatorname{id}_ X \otimes u} X \otimes Y \xrightarrow {\operatorname{ev}} {\bf 1}$. Since $\operatorname{ev}'$ exhibits $Y'$ as a weak right dual of $X$, the morphism $u$ must be an isomorphism, so that the morphism $\operatorname{ev}$ is also a duality datum.
$\square$