Recall that a monoidal category $\mathcal C$ is rigid if every object $X\in \mathcal C$ has both left and right duals, i.e. objects $X^l$ and $X^r$ with maps $X^l \otimes X \to \mathbf 1 \to X \otimes X^l$ and $X \otimes X^r \to \mathbf 1 \to X^r \otimes X$ satisfying certain equations. It is a fundamental fact about monoidal categories that "having a left dual" and "having a right dual" are properties of an object, not data: given $X$, the objects $X^l$ and $X^r$, if they exist, are uniquely determined up to unique isomorphism. As such, for a monoidal category itself to be rigid is also property --- the only data is the data of being monoidal.
This should remind you of groups. Given a monoid $G$, an element $x\in G$ is invertible if there are elements $x^l$ and $x^r$ and equalities $x^l x = 1 = x x^r$. (Undergraduate exercise: $x^l = x^r$.) Being invertible is property. A monoid $G$ is a group if every element therein is invertible. Thus this too is a property.
In some cases (e.g. algebraic geometry) you don't always want to think about the elements of a group $G$. Fortunately, there is a very nice way to say when a monoid is a group that does not directly refer to invertibility of elements. Let $G$ be a monoid and consider the map $G \times G \to G\times G$ (a map of underlying spaces, not of groups) that takes $(x,y)$ to $(x,xy)$. The monoid $G$ is a group iff this map is an isomorphism (of underlying spaces).
Is there a similar characterization of when a monoidal category is rigid? Something like "consider the map $\mathcal C \times \mathcal C \to \mathcal C \times \mathcal C$, and ask that it be a left adjoint"?