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"?

isa map to the "dual" (co)algebra, where "dual" here means in the Morita bicategory. Similarly, for a category $\mathcal C$, the opposite category $\mathcal C^\op$, which is the recipient of $X \mapsto X^*$, is the dual object in the "Morita" bicategory of categories, profunctors, and natural transformations. I don't know if that is a useful similarity or not. $\endgroup$ – Theo Johnson-Freyd Dec 6 '15 at 16:41Lectures on Tensor Cateogries and Modular Functors. Do you understand the argument here? $\endgroup$ – Chris Schommer-Pries Dec 7 '15 at 13:04