TITLE: The dual of a dual in a rigid tensor category
QUESTION [8 upvotes]: For a rigid tensor category $\cal{C}$, can it happen that, for some $X \in {\cal C}$, we have that $X$ is not isomorphic to $(X^{*})^*$, for $*$ denoting dual? If so, what is a good example.

REPLY [4 votes]: Another way of showing this isn't true in general (and this is a fairly systematic way to handle this kind of questions) is to look at string diagrams, ie at the free rigid (strict, say) monoidal category generated by one object $X$. It is not hard to see that morphisms in that category can be represented by "planar open oriented tangles", i.e. collections of intervals embedded in a rectangle so that the endpoints of each interval are attached to the top or bottom edge of the rectangle, modulo planar isotopy.
In that category, $Hom(X,X^{**})$ is simply empty. ANother way to say this is that if this statement was always true, then it would be a formal consequence of the axioms of rigid monoidal categories, and one would be able to write down such an isomorphism purely in terms of the duality data, and it's easy to see there is'nt even a candidate map (let alone an isomorphism).
It might be worth mentionning that the statement that does generalize to any rigid category, is the fact that there are canonical isomorphism
$${}^*(X^*)\cong X \cong ({}^*X)^*.$$