Is the category of sup-lattices rigid? In An Extension of the Galois Theory of Grothendieck Joyal and Tierney show that the category $\mathsf{SupLat}$ of sup-lattices is star-autonomous. By listing compact closed categories and $\mathsf{SupLat}$ seperately the nLab entry on star-autonomous categories makes it seem as though $\mathsf{SupLat}$ is not rigid. However, some sources claim this to be the case (e.g. 1 and 2). Let the duality functor on objects send each sup-lattice to its opposite poset. Take the usual tensor product of sup-lattices (for an explicit construction see 3).

What are the evaluation and coevaluation morphisms making $\mathsf{SupLat}$ into a rigid monoidal category?

Is the two-element lattice $\{ 0 < 1\}=\mathcal{P}(\{\ast\})$ the monoidal unit with respect to the given tensor product? If so, the right evaluation has to be a map $M \otimes M^* \cong \text{Hom}(M,M)^* \rightarrow \mathcal{P}(\{\ast\})$? What are natural candidates?
 A: The category of suplattices is not rigid.  If it were rigid, then in particular, for any suplattice $M$, the suplattice $\operatorname{Hom}(M,M)$ would be self-dual (since we would have $\operatorname{Hom}(M,M)\cong M^*\otimes M\cong(M\otimes M^*)^*\cong (M^*\otimes M)^* \cong \operatorname{Hom}(M,M)^*$).  But this is not true.
For instance, let $M=\{0,a,b,c,1\}$, with $a,b,$ and $c$ all incomparable.  A morphism $f:M\to M$ can be described as a triple $(x,y,z)$ with $x\vee y=x\vee z=y\vee z$, where $x=f(a),y=f(b)$, and $z=f(c)$.  There are only $9$ coatoms of $\operatorname{Hom}(M,M)$, namely the elements of the form $(x,1,1)$ or a permutation thereof, where $x$ is $a,b,$ or $c$.  Indeed, it is easy to see that any triple $(x,y,z)\neq (1,1,1)$ is less than or equal to one of these $9$ coatoms.  However, there are more than $9$ atoms of $\operatorname{Hom}(M,M)$: there are $9$ atoms of the form $(x,x,0)$ or a permutation thereof where $x$ is $a,b,$ or $c$, but $(a,b,c)$ is also an atom.  So $\operatorname{Hom}(M,M)$ is not isomorphic to its dual.
