Does $\mathcal{KL}(D)$ admit the "yanking" axiom

Question

Bob Coecke made the "yanking" axiom famous as he applied it to teleportation in Quantum Computing:

This is normally presented on the category of Hilbert spaces, and so here is a derivation of the yanking rule on Hilb:

I am interested to know if the yanking axiom is admitted by the Kleisli Category of the Distribution monad, $\mathcal{KL}(D)$. A good description of this category can be found in Jacobs and Furber "Towards a categorical account of conditional probability.".

Much work has been done to find the axioms for $\mathcal{KL}(D)$ and categories like it, as you can see in Fritz, Tobias. "A synthetic approach to Markov kernels, conditional independence and theorems on sufficient statistics." as well as Jacobs, Bart, A. Palmigiano, and M. Sadrzadeh. "Multisets and distributions, in drawing and learning."

What I know so far is that $\mathcal{KL}(D)$ is symmetric monoidal and has cups like this $I \rightarrow X \otimes Y$, it might even have a dagger which would give it caps like this $X \otimes Y \rightarrow I$. If $\mathcal{KL}(D)$ had conditionals, then it would have a functorial dagger. Even if it did, though, you have to then prove that these interact in the right way to get yanking.

Answer 1

You added a proof that a certain category satisfies the "yanking axiom", and I think I can infer that it simply says that every object in your symmetric monoidal category is dualizable. I'm not sure, though. If the "yanking axiom" means something different, then the following may be moot.

Anyway, if that is what the "yanking axiom" says, then I believe the answer to your question is no. The symmetric monoidal structure on $KL(D)$ which you have in mind presumably is the one lifting the cartesian monoidal structue on $Set$, where $X \otimes_{KL(D)} Y = X \times Y$. Under this monoidal structure, the monoidal unit is the one-element set $1$. Moreover, $1$ is terminal in $KL(D)$. If $X$ is dualizable, then its dual $X^\vee$ is given by the internal hom $X^\vee = [X, 1]$. I don't think that internal homs exist in general in $KL(D)$ (although I believe that $[X,Y]$ exists when $X$ is a finite set), but since $1$ is terminal, the internal $[X,1]$ certainly exists and is given by $1$. So if $X$ is dualizable, then it dual is the terminal object $1$. That is, there is at most one dualizable object up to isomorphism. The unit of a monoidal category is always dualizable, so in this case we see that $1$ is the unique dualizable object of $KL(D)$.