Conjecture [efr-4LCO]

October 22, 2025
Eigil Fjeldgren Rischel

Consider a category with the following structure and properties:

Markov
Finite coproducts and $\otimes$ distributes
A natural transformation $A \otimes A \to A$
The resulting binary operation on $\mathcal {C}(X,A)$ makes all these into abstract convex bodies in the sense of Escardo-Simpson.

Call such a category a iterable choice Markov category

Then I conjecture that $\mathsf {FinStoch}$ with the obvious "coinflip" operation is the initial iterable choice Markov category.

Conjecture 2: $\omega \mathsf {Stoch}$ , the category of countable sets and discrete probability measures, is the initial Markov category with distributive countable coproducts and an iterable coinflip structure.

Conjecture 3: $\mathsf {BorelStoch}$ is the initial category with countable distributive coproducts, countable Kolmogorov products, and an iterable coinflip structure.

Conjecture 4: $\mathsf {FinStoch}_{\leq 1}$ or possibly $\omega \mathsf {Stoch}_{\leq 1}$ is initial countably distributive CD-category with trace on plus and [some axioms]

Subconjecture: $\mathsf {Borel} \hookrightarrow \mathsf {BorelStoch}$ is the inital functor (preserving coproducts, tensor, and Kolmogorov products) to a iterable coinflip category. Could be true! Seems enough to check maps into countable sets (clear) and bitstreams.

Remark

October 22, 2025
Eigil Fjeldgren Rischel

It is not necessarily clear that one can always add a free iterable coinflip structure to a given Markov category, but we can at least ask whether a given functor $\mathcal {C} \to \mathcal {D}$ is the free such.

Given $\mathsf {Borel} \to \mathcal {C}$ where $\mathcal {C}$ has Kolmogorov products and iterable coinflips, a measure $* \to 2^\omega$ is simply carried to the pairing/limit of the image of the marginals $* \to 2^N$ . The remaining question is how to deal with kernels $2^\omega \to X$ . That is, we know the "pointwise" value of such maps, but that doesn't necessarily suffice.

To do this it suffices to define an operation on measurable maps $2^\omega \to GX$ ( $G$ being the Giry monad), and show it well-defined. Since $\mathsf {BorelStoch}$ is a model, we certainly can't define any non-measurable maps in the initial thing. Note that $GX \cong 2^\omega$ for every $X$ , and moreover we can choose a suitable representation of this (at least when $X$ is finite, which suffices by the filtration argument). For {2}, this is the map $2^\omega \to [0,1]$ . This is surjective, but not injective, so every kernel $2^\omega \to 2$ has a representation by lifting over this. The kernel $2^\omega \to 2$ is of course the canonical midpoint iteration map, and thus its image is determined. Then we just need to show given two measurable $2^\omega \to 2^\omega$ , if they have the same sampling probability at each point, this is provable "using the logic available to us".

For larger X this should be doable in a similar way (not totally clear yet)