Idempotent splitting for Markov kernels Let $X$ be a standard Borel space and $e : X \to X$ a Markov kernel. Suppose that $e$ is idempotent, that is $e \circ e = e$, or written out using the Chapman-Kolmogorov equation,
$$e(A|x) = \int_X e(A|y) \, e(dy|x) \qquad \forall x \in X, A \in \Sigma_X.$$
Does this imply that $e$ splits? That is, do there exist another standard Borel space $Y$ and Markov kernels $p : X \to Y$ and $i : Y \to X$ such that $i \circ p = e$ and $p \circ i = \mathrm{id}_Y$?
Two remarks:

*

*Note that the splitting easily implies the idempotency, but the converse is not so clear.


*We have a preliminary proof that splitting is possible based on an old result of Blackwell on idempotent Markov kernels combined with some category-theoretical machinery. So my main question is really: is this new? If not, where was this done? We haven't seen the problem mentioned anywhere in the literature so far.
 A: I do not know if the result is new or not. But I believe the problem is easy if you think in terms of the Eilenberg-Moore category of the Giry monad. Since that category is isomorphic to a certain subcategory of superconvex spaces let me work in that category. (If you're not familiar with superconvex spaces, resort to EM, just notice that all the measurable maps happen to be countably affine functions also!)
So let $\mathcal{P}X$ be the set $\mathcal{G}X$ endowed with the natural superconvex space structure which is defined pointwise.  Similarly, for $f:X \rightarrow Y$ a morphism in standard Borel let $\mathcal{P}f: \mathcal{P}X \rightarrow \mathcal{P}Y$ denote the countably affine map in superconvex spaces.
Let $k:X \rightarrow \mathcal{G}X$ be your kernel viewed in the Kleisi category, which, in the category of superconvex space, corresponds to the countably affine map $\mu_X \circ \mathcal{P}k:\mathcal{P}X \rightarrow \mathcal{P}X$.  To say it is idempotent just says $\mu_X \circ \mathcal{P}k \circ \mu_X \circ \mathcal{P}k = \mu_X \circ \mathcal{P}k$.  In the EM category (or Superconvex category) all colimits exist.  So to find the splitting just take the coequalizer $q$ of the two parallel arrows,
$\mathbf{id}: \mathcal{P}X \rightarrow \mathcal{P}X$ and $\mu_X \circ \mathcal{P}k: \mathcal{P}X \rightarrow \mathcal{P}X$.
Now, by definition, the superconvex space structure on the quotient space is defined by $\sum_{i \in \mathbb{N}}p_i q[P_i] := q(\sum_{i \in \mathbb{N}} p_i P_i)$ where $\mathbf{p} \in \mathcal{G}(\mathbb{N})$ with components $p_i$.
The definition of the structure on the quotient space implies the insertion map $\iota: \mathcal{P}X/\sim \rightarrow \mathcal{P}X$ is a countably affine map, i.e., take a representative from each equivalence class and map $[P] \mapsto P$.   That's your splitting which, incidentally, is a nice example of a $\mathcal{G}$-algebra arising in practice.
