Is there a coupling that induces a given coupling via a transition kernel?

Question

Let $X,Y$ be two measurable spaces, $\mu,\nu$ two probability measures on $X$, and $\kappa$ a transition kernel from $X$ to $Y$. Define $\tilde\mu(dy)=\int_X\kappa(dy|x)\mu(dx)$ and $\tilde\nu(dy)=\int_X\kappa(dy|x)\nu(dx)$.

Denote by $\Gamma(\mu,\nu)$ the set of couplings of $\mu$ and $\nu$ and by $\Gamma(\tilde\mu,\tilde\nu)$ the corresponding set for $\tilde\mu$ and $\tilde\nu$.

Furthermore, let $S_{\kappa}:\Gamma(\mu,\nu)\to\Gamma(\tilde\mu,\tilde\nu)$ be the mapping defined by

$$S_{\kappa}(\pi)(dy_1,dy_2)=\int_{X\times X}\kappa(dy_1|x_1)\kappa(dy_2|x_2)\pi(dx_1,dx_2),\quad \pi\in\Gamma(\mu,\nu).$$

Question: Is $S_{\kappa}$ onto $\Gamma(\tilde\mu,\tilde\nu)$?

Answer 1

It's not true. If $\mu \mapsto \tilde\mu$ is injective and $X$ has at least two points then the transition kernels have to be deterministic, i.e. there is a measurable map $T:X\to Y$ such that $\kappa(dy|x)=\delta_{T(x)}$.

To illustrate this take the trivial example $X = \{1,2\}$, $Y=\{(1,1),(1,2),(2,1),(2,2)\}$ and $\kappa(dy|i) = \frac{1}{2}\left(\delta_{(i,1)} + \delta_{(i,2)}\right).$ Now choose $\mu = \delta_1$ and $\nu = \delta_2$. Observe that $\tilde\mu = \frac{1}{2}\left(\delta_{(1,1)} + \delta_{(1,2)}\right)$ and $\tilde\nu = \frac{1}{2}\left(\delta_{(2,1)} + \delta_{(2,2)}\right)$ However, there is only one coupling between $\mu$ and $\nu$ but there are uncountably many couplings between $\tilde\mu$ and $\tilde\nu$.

This construction generalizes to arbitrary $X$ and $Y$ by observing that there are uncountably many couplings between two transition kernels $\kappa(dy|x_1)$ and $\kappa(dy|x_2)$ but the map $S_\kappa$ only picks the product coupling (choose $\mu=\delta_{x_1}$ and $\nu=\delta_{x_2}$ for $x_1\ne x_2$). If $\mu$ and $\nu$ are non-atomic then the construction of the counterexample is a bit messier but still works.