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 $\Gamma(\tilde\mu,\tilde\nu)$?



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


