Consider a measurable function $g$ mapping a probability space $\left(\Omega,\mathcal{F},\mu\right)$ to a product measurable space $\left(T,\mathcal{T}\right)$ with cartesian product $T = X \times Y$ and product sigma-algebra $\mathcal{T} = \mathcal{X} \otimes \mathcal{Y}$; $g$ thus induces a push-forward joint probability measure $\mu_T$ on $\mathcal{T}$ with marginals $\mu_X$, $\mu_Y$ respectively on $\mathcal{X}$, $\mathcal{Y}$.
Let $g, \mu_X, \mu_Y$ be fixed and define $\Pi\left(\mu_X,\mu_Y\right)$ as the class of couplings of $\mu_X$, $\mu_Y$ on $\mathcal{T}$, i.e. all joint probability measures with marginals $\mu_X$, $\mu_Y$.
Is it true that any of such coupling is obtainable as the push-forward via $g$ of an appropriate probability measure $\nu$ on $\left(\Omega,\mathcal{F}\right)$, i.e. if $\Gamma \in \Pi\left(\mu_X,\mu_Y\right)$ then $\Gamma = \nu g^{-1}$ for some $\nu$?
No invertibility hypothesis are made on $g$. Applications are on $T = \mathbb{R}^{k}$ but I am mostly interested in the general case.
Best Answer
The statement as presented is false. For example if $g$ is defined by $x = \omega$, $y = \omega$, then the only attainable coupling via $\nu g^{-1}$ is the maximal one no matter the $\nu$.