Consider two product measurable spaces $\left(X \times Y,\mathcal{X} \otimes \mathcal{Y}\right)$, $\left(X' \times Y',\mathcal{X'} \otimes \mathcal{Y'}\right)$ with the usual product sigma-algebra, and a measurable function $g$ mapping from $X \times Y$ to $X' \times Y'$ defined by $x' = g_1(x), y' = g_2(y)$. Notice the assumption of no cross-dependence.
If $\mu$ is a probability measure on $\mathcal{X} \otimes \mathcal{Y}$ then $g$ induces a push-forward joint probability measure $\mu' := \mu g^{-1}$ on $\mathcal{X'} \otimes \mathcal{Y'}$ with marginals $\mu'_X$, $\mu'_Y$.
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 $\mathcal{X'} \otimes \mathcal{Y'}$, 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 $\nu$, i.e. if $\Gamma \in \Pi\left(\mu'_X,\mu'_Y\right)$ then $\Gamma = \nu g^{-1}$ for some $\nu$?
If it helps, in applications all spaces are Euclidian of varying dimensionality.
Best Answer
The answer is yes in the Euclidean case. This follows from Lemma 2.2 in
Varadarajan, Veeravalli S. "Groups of automorphisms of Borel spaces." Transactions of the American Mathematical Society 109.2 (1963): 191-220.
Here is the statement of the lemma:
Lemma: Let $(X,\mathcal{B})$ and $(Y,\mathcal{C})$ be analytic Borel spaces and let $\pi$ be a measurable map of $X$ onto $Y$. If $\nu$ is any measure on $\mathcal{C}$ there exists a measure $\mu$ on $\mathcal{B}$ such that $\nu(A)=\mu\big(\pi^{-1}(A)\big)$ for all $A\in\mathcal{C}$.
To apply it, note that Euclidean spaces with their Borel $\sigma$-algebra are analytic Borel spaces, and so are their images in Euclidean spaces (with the relative Borel $\sigma$-algebra) under measurable functions. The assumption of no cross-dependence guarantees that the range of $g$ is a rectangle on which all couplings are supported.