Let $\mu$ be a probability measure on $\mathbb{R}^n$ that is absolutely continuous w.r.t Lebesgue measure, and $T_1$ is a push-forward map such that $T_1\# \mu= \mu_1$. Let $\mu_2$ be any probability measure on $\mathbb{R}^n$ or on more general $\mathbb{R}^m$.
If there is a joint distribution $\pi$ whose first marginal is $\mu_1$, and the second marginal is $\mu_2$. Does it always exists a map $T_2$ such that ${{(T_1,T_2)\# \mu = \pi}}$?
<strike>${(T_1,T_2)\# \mu_1 = \pi}?$</strike> (which is a typo)
Recall, if $\nu_1,\nu_2$ are probability measures on $\mathbb{R}^n$ and $\nu_1$ absolutely continuous w.r.t Lebesgue measure, then there exists at least a map such that $T\#\nu_1=\nu_2$. This can be found Existence of a push forward map or Lemma 1.28 in Santambrogio's book.
In general, given a $n$-dim joint distribution $\pi\in \mathbb{P}(X_1,X_2,\cdots,X_n)$ where $X_i\subset\mathbb{R}^n$, can one always find maps $(T_i)_{i=1}^n$ such that $\pi=(T_1,\cdots,T_n)\#\mathcal{L^n}$, where $\mathcal{L^n}$ is the Lebesgue measure on $\mathbb{R}^n$. From the view of change of variables formula, if this is true, it seems that one may find a universal parameterization for $X_i$, which is doubtful especially on some multi-valued cases.
