Existence of a measurable function such that the pushforward of a measure is equal to another measure Suppose we have two probability spaces $(\mathcal{X},\Sigma_\mathcal{X},\mathbb{P}_\mathcal{X})$ and $(\mathcal{Y},\Sigma_\mathcal{Y},\mathbb{P}_\mathcal{Y})$. What are the conditions on the probability spaces such that there exists a measurable function $f:\mathcal{X}\rightarrow\mathcal{Y}$ such that $f_\#\mathbb{P}_\mathcal{X} = \mathbb{P}_\mathcal{Y}$? 
For instance if we had $\mathbb{P}_\mathcal{X} = \delta_x$ for some $x\in\mathcal{X}$ and if we had $\mathbb{P}_\mathcal{Y} = \frac{1}{2}(\delta_y + \delta_z)$ for some $y,z\in\mathcal{Y}$, there exists no measurable function $f:\mathcal{X}\rightarrow\mathcal{Y}$ such that $f_\#\mathbb{P}_\mathcal{X} = \mathbb{P}_\mathcal{Y}$. This is because we can't split the atom $\{x\}$. 
My intuition says that if $\mathcal{X}$ and $\mathcal{Y}$ have no atoms then we can guarantee the existence of such an $f$. However it may be more complicated than this and rely on more regularity in the structure of $\mathcal{X}$ and $\mathcal{Y}$ (e.g. they may have to be Borel spaces generated by diffeomorphic topological spaces say).
 A: I’m not familiar with your notations, but it seems that Theorem 17.41 from [Kech] might be useful for you.

We recall some definitions used in this theorem. 



(17.4) A measure $\mu$ on a measurable space $X$ is called continuous if $\mu(\{x\})=0$ for all $x$.
(17.A) A measure $\mu$ on a measurable space $(X,\mathcal S)$ is a probability measure, if $\mu(X)=1$. 
(17.E) Let $X$ be a separable metrizable space. We denote by $P(X)$ the set of probability Borel measures on $X$. 
Let $X, Y$ be topological spaces. A map $f:X\to Y$ is Borel (measurable), if the inverse image of a Borel (equivalently: open or closed set is Borel). If $Y$ has a countable subbasis $\{V_n\}$, it is enough to require that $f^{-1}(V_n)$ is Borel for each $n$. We call $f$ a Borel isomorphism if it is a bijection and both $f$, $f^{-1}$ are Borel, i.e. for $A\subseteq X$, $A\in \mathbf{B}(X)\Leftrightarrow f(A)\in \mathbf{B}(Y)$.
References
[Kech] A. Kechris, Classical Descriptive Set Theory, – Springer, 1995.
A: Already in the real line you to not have a map for which the pushforward of the dirac delta in $0$ is absolutely continuous with respect to the Lebesgue measure. I suggest you to study the notes:
http://cvgmt.sns.it/media/doc/paper/195/users_guide-final.pdf
There are a lot of things that may interest you in section 1 of the aforementioned notes.
If you are looking an absolutely general characterisation, I am afraid you are quite out of luck.
You may think to replace $f$ with a probability measure on $X\times Y$ whose marginals are the probability measures you are looking for.
