Surjectivity of pushforward on image

Question

Let $\mathcal X\subseteq\mathbb R^m$ be a Borel measurable set. $\Phi:\mathcal X\to\mathbb R^n$ be a continuous mapping and $\mathcal Y = \Phi(\mathcal X)\subseteq\mathbb R^n$ its image. Let $\mathcal P_{\mathcal X}$ denote the space of all Borel probability measure on $\mathcal X$ (under the standard topology) and $\mathcal P_{\mathcal Y}$ be the set of Borel probability measures on $\mathbb R^n$ supported in $\mathcal Y$ (namely whose support is included in $\mathcal Y$). Is it true that for any $P\in\mathcal P_\mathcal Y$ there is $P'\in\mathcal P_\mathcal X$ such that $P = \Phi^\# P'$, the push-forward of $P'$ under $\Phi$?

I would guess it is the case, but proving it does not seem trivial... If the support of $P$ is countable, then for each $y$ the set $\xi_y=\Phi^{-1}(\{y\})$ is Borel and all these sets are distinct. So one can find a $P'$ such that $P'(\xi_y) = P(\{y\})$ (for instance one picks a single point from each $\xi_y$ and puts there all the mass). This is a probability and $P = \Phi^\# P'$. However, if the support of $P'$ is uncountable things are definitely trickier.

Answer 1

It is true; the trick is to use the von Neumann-Jankov measurable selection theorem to construct a right inverse of $\Phi$ on $\Phi(\mathcal X)$.

The result is essentially Lemma 2.2. of [Varadarajan, V. S. "Groups of automorphisms of Borel spaces." Transactions of the American Mathematical Society 109.2 (1963): 191-220.]