# Does a surjective measurable map induce a surjective pushforward operator?

I hope it is OK to post a question that is basically the same as the months old currently unanswered question at math stackexchange

Suppose X, Y are Polish spaces (without loss of generality, we may assume that X, Y are both the unit interval for the purpose of this question). Let $f: X \to Y$ be a measurable surjective function. Is it true that the pushforward operator $f_*: P(X) \to P(Y)$, where $P(X)$ denotes the set of all (Borel) probability measures on $X$, is also surjective? In other words, does an arbitrary probability measure $\nu \in P(Y)$ lift to some preimage $\mu \in P(X)$ such that $\nu$ is the pushforward measure of $\mu$ under $f$?

Remark 1: If we additionally assume that $f$ is continuous and that $X$ is compact, then the answer is positive for this special case, as is shown in: https://math.stackexchange.com/questions/593821/prove-the-existence-of-a-measure-mu

Remark 2: As tomasz pointed out in the original thread, we may assume that $f$ is continuous by replacing the topology of X by adding more open sets so that the resulting topology is still Polish but has the same Borel sets. I don't think the resulting topology is guaranteed to be locally compact, let alone, compact or being the usual topology of the unit interval.

Remark on the selected answer: The proof in the referenced book proceeds by using the existence of a universally measurable selector (w.r.t. $f$) in the obvious way.

This holds even for $X$ an analytic space and $Y$ separable metric space. One reference for this result is this book by Doberkat, in Proposition 1.101. Actually, he proves it for subprobabilities, but I think this should go through.