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Let $X$ and $Y$ be compact metric spaces, $\pi: X \to Y$ a continuous onto map, and $\nu$ a Borel probability measure on $Y$ such that $\nu$-almost every $y \in Y$ has unique $\pi$-preimage. Does there exist a measurable map $s: Y \to X$ such that for $\nu$-almost every $y \in Y$, $(\pi \circ s)(y) = y$?

For the application I'm interested in, $X$ and $Y$ are shift spaces with finite alphabet over a countable group, if that helps at all.

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Every continuous onto map $\pi: X \to Y$ between compact metric spaces has a Borel measurable section, i.e. there is a Borel measurable map $s: Y \to X$ such that $\pi \circ s (y) = y$ for all $y$. The assumptions about the measure $\nu$ and the preimages of $\pi$ are superfluous.

For a proof of this, you can look up Theorem 6.9.7 in Bogachev's Measure Theory, Volume 2. The idea of the proof is to start with the case where $X \subset [0,1]$. In this case, for a given $y \in Y$ you can select a preimage of $y$ by taking the smallest one, i.e. by setting $$ s(y) = \inf \; \pi^{-1}(y). $$ In the general case, the trick is to use the fact that for any compact metric space $X$ there is some compact set $K \subset [0,1]$ and a surjective continuous map $f: K \to X$ (in fact you can take $K$ to be the Cantor set). Then you can apply the previous case to the composition $\pi \circ f$.

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  • $\begingroup$ Thanks! The assumptions about the measure $\nu$ and the preimages of $\pi$ are needed for my application, but it's good to know they aren't needed for the section to exist. $\endgroup$ – Sophie MacDonald Mar 27 at 16:57
  • $\begingroup$ My pleasure! Of course under these assumptions any two sections will agree $\nu$-almost everywhere, whereas in general one can't say much. $\endgroup$ – Dominique R.F. Mar 27 at 17:12
  • $\begingroup$ Yes, and in fact my situation is that $\nu = \pi_* \mu$ where $\mu$ is a fully supported probability measure on $X$; this means that $s$ is also a left inverse of $\pi$ $\mu$-almost everywhere. Moreover, $\mu$ and $\nu$ are both invariant under the continuous action of a countable group $G$, and $\pi$ is $G$-equivariant; the assumptions thus imply that $s$ is also $G$-equivariant almost everywhere. $\endgroup$ – Sophie MacDonald Mar 27 at 17:28

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