Finite-to-one continuous factor map preserves entropy? Let $X$ and $Y$ be compact metric spaces with Borel $\sigma$-algebras $\mathcal{B}$ and $\mathcal{C}$. Let $S: X \to X$ and $T: Y \to Y$ be homeomorphisms. Let $\pi: X \to Y$ be a bounded-to-one continuous factor map, i.e. $\pi$ is surjective, $\pi \circ S = T \circ \pi$, and for some $1 \leq M < \infty$ and all $y \in Y$, $\mathrm{card} \,\pi^{-1}(\{ y \}) \leq M$. Let $\mu$ be an $S$-invariant Borel probability measure on $(X, \mathcal{B})$, and equip $(Y, \mathcal{C})$ with the $T$-invariant measure $\nu = \pi_* \mu$.
Is it the case that $h_{\nu}(T) = h_{\mu}(S)$? If so, how much can my hypotheses be weakened?
I have tried to apply the Abramov-Rokhlin formula, which states that
$$
h_{\mu}(S) = h_{\nu}(T) + h_{\mu}(S \, | \, T)
$$
where $h_{\mu}(S \, | \, T) = h_{\mu}( S \, | \, \pi^{-1} \mathcal{C} )$ is the relative entropy. I think I should be able to use the bounded-to-one hypothesis to show that this relative entropy is $0$, but I don't know how to do this.
 A: The answer is yes, by an Abramov-Rohlin type formula of Ledrappier and Walters ("A relativised variational principle for continuous transformations", p.569, Thm.2.1; they prove the statement for pressure).
Theorem (Ledrappier-Walters): Let $X,Y$ be compact metric, $S:X\to X$, $T:Y\to Y$ be continuous, $\pi:X\to Y$ be a continuous surjection such that $\pi\circ S=T\circ \pi$ (so that $T$ is a topological factor of $S$). Then for any $\nu\in\operatorname{Prob}(Y,T)$:
\begin{align*}
\operatorname{ent}_{\nu}(T) 
&\leq \inf_{\substack{\mu\in\operatorname{Prob(X,S)}\\\pi_\ast(\mu)=\nu}} \operatorname{ent}_{\mu}(S) \\
&\leq \sup_{\substack{\mu\in\operatorname{Prob(X,S)}\\\pi_\ast(\mu)=\nu}} \operatorname{ent}_{\mu}(S)
= \operatorname{ent}_{\nu}(T) + \int_Y \operatorname{topent}(S;\pi^{-1}(y))\, d\nu(y).
\end{align*}
Here $\operatorname{topent}(S;\pi^{-1}(y))$ is defined using $(S,n,\epsilon)$-separated subsets of $\pi^{-1}(y)$ (or sets that $(S,n,\epsilon)$-span $\pi^{-1}(y)$), and $y\mapsto \operatorname{topent}(S;\pi^{-1}(y))$ is measurable. When a fiber $\pi^{-1}(y)$ is finite it's straighforward that $\operatorname{topent}(S;\pi^{-1}(y))=0$; when each fiber is finite (not necessarily uniformly so) the integral on the RHS vanishes.

Let me also mention that this theorem builds on an analogous theorem on topological entropy of Bowen:
Theorem (Bowen): With $X,Y,S,T,\pi$ as in the previous theorem,
\begin{align*}
\operatorname{topent}(T) 
&\leq \operatorname{topent}(S) \\
&\leq \operatorname{topent}(T) + \sup_{y\in Y}  \operatorname{topent}(S;\pi^{-1}(y)).
\end{align*}
Bowen's proof of this is combinatorial, and the theorem gives the topological version of what you are asking. Of course via the variational principle the Ledrappier-Walters formula also gives it. Robinson's Dynamical Systems: Stability, Symbolic Dynamics, and Chaos has a proof of the uniformly finite fibers case for the topological entropy (p.340,Thm.1.8 of the first edition). I haven't read it but one might be able to adapt it to bypass Ledrappier-Walters.
