A similar question to mine was asked before at the address below but it was not answered there so I am asking it again. Also there is a more specific case I am interested in.
Topological entropy of isometric extension
This is Exercise 2.5.6 in 'Introduction to Dynamical Systems' by Michael Brin and Garrett Stuck. Let $Y,Z$ be compact metric spaces and let $X = Y \times Z$. Let $d$ be the metric on $X$. Write $\pi:X \to Y$ be the projection onto the $Y$-coordinate. Suppose $f: X \to X$ is an isometric extension of $g: Y \to Y$; that is to say $\pi \circ f = g \circ \pi$ and $d(f(x_1),f(x_2)) = d(x_1,x_2)$ for all $x_1,x_2 \in X$ with $\pi(x_1) = \pi(x_2)$. Prove that $h(f) = h(g)$, where $h$ is the topological entropy of a map.
I understand that points in the same fiber do not diverge, since $f$ acts isometrically there. However, in order to show entropy does not increase we need to control the divergence of points in nearby fibers. I don't see how to do this.
I've been focusing on trying to understand the following specific case. Let $Y = Z = S^1$ so that $X$ is the standard two-dimensional torus. Let $g$ be the identity map on $Y$, so clearly $h(g) = 0$. Take any continuous function $\phi:S^1 \to S^1$ and define a homeomorphism $f_\phi: X \to X$ by letting $f_\phi$ leave every fiber setwise invariant and rotating the fiber over $y \in Y$ by an angle of $\phi(y)$. Then clearly $f_\phi$ is an isometric extension of $g$. However in the case $\phi(y)$ is not Lipschitz I am having a hard time proving that $h(f_\phi) = 0$.
