L.s.,
This is a homework question some of my fellow students and I are having great difficulty with.
Let $Y,Z$ be compact metric spaces, $X = Y \times Z$, and $\pi$ the projection to $Y$. Denote $h$ the topological entropy. Suppose that $f: X \rightarrow X$ is an isomorphic extension of $g: Y \rightarrow Y$. Show that $h(f) = h(g)$.
What we have come up with:
We know that for an isomorphic extension, $\pi \circ f = g \circ \pi$. (semi-conjugacy). Also we know that for all $x_1, x_2 \in X$ for which $\pi(x_1) = \pi(x_2)$ we have that $ d(f(x_1), f(x_2))= d(x_1, x_2).$
And we know that $\pi(x_1) = \pi(x_2)$ if the $y$-coordinate in $X$ is the same. So we expect that in a way the $z$-coordinate 'doesn't matter'.
We can use a theorem that says that if $(X, g)$ a factor of $(X, f)$, that then $h(f) \geq h(g)$. So we only need to show the reversed inequality. So we tried to show that the 'sep' of $(X, f)$ is always smaller than the 'sep' of $(X,g)$. But here we're having trouble, because for every separating set we define on $(X, f)$, we don't know if you can also say it separates $(X, g)$! This is because we don't know what $f$ and $g$ precisely do..
Furthermore we don't have any theorems or propositions we can use, so we probably overlook something and everything follows from the definitions?
Anyways, any hints, directions would be greatly appreciated!
Kind regards,
Teuntje