I'm studying a map $f(x)=\frac{x - x \bmod b}{b} + k(x \bmod b)$ that has two parameters $b, k \in \mathbb{Z}, b ≠ 0$. I'm trying to show that it's semiconjugate/conjugate to the beta transformation but with parameter $k$, i.e. $B(x) = kx \bmod 1$. This is on intervals $[0, bk-1)$ for $f(x)$ and $[0, 1)$ for $B(x)$.
There is indeed a map $h(x)$ such that $h(f(x)) = B(h(x))$, it even is a homeomorphism. But since these functions are parameter-dependent, Scholarpedia also indicated that there also has to be a homeomorphism between their parameter spaces, though that article was on full conjugacy, so I assume it only requires to be a surjective map for semiconjugacy.
Since $f(x)$ has parameters $b, k$ and $B(x)$ with just $k$, I'd have to have a map $p: (b, k) ↦ k$, which I assume is just surjective.
Now $f(x) = h^{-1}(B(h(x)))$ would be desirable of course, I kinda have proof it holds for [0, bk-1) (it relies on the fact that x mod bk-1 = x in the interval), but in the context of conjugacy things, it requires $h(x)$ to be invertible, which in this case it is, but I'm not sure whether the mapping between parameter spaces being just surjective stops me from being able to conclude that. I at least assume it doesn't let me consider them ($f(x)$ and $B(x)$) fully conjugate. I mean, $h^{-1}$ is gonna have both parameter variables anyway so $B(x)$ only depending on $k$ I assume wouldn't matter as much in getting $f(x)$ back.
I'm just kinda lost, as I found no other source on conjugacy of parameter-dependent maps.