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Measure theoretic entropy of General Tent maps

The linked question made me wonder how to calculate the topological entropy of a general tent map.

Let $I=[0,1]$ and $\alpha \in ( 0,1)$. Define $T: I \rightarrow I$ by

$T(x)= x/\alpha$ for $x \in [0,\alpha]$ and $(1-x)/(1-\alpha)$ for $x \in [\alpha,1]$

What is the topological entropy of $T$ and how does one prove it?

I suspect it is $\log 2$ regardless of value of $\alpha$, unlike its metric entropy.

I suspect it is $\log 2$ because $T$ seems somewhat conjugate to the shift map on $2^{\mathbb N}$ except it can't be because the interval is not homeomorphic to $2^{\mathbb N}$.

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Le Curious
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1 Answers1

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Since the peak of the skewed tent touches ocurrs at $(\alpha, 1)$, the pre-image of this point will consist of two points, therefore its second iterate will have four laps. But now we can apply this argument to both these new points (only because of the surjectivity of the function ensured by the peak of the tent "touching the top"). In general the critical point will have $2^n$ $n$th pre-images. Therefore the number of laps will be $l(T^{(n)})=2^n$. So its growth number will be $s(T)=\lim_{n\rightarrow + \infty}l(T^{(n)})^{1/n}=2$. Finally by a known result of Misiurewicz $h(T)=\log s(T)$. Therefore the entropy is $\log 2$.

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