properties of a truncated tent maps on unit interval

Question

given the following truncated tent map:

$T_h: [0,1] \rightarrow [0,1], x \mapsto min(h, 1-2|x -1/2|)$ for $0 \leq h \leq 1$

my script in dynamical systems states the following properties:

1. $T_1$ has only one periodic point (the fixed point 0) while the tent map $T_1$ has a 3-cycle {2/7, 4/7, 6/7} and hence has all natural numbers as periods by the Sharkovsky Forcing Theorem.
2. Any cycle $\mathcal{O} \subset [0, h)$ of $T_h$ is a cycle for $T_1$, and any cycle $\mathcal{O} \subset [0, h]$ of $T_1$ is a cycle for $T_h$.

Since the script does not provide any proofs on these claims I tried showing them myself, but got stuck at some point. Can someone help me?

Proof

1. My problem starts with the definition already. What is the difference between $T_1$ and tent map $T_1$? I thought $T_1$ was my tent map? So, for the tent map: $T_1 = 1-2|x-1/2|$ for $0 \leq x \leq 1$. $T_1(0) = 0$, thus $x = 0$ is a fixed point and therefore of period one. Now, $T_1(2/3) = 2/3$, thus $x = 2/3$ is also a fixed point and of period one. What am I missing?
   It is easy to check the second part, since $f(f(f(2/7))) = f(f(4/7)) = f(6/7) = 2/7$
2. I know how to show that the points $s^2/m (m,s \in \mathbb{N}, with 0<s\leq 2m)$ are eventually fixed points for T, s.d. $T^n(s/2^m) = p$. But I'm not sure that this is of any help here.

Any input is much appreciated!

Answer 1

In case onyone ever comes across this question again, here are some answers:

1. There was a typing error in the script. It was supposed to say $T_0$ only has one fixed point. Which is obviously true, because $T_0 (x) = 0$. Thus $T_0(0) = 0$ is the only fixed point. The second part of the statement is correct.
2. Let $\mathrm(F)$ denote the set of fixed points for a function on an interval. Then, $$\mathrm(F)(T_h) \subset \mathrm(F)(T_1)$$ on the interval $[0,h)$. Notice that this only valid for the half open interval, since $T_h(h) = h$ is a fixed point for every $T_h$, but not for $T_1$. Likewise, $$\mathrm(F)(T_1) \subset (F)(T_h)$$ on $[0,h]$.