For trying to prove that the tent map
$$T(x)=
\begin{cases}
2x &\text{ if } x\in[0,\frac{1}{2}]\\
2-2x &\text{ if } x\in[\frac{1}{2},1]
\end{cases}
$$
is ergodic, I have already shown that the dyadic map (period-doubling map) given by $E(x)=2x$ mod $1$ is ergodic using a Fourier transform of $f$ and comparing coefficients to show that any measurable $f:X\to\mathbb{R}$ with $\ f \circ E = f$ almost everywhere implies $f$ is constant almost everywhere.
Anyways, to show that the tent map is ergodic, I tried topological conjugation with the dyadic map (which is ergodic); I've done the following:
Lemma.
The tent map $T$ is topologically semi-conjugate to the dyadic map $E(x)=2x$ mod 1.
Proof.
Let $E: [0,1] \to [0,1]$ be the dyadic map $E(x) = 2x$ mod 1. Let $T$ be the tent map as before.
Let $\varphi: [0,1] \to [0,1]$ also be the tent map, the same as $T$; i.e. $\varphi\equiv T$. Since
\begin{align*}
\varphi\circ E(x)=T(2x\text{ mod 1})
&=\begin{cases}
2(2x\text{ mod 1})\ &\text{ if }0\leqslant2x\text{ mod } 1\leqslant\frac{1}{2}\\
2-2(2x\text{ mod 1})\ &\text{ if }\frac{1}{2}\leqslant2x\text{ mod } 1\leqslant1
\end{cases}\\
&=
\begin{cases}
4x\ &\text{ if }x\in[0,\frac{1}{4}]\cup[\frac{1}{2},\frac{3}{4}]\\
2-4x \ &\text{ if }x\in[\frac{1}{4},\frac{1}{2}]\cup[\frac{3}{4},1]
\end{cases}\\
&=T^2(x)=T\circ\varphi(x), %%Do not change, this is best way to write down, I noticed by trial and error
\end{align*}
we have that $\varphi\circ E = T\circ\varphi$; i.e. $T$ is a factor of $E$ (or $E$ is an extension of $T$). $\Box$.
I know that this is only semi-conjugation for $\varphi$ is not invertible. I think the argument for ergodicty does not go wrong only using semi-conjugation, but I would like to have "full" conjugation. This is the ergodicity argument assuming ergodicity of $E$:
Theorem.
The tent map $T$ is ergodic.
Proof.
Let $A$ be an invariant set in $[0,1]$ for $T$; i.e. $T^{-1}(A)=A$. Since $\varphi\circ E = T\circ\varphi$ with $\varphi,\ E$ and $T$ as in the lemma above, it follows that
\begin{align*}
(\varphi\circ E)^{-1}&=(T\circ\varphi)^{-1}\\ E^{-1}\circ\varphi^{-1}&=\varphi^{-1}\circ T^{-1}
\end{align*}
which, after plugging in $A$, gives
\begin{equation*}
E^{-1}(\varphi^{-1}(A))=\varphi^{-1}(T^{-1}(A))=\varphi^{-1}(A);
\end{equation*}
so $\varphi^{-1}(A)$ is invariant for $E$. Now since $E$ is ergodic, we have that $\varphi^{-1}(A)$ has zero or full Lebesgue measure. As $\varphi=T$ (and $\varphi^{-1}=T^{-1}$), we have that $\varphi^{-1}(A)=T^{-1}(A)=A$
and hence also $A$ has zero or full Lebesgue measure; i.e. $T$ is ergodic.$\Box$.
Question 1A: Is the proof of the theorem correct?
Question 1B: Does this last theorem proof that ergodicity is preserved under topological semi-conjugation?
Question 2: How does one prove topological conjugation between the tent map and the dyadic map?
Thanks in advance for time and help!
Best Answer
1.A. The proof is correct, although in the computation of $\varphi \circ E(x)$, the cases $1/2\leq x\leq 3/4$ and $x\geq 3/4$ were forgotten.
1.B. Yes, ergodicity of a system implies ergodicity of its factors. However, it must be pointed out that 'factor' must be understood as factor as for a measure preserving transformation (not topological semi-conjugacy): a system $(Y,\nu,S)$ is a factor of $(X,\mu,T)$ if there is a map $f:X\to Y$, measurable, such that $f_*\mu=\nu$, and $f\circ T=S \circ f$.
As a matter of fact, in your proof, you did not use the fact that $\varphi$ is continuous. However, at the end, you used the specific of the situation ($\varphi=T$) to conclude, but could easily have used the fact that $Leb(\varphi^{-1}A)=Leb(A)$, i.e. $\varphi_*(Leb)=Leb$, instead, and that would have been the general proof of the statement that a factor map is ergodic if the extension is.
model $E_1$ : $E_1:[0,1]\to [0,1]$,
model $E_2$ : $E_2:\mathbb{R}/\mathbb{Z} \to \mathbb{R}/\mathbb{Z}$,
model $T_1$ : $T_1:[0,1]\to [0,1]$,
model $T_2$ : $T_2:\mathbb{R}/\mathbb{Z}\to \mathbb{R}/\mathbb{Z}$.
$E_1$ and $T_2$ cannot be topologically conjugate as a circle is not homeomorphic to a closed interval. Same for $T_1$ and $E_2$. $E_1$ cannot be conjugate to $T_1$, because $T_1$ is continuous but $E_1$ is not. $E_2$ and $T_2$ cannot be conjugate, because the degree of $E_2$ is two, and the degree of $T_2$ is zero.
So unless one considers another topological model for these maps, they do not seem to be topologically conjugated.