Topological conjugacy between dyadic map and tent map 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!
 A: 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.


*The answer depends on the precise topological model you choose.

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.
