[proofplan]
We prove the stronger cover-level statement that every finite open cover of $Y$ has the same entropy for $S$ as its pullback has for $T$. The conjugacy identity implies that pulling back by $\varphi$ commutes with all iterated inverse images appearing in the dynamical join. Since $\varphi$ is a surjective homeomorphism, a subcollection covers $Y$ exactly when its pullback covers $X$, so the minimal covering numbers agree for every iterate. Taking suprema over finite open covers on both spaces gives the equality of topological entropies.
[/proofplan]
[step:Pull finite open covers back through the conjugacy]
Let $\mathcal{V}$ be a finite open cover of $Y$. Define its pullback cover on $X$ by
\begin{align*}
\varphi^{-1}\mathcal{V} := \{\varphi^{-1}(V) : V \in \mathcal{V}\}.
\end{align*}
Because $\varphi: X \to Y$ is continuous, each $\varphi^{-1}(V)$ is open in $X$. Because $\varphi$ is surjective and $\mathcal{V}$ covers $Y$, the family $\varphi^{-1}\mathcal{V}$ covers $X$.
For a finite open cover $\mathcal{A}$ of a [compact space](/page/Compact%20Space), let $N(\mathcal{A})$ denote the least cardinality of a subcover of $\mathcal{A}$. This number is finite because the cover is finite.
[guided]
We begin with an arbitrary finite open cover $\mathcal{V}$ of $Y$ because topological entropy is defined by taking a supremum over such covers. The conjugacy map
\begin{align*}
\varphi: X \to Y
\end{align*}
allows us to transfer $\mathcal{V}$ to a cover of $X$ by taking inverse images:
\begin{align*}
\varphi^{-1}\mathcal{V} := \{\varphi^{-1}(V) : V \in \mathcal{V}\}.
\end{align*}
This is a finite open cover of $X$. It is finite because $\mathcal{V}$ is finite. It is open because $\varphi$ is continuous. It covers $X$ because for every $x \in X$, the point $\varphi(x)$ belongs to some $V \in \mathcal{V}$, and therefore $x \in \varphi^{-1}(V)$.
We also fix notation for minimal covering numbers. If $\mathcal{A}$ is a finite open cover of a compact space, then $N(\mathcal{A})$ denotes the smallest number of elements of $\mathcal{A}$ needed to cover the space. This is the quantity that appears in the open-cover definition of topological entropy.
[/guided]
[/step]
[step:Show that dynamical joins commute with pullback]
For each integer $n \geq 1$, define the $n$-fold dynamical joins
\begin{align*}
\mathcal{V}_S^{(n)} := \bigvee_{k=0}^{n-1} S^{-k}\mathcal{V}
\end{align*}
and
\begin{align*}
(\varphi^{-1}\mathcal{V})_T^{(n)} := \bigvee_{k=0}^{n-1} T^{-k}(\varphi^{-1}\mathcal{V}).
\end{align*}
Thus $\mathcal{V}_S^{(n)}$ consists of all sets of the form
\begin{align*}
V_0 \cap S^{-1}(V_1) \cap \cdots \cap S^{-(n-1)}(V_{n-1})
\end{align*}
with $V_0,\dots,V_{n-1} \in \mathcal{V}$.
The conjugacy relation $\varphi \circ T = S \circ \varphi$ implies by induction that
\begin{align*}
\varphi \circ T^k = S^k \circ \varphi
\end{align*}
for every integer $k \geq 0$. Hence, for every $V \in \mathcal{V}$ and every $k \geq 0$,
\begin{align*}
T^{-k}(\varphi^{-1}(V)) = \varphi^{-1}(S^{-k}(V)).
\end{align*}
Therefore
\begin{align*}
(\varphi^{-1}\mathcal{V})_T^{(n)} = \varphi^{-1}(\mathcal{V}_S^{(n)}).
\end{align*}
Here the right-hand side means the cover obtained by taking the inverse image under $\varphi$ of every member of $\mathcal{V}_S^{(n)}$.
[guided]
The key point is that the conjugacy identity persists under iteration. Since
\begin{align*}
\varphi \circ T = S \circ \varphi,
\end{align*}
an induction gives
\begin{align*}
\varphi \circ T^k = S^k \circ \varphi
\end{align*}
for every integer $k \geq 0$. This identity is exactly what is needed to compare inverse images. Fix $V \in \mathcal{V}$ and $k \geq 0$. For $x \in X$, we have
\begin{align*}
x \in T^{-k}(\varphi^{-1}(V)) \iff \varphi(T^k x) \in V \iff S^k(\varphi(x)) \in V \iff x \in \varphi^{-1}(S^{-k}(V)).
\end{align*}
Thus
\begin{align*}
T^{-k}(\varphi^{-1}(V)) = \varphi^{-1}(S^{-k}(V)).
\end{align*}
Now take intersections. An element of $(\varphi^{-1}\mathcal{V})_T^{(n)}$ has the form
\begin{align*}
\varphi^{-1}(V_0) \cap T^{-1}(\varphi^{-1}(V_1)) \cap \cdots \cap T^{-(n-1)}(\varphi^{-1}(V_{n-1}))
\end{align*}
with $V_0,\dots,V_{n-1} \in \mathcal{V}$. Using the inverse-image identity term by term, this set equals
\begin{align*}
\varphi^{-1}(V_0) \cap \varphi^{-1}(S^{-1}(V_1)) \cap \cdots \cap \varphi^{-1}(S^{-(n-1)}(V_{n-1})).
\end{align*}
Because inverse images commute with finite intersections, this is
\begin{align*}
\varphi^{-1}(V_0 \cap S^{-1}(V_1) \cap \cdots \cap S^{-(n-1)}(V_{n-1})).
\end{align*}
These are exactly the inverse images under $\varphi$ of the elements of $\mathcal{V}_S^{(n)}$. Therefore
\begin{align*}
(\varphi^{-1}\mathcal{V})_T^{(n)} = \varphi^{-1}(\mathcal{V}_S^{(n)}).
\end{align*}
[/guided]
[/step]
[step:Compare the minimal covering numbers for every iterate]
Let $\mathcal{C}$ be any finite open cover of $Y$. Since $\varphi$ is surjective, a subcollection $\mathcal{C}_0 \subset \mathcal{C}$ covers $Y$ if and only if the subcollection
\begin{align*}
\varphi^{-1}\mathcal{C}_0 := \{\varphi^{-1}(C) : C \in \mathcal{C}_0\}
\end{align*}
covers $X$. Hence
\begin{align*}
N(\varphi^{-1}\mathcal{C}) = N(\mathcal{C}).
\end{align*}
Applying this to $\mathcal{C} = \mathcal{V}_S^{(n)}$ and using the previous step gives
\begin{align*}
N((\varphi^{-1}\mathcal{V})_T^{(n)}) = N(\mathcal{V}_S^{(n)})
\end{align*}
for every integer $n \geq 1$.
[/step]
[step:Pass from cover entropy to topological entropy]
By the open-cover definition of topological entropy for a finite open cover,
\begin{align*}
h_{\mathrm{top}}(T,\varphi^{-1}\mathcal{V}) = \lim_{n \to \infty} \frac{1}{n}\log N((\varphi^{-1}\mathcal{V})_T^{(n)})
\end{align*}
and
\begin{align*}
h_{\mathrm{top}}(S,\mathcal{V}) = \lim_{n \to \infty} \frac{1}{n}\log N(\mathcal{V}_S^{(n)}).
\end{align*}
The existence of these limits is part of that definition: it follows from the standard subadditivity of the logarithmic minimal covering numbers and Fekete's lemma.
The equality of minimal covering numbers for every $n \geq 1$ gives
\begin{align*}
h_{\mathrm{top}}(T,\varphi^{-1}\mathcal{V}) = h_{\mathrm{top}}(S,\mathcal{V}).
\end{align*}
Since $\mathcal{V}$ was an arbitrary finite open cover of $Y$, taking the supremum over all such $\mathcal{V}$ yields
\begin{align*}
h_{\mathrm{top}}(S) \leq h_{\mathrm{top}}(T).
\end{align*}
Apply the same argument to the inverse homeomorphism
\begin{align*}
\varphi^{-1}: Y \to X.
\end{align*}
The conjugacy identity $\varphi \circ T = S \circ \varphi$ is equivalent to
\begin{align*}
\varphi^{-1} \circ S = T \circ \varphi^{-1}.
\end{align*}
Therefore the same cover-level comparison gives
\begin{align*}
h_{\mathrm{top}}(T) \leq h_{\mathrm{top}}(S).
\end{align*}
Combining the two inequalities,
\begin{align*}
h_{\mathrm{top}}(T) = h_{\mathrm{top}}(S).
\end{align*}
This proves that topological entropy is invariant under topological conjugacy.
[/step]
