Proof of the Curtis-Hedlund Theorem: Why is there a function $\mu\colon A^S\to A$ such that $\tau(x)(1_G)=\mu(x_{|S})$ for all $x\in A^G$?

Question

Here is the Curtis-Hedlund Theorem and its proof [the sets $V(\cdot,\cdot)$ used in this proof are explained below.]:

My problem is I am not sure that I have understand that correctly. So I paraphrase it and rewrite it in my words in order that you can see if I got it. Could you please tell me if I got it?

Here is how I do understand the proof:

Let $x\in A^G$ be arbitrary. $\varphi$ is continious, meaning that for the neighbourhood $U=\left\{\varphi(x)\right\}$ in $A$, there is a neighbourhood $V$ of $x$ in $A^G$ such that $\varphi(V)\subset U$. Because the sets $V(\cdot,\cdot)$ as defined below are a neighbourhood base, there is a finite $\Omega_x\subset G$ such that $V(x,\Omega_x)\subset V$. So for $y\in V(x,\Omega_X)$ (meaning $y_{|\Omega_x}=x_{|\Omega_x}$) it is $\varphi(y)\in U$, i.e. $\varphi(y)=\varphi(x)$. Keep this in mind. By compactness it is $A^G=\bigcup_{x\in F}V(x,\Omega_x)$ for a finite $F\subseteq A^G$. Set $S:=\bigcup_{x\in F}\Omega_x$.

I think I did understand it up to here?

The next part is not that clear to me; anyway, here is how I see it:

We want to show that there is a function $\mu:\colon A^S\to A$ such that $$ \varphi(x)=\mu(x_{|S})~\forall~x\in A^G. $$ I think this means we have to show that for any configuration $x\in A^G$ the value $\varphi(x)$ only depends on the restriction of $x$ on $S$. So the author of that proof above takes two configurations $y,z\in A^G$, supposes that they coincide on $S$ and shows that for the resulting values it is $\varphi(y)=\varphi(z)$ (using that argument from above that if $y$ and $z$ coincide on the finite set $\Omega_{x_0}$ then $\varphi(y)=\varphi(x_0)=\varphi(z)$).

I ask myself: Isn't it a restriction to suppose that $y$ and $z$ coincide on $S$?

Maybe you can say me if my understanding of the proof is allright.

---

Notation:

Answer 1

I think you got it mostly right. The author wants to show that $S$ is the memory set of the local map $\mu$ of the cellular automaton $\tau$.

To prove existence of the local function $\mu$ it is enough to show that any two configurations $y,z$ which coincide in the finite set $S$ will produce the same symbol at coordinate $1_G$ when the map $\phi$ is applied. This then allows to prove the local map $\mu$ is in fact well-defined.

With this procedure you get the local map which produces the symbol at coordinate $1_G$ and here you basically used compactness of the space and continuity of the map. To extend this to a local map you use $G$-equivariance, i.e. the same map is used everywhere to produce the symbols at all other coordinates.