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A synchronized system is a transitive shift space $X$ which has a synchronizing block $v$, that is $v$ is an admissible block for $X$ and whenever $vw$ and $uv$ are admissible blocks in $X$, then $uvw$ is also admissible. Let $\Psi\colon X\to Y$ be a conjugacy of synchronized systems.

Since $\Psi$ is continuous and shift-commuting there is an $n\in\mathbb{N}$, a map from admissible $n$-bloks in $X$ to $1$-blocks in $Y$, denoted $\psi\colon \mathscr{B}_n(X ) \to \mathscr{B}_1(Y)$, and an $m \in\mathbb{Z}$ such that, in standard notation, $(\Psi \circ \sigma^m(x))_i = \psi(x_{[i,i+n)})$, for all $i \in \mathbb{Z}$ and all $x \in X$, cf. Lind and Marcus book. The map $\psi$ extends to maps $\psi^{(k)} \colon \mathscr{B}_k(X ) \to \mathscr{B}_{k-n+1}(Y)$, $k \ge n$, in the natural way.

Is it true that a $k$-block $v$ ($k\ge n$) is a synchronizing block for $X$ if and only if $\psi^{(k)}(v)$ is a synchronizing block for $Y$?

In general, a factor of a synchronized system may not be synchronized.

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  • $\begingroup$ It turns out that the answer to the above question is negative (see below). Still one can prove that for every synchronizing block $v$ for $X$ there is a block $w$ such that $vw$ is admissible in $X$ and $\psi^{(|vw|)}(vw)$ is synchronizing for $Y$. $\endgroup$
    – Dominik Kwietniak
    Commented Feb 11, 2014 at 22:18

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No! If so, you would get a contradiction by considering the length of the shortest synchronizing blocks of $X$.

Namely, suppose that the synchronizing blocks of $X$ all have length at least $2$. Let $v$ be a synchronizing block of $X$ having length $n\geq 2$. Choose a subshift $Y$ that is conjugate to $X$ via an $n$-block map $\psi$. (For example, $Y$ could be the $n$-th higher block presentation of $X$.) The inverse map from $Y$ to $X$ is induced by another block map $\varphi$, which has window size $m\geq 1$.

If the claim is true, $Y$ must have a synchronizing block $c=\psi(v)$ of length $1$. If $c$ is synchronizing for $Y$, so is every extension of $c$ to an admissible block. In particular, $Y$ has a synchronizing block $u$ of length $m$. Applying the claimed statement again we obtain that $X$ must have a synchronizing block $\varphi(u)$, which has length $1$, hence a contradiction.

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  • $\begingroup$ But it may be not true that all synchronizing blocks have length at least two - there are plenty of synchronizing shifts with a synchronizing symbol (one letter word)... Moreover, identity map is always a conjugacy and it clearly preserves the set of synchronizing words. If I understand your claim correctly, you have proved that if the minimal length of a synchronizing word is at least 2, then no $n$-block map for $n\ge2$ can preserve the set of synchronizing words. $\endgroup$
    – Dominik Kwietniak
    Commented Feb 11, 2014 at 15:19
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    $\begingroup$ I have shown how you could find a counter-example that refutes the statement "a $k$-block $v$ ($k\ge n$) is a synchronizing block for $X$ if and only if $\psi^{(k)}(v)$ is a synchronizing block for $Y$". For that, you only need one synchronizing shift $X$ whose synchronizing blocks are all of length at least $2$. An example would be the shift space of all sequences in $\{0,1,2\}^\mathbb{Z}$ in which between every two consecutive occurrences of $11$, there is at most one occurrence of $2$. $\endgroup$
    – Algernon
    Commented Feb 11, 2014 at 16:22
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    $\begingroup$ You are right, I only commented to clarify and better understand your statement, since your first sentence may be read as: It is never the case that... $\endgroup$
    – Dominik Kwietniak
    Commented Feb 11, 2014 at 22:14

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