3
$\begingroup$

Let $\pi\colon(X,T)\to (Y,T)$ be a factor map between minimal subshifts. Suppose there exists $\tilde{Y} \subseteq Y$ such that

  1. $\# \pi^{-1}(y) = 1$ for all $y \in \tilde{Y}$.
  2. $\tilde{Y}$ is a residual subset of $Y$ i.e. $\tilde{Y} = \bigcap_{n=1}^\infty\tilde{Y}_n$ for some collection $\{\tilde{Y}_1,\tilde{Y}_2,\dots\}$ of dense open subsets of $Y$.
  3. $\mu(\tilde{Y}) = 1$ for every $T$-invariant probability measure in $Y$.

Is it true that $\pi$ is injective?

$\endgroup$
5
$\begingroup$

No. Consider an irrational rotation $R$ of the circle (which I identify with [0,1)) by an angle $\alpha$. Let $\alpha<\beta<1$ be a point not lying in the orbit of 0 under $R$. Set $A_1=[0,\alpha)$, $A_2=[\alpha,\beta)$, $A_3=[\beta,1)$ also $B_1=A_1$ and $B_2=A_2\cup A_3$. Consider the partitions $P=\{A_1,A_2,A_3\}$ and $Q=\{B_1,B_2\}$. For $x\in S^1$, let $\epsilon_1(x)\in \{1,2,3\}^{\mathbb Z}$ be such that $R^n(x)\in A_{\epsilon_1(x)_n}$ and similarly let $\epsilon_2(x)\in \{1,2\}^{\mathbb Z}$ be such that $R^n(x)\in B_{\epsilon_2(x)_n}$.

Let $X$ be the (minimal) subshift of $\{1,2,3\}^{\mathbb Z}$ consisting of the orbit closure of $\{\epsilon_1(x)\colon x\in [0,1)\}$ and $Y$ be the orbit closure of $\{\epsilon_2(x)\colon x\in [0,1)\}$. There is an obvious factor map from $X$ to $Y$ (replacing 3 symbols by 2's). This factor map is 1-1 off a countable set (corresponding to orbits passing through $\beta$).

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Nothing wrong with Anthony Quas's answer, and indeed the answer is "no", but I'll just remark that you already asked essentially the same question before, and the same answer works directly, with $X = Y$. Let me recall my answer to this question of yours.

"The answer is yes [the previous question was phrased differently]. In this paper Downarowicz proves the following theorem

There exists a regular Toeplitz sequence $\omega$ such that the induced Toeplitz flow $(\bar O(\omega), S)$ is noncoalescent, more precisely, it admits an endomorphism $\pi : \bar{O}(\omega) \to \bar{O}(\omega)$ of the first type.

Here, $\bar O(\omega)$ is the orbit closure (so a Toeplitz subshift since $\omega$ is Toeplitz), $S$ the shift map, noncoalescent means not injective, and first type means every Toeplitz point has a unique preimage, in particular some point does."

To see that this answers your question, set $X = Y$, and $\tilde Y$ the Toeplitz points of $Y$, so that every point in $\tilde Y$ has unique preimage. What we need is that the Toeplitz points are a residual set, and that their measure is $1$ in every invariant probability measure. Both facts are well-known. The first fact is mentioned e.g. on page 23 of this paper (it works for any Toeplitz subshift), and the second is Theorem 13.1. in the same reference (it works for any regular Toeplitz subshift).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.