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Let $(X,T)$ be a minimal subshift. Can it happen that an endomorphism $\varphi\colon (X,T) \to (X,T)$ is almost 1-to-1 but not 1-to-1? Can it happen that a factor $\pi\colon (X,T) \to (Y,T)$ between minimal subshifts is almost 1-to-1 but not 1-to-1? I know that, for example, Toeplitz subshifts are almost 1-to-1 extensions of a Cantor system (some odometer), but in the subshift case I couldn't find anything on internet.

PD. Here, a factor $\pi\colon(X,T)\to(Y,T)$ is almost 1-to-1 if $\exists y \in Y$ such that $\#\pi^{-1}(y) = 1$. From this, one can prove that $\#\pi^{-1}(y) = 1$ for all $y$ in a residual set.

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  • $\begingroup$ @YCor I added the definition! $\endgroup$ Commented Apr 24, 2020 at 17:40
  • $\begingroup$ Isn't Downarowicz' royal couple an example of this? $\endgroup$
    – Ville Salo
    Commented Apr 24, 2020 at 18:10
  • $\begingroup$ IIRC only a special point contains a royal couple (and has multiple preimages), others have only normal couples and thus unique preimages. $\endgroup$
    – Ville Salo
    Commented Apr 24, 2020 at 18:14

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The answer is yes. 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.

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