# Almost one-to-one endomorphism of minimal subshift?

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.

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

## 1 Answer

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.