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My question is as follows.

QUESTION. Is there a topological description of the class of arc-like continua that arise as inverse limits of $[0,1]$ with a single continuous surjective bonding map $f\colon [0,1]\to [0,1]$ that satisfies $f(x)<x$ for $0<x<1$?

Background and further details. Let $f$ be a function as above; recall that its inverse limit $X$ is the set of all backward orbits under $f$, endowed with the product topology. Let $x_0,x_1\in X$ be the backward orbits $(0,0,0,\dots)$ and $(1,1,1,\dots)$ (observe that $0$ and $1$ are both fixed by $f$ from the assumption). Observe also that the one-sided shift $\sigma\colon X\to X$ is a homeomorphism.

Such continua appear naturally in recent work on invariant continua in the Julia sets of certain entire functions. It also turns out that they were studied in a paper of Rogers in 1970 (Decomposable inverse limits with a single bonding map on [0,1] below the identity, Fund. Math. 1970, eudml); hence I shall refer to them as Rogers continua. By a result of Henderson, the pseudo-arc is a Rogers continuum.

By definition, $X$ is arc-like (an inverse limit of arcs). Furthermore,

Property (S). There are terminal points $x_0$ and $x_1$ and a homeomorphism $\sigma\colon X\to X$ such that $X$ is irreducible between $x_0$ and $x_1$, such that $\sigma$ fixes $x_0$ and $x_1$, and such that $\sigma^n\to x_1$ uniformly on compact subsets of $X\setminus x_0$.

(Here $x$ is terminal if the subcontinua of $X$ containing $x$ are ordered linearly by inclusion, and $X$ is irreducible between $x_0$ and $x_1$ if there is no proper subcontinuum containing both.)

Rogers' paper establishes the following partial converse.

Theorem. If $X$ is a decomposable arc-like continuum with Property (S), then $X$ is a Rogers continuum.

This suggests the question:

Question 1. Is there an indecomposable arc-like continuum with Property (S) such that $X$ is not Rogers?

If not, then the answer to my original question above is positive. Otherwise:

Question 2. Is there a "topological" property (S') similar to (S) such that an arc-like continuum is Rogers iff it satisfies (S')?

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