It is known that minimality does not imply unique ergodicity (Furstenberg example). I ask whether the implication holds in following particular situation:



> 
> Suppose $X$ is a compact space, $f:X \to X$ is a minimal and uniquely ergodic homeomorphism. Suppose $p: Y \to X$ is a **2-to-1 covering map**, and that $g:Y \to Y$ is a homeomorphism covering $f$, i.e., $p \circ g = f \circ p$. Suppose $g$ is minimal. *Does it follow that $g$ is uniquely ergodic?*
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Trivial example: If $f$ is an irrational rotation of the circle then it is easy to see that so is $g$, and so in this case the question has a positive answer.


A less general but already interesting situation is when the covering is trivial, i.e., $Y = X \times \{0,1\}$. In this case, I think that my question is equivalent to the following **coboundary rigidity** question: 



> 
> Let $f$ be as above. Let $\phi:X \to \mathbb{Z}\_2$ be a continuous map **on the group with two elements**. Suppose it is a measurable-coboundary, i.e., there exists a measurable map $\psi:X \to \mathbb{Z}\_2$ such that $\phi = \psi \circ f - \psi$ almost everywhere (w.r.t. the unique $f$-invariant probability measure). *Does it follow that $\phi$ is a continuous-coboundary?* (I.e., can $\psi$ be chosen continuous?)
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Rem.: To relate the two questions, notice that $g(x,t)=(f(x),t+\phi(x))$ is uniquely ergodic (resp., minimal) iff $\phi$ is not a measurable-coboundary (resp., not a continuous coboundary).



Getting unique ergodicity from minimality
I think there is a counter-example.


We'll build a subshift on 2 symbols, 0 and 1. Given a word $W$ with symbols 
0
and 1, $\overline W$ will denote the word with all symbols
flipped. Let $W\_0=1$ and $W\_{n+1}=W\_n^{2^n}\overline{W\_n}W\_n^{2^n}$.


The orbit closure of $W\_\infty$ gives a minimal, but not uniquely
ergodic subshift: The orbit closure of $W\_\infty$ is the same as the
orbit closure of $\overline{W\_\infty}$, but the frequency of 1's in
$W\_\infty$ is greater than $\frac 12$. Let $X$ be the subshift consisting of the orbit closure
and consider the map $\Phi\colon X\to \{0,1\}^\mathbb Z$ defined by
$\Phi(x)\_n=(x\_n+x\_{n+1})\bmod 2$. Under this map, $W\_\infty$ and
$\overline{W\_\infty}$ map to the same word.


The image is the subshift generated in the following way: Let $V\_0=0$
and for any word $v$, let $v^\*$ denote $v$ with the last digit
flipped. Then define $V\_{n+1}=V\_n^{2^n-1}V\_n^\*V\_n^\*V\_n^{2^n}$. Now
the image of $\Phi$ applied to a word starting $W\_n^k$ or
$\overline{W\_n}^k$ starts with $V\_n^k$ or $V\_n^{k-1}V\_n^\*$ depending
on the symbol that follows.


The image shift is therefore the orbit closure of $V\_\infty$. This is
uniquely ergodic because it's rank 1. 


