This is a homework question I have been crunching my brains on for a lot of time, but unfortunately I'm stuck. I would greatly appreciate any help!
The problem is as follows:
We have some continuous map $f: A \rightarrow A$ of a compact metric space $(X, d)$. Denote $\tilde{A}$ the space $\{a \in A^{Z} | f(a)_n = (a)_{n+1} \}$.
Then we can extend $f$ with $ \tilde{f} : \tilde{A} \rightarrow \tilde{A}$, where $\tilde{f}(a_{n}) = a_{n+1}$ for all $n$. (so a shift)
$\tilde{A}$ has metric $\tilde{d}(a, b) = \sum_{Z} 2^{|-i|}d(a_i, b_i)$
Write $\pi : \tilde{A} \rightarrow A$ for the map sending $a$ to $a_0$.
Now show that $\pi_* M(\tilde{f}(\tilde{A})) = M(f(A)$, where $M$ stands for the set of Borel probability measures on $A$.
my thoughts:
to start, I understand that $\pi$ is a topological factor map and that $\tilde{A}$ is some kind of an orbit space. Also I think that I have to construct a measure on $\tilde{A}$ given a measure on $A$that is compatible with $\pi$
So I thought this would be a good cylinder-like construction.
Let $\mu$ be some measure on $A$. Now let $B \subset \tilde{A}$. Then define measure $\tilde{\mu}(B) := \prod_Z \mu(\cup_{b\in B}(b_i))$.
because this measure is I think a probability measure, and also it is easy to go from a measure of this form to a measure on $A$.
But how can I possibly show that measures are of such a product form? or is that even true? And how can I now show that this is an actual measure on all open sets? Now I only know how it behaves on cylinders.
Or don't I even need an explicit construction?
Please let me know if you have any thoughts, directions, hints etc!
Thanks and kind regards,
Derk van Willekenburg