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Let $f:X\to X$ and $g:Y \to Y$ be two homeomorphisms with $X$ and $Y$ compact metric spaces. Suppose there is a continuous surjective function $h:X \to Y$ which satisfies $h \circ f = g \circ h$.

A function like this is called a semiconjugation. I am trying to see that $h(\Omega(f)) = \Omega(g)$, being $\Omega(f)$ the non-wandering set of $f$ defined by $\{x \in X: \forall U \text{ open }, \text{with } x \in U, \exists \ n>0 \ s.t \ f^n(U)\cap U \neq \emptyset \}$.

It is easy to see that $h(\Omega(f)) \subset \Omega(g)$ but I can't get the other inclusion. Indeed if $Y$ doesn't have enough properties I could see this is false.

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  • $\begingroup$ Just a guess, but, it would seem to hinge on obvious equality $h\circ f^n=g^n\circ h.$ $\endgroup$ May 26 '21 at 1:06
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    $\begingroup$ @ThomasAndrews thanks! it was a typo. I edited the question to clarify that. Also, I had forgotten to mention that $h$ is surjective and continuous. $\endgroup$
    – HFKy
    May 26 '21 at 1:16
  • $\begingroup$ Yeah, if true, you’ll definitely need compactness. $\endgroup$ May 26 '21 at 2:03
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An idea, too long for a comment.

For any $y\in X,$ $h^{-1}(y)$ is closed, and hence compact subset of $X.$

If $y\in \Omega(g),$ if you want $y\notin h(\Omega(f)),$ you need all $x\in h^{-1}(y)$ to have a neighborhood $U_x$ such that $f^n(U_x)\cap U_x=\emptyset$ for all $n.$

Since it is compact, $f^{-1}(y)$ must have a finite sub-cover, $V_1=U_{x_1},\dots,V_m=U_{x_m}.$

Not sure where to go from there, because $f^n(V_i)$ and $f^m(V_j)$ are are not necessarily disjoint.

But if your statement is true, it requires compactness of $X$ - it’s easy to come up with examples $X=Y\times \mathbb R$ with $f(y,r)=(g(y),r+1)$ where $\Omega(X)=\emptyset.$

There is a homeomorphism $\phi:S^1\to S^1$ with one fixed point $1.$ You can show $\Omega(\phi)=\{1\}.$ Then $X=S^1\times Y$ with $f:(x,y)\mapsto(\phi(x),g(y))$ has only one point $x\in f^{-1}(y)$ such that $x\in\Omega(f)$ for every $y\in \Omega(g).$

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  • $\begingroup$ Yeah, I explored this idea but cant manage to do it. Another similar idea is considering a neighborhood $U$ of $f^{-1}(y)$. Hence $U^c$ is closed, hence $h(U^c)^c$ is open and since $h(U)\supset h(U^c)^c$ then $h(U)$ is a neighborhood of $y$. If $y \in \Omega(g)$ then there is a natural number $n>0$ with $g^n(h(U))\cap U \neq \emptyset$, but not sure where to go from there $\endgroup$
    – HFKy
    May 26 '21 at 2:48
  • $\begingroup$ It requires even more than just compactness. To see this, if $X=S^1$ and $f$ a dynamical system with just one fixed point, and if $Y = S^1/\sim$ with $x \sim y$ iff $y = f^n(x) $ for some $n$, then taking $g: S^1/\sim \to S^1/\sim$ the identity gives a counterexample. What I suspect is that $Y$ is not metrizable. $\endgroup$
    – HFKy
    May 26 '21 at 2:51
  • $\begingroup$ At least with the $f$ I was thinking of, $X/\sim$ isn’t Hausdorff. Every point is in any neighborhood of the fixed point. $\endgroup$ May 26 '21 at 4:26
  • $\begingroup$ Basically: $f:e^{2\pi i t }\mapsto e^{2\pi i \sqrt t }$ was the example I was thinking of. So $f^n(x)\to 1$ for all $x$ $\endgroup$ May 26 '21 at 4:30
  • $\begingroup$ You are right that $X/\sim$ is not hausdorff, I was thinking about the same dynamical system. Now, what about your last sentence in your answer? Is a good observation but does not resolve the problem, does it? $\endgroup$
    – HFKy
    May 26 '21 at 12:31
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By contradiction, suppose $h(\Omega(f))\subsetneq \Omega(g)$. We can assume $Y = \Omega(g)$ since $K = h^{-1}(\Omega(g))$ is compact and $f-$invariant, so we have $h:K\to \Omega(g)$ surjective continuous function which satisfies $hf = gh$.

Now since $h(\Omega(f))$ is closed in $X$ we can find an element $w$ and an open set $U$ containing $w$ which does not intersect $h(\Omega(f))$. Since there is a dense subset of recurrent points (because $Y = \Omega(g)$), this shows we can find an element $y \in U$ such that $y$ is recurrent, i.e, there exists $n_k \to +\infty$ such that $g^{n_k}(y) \to y$. Take $x$ any preimage of $y$. Taking subsequences we can assume that $f^{n_k}(x)$ is convergent to some element $z$. Since $h$ is a semi-conjugacy then $h(z)=y$. But $z \in \omega(x)$, then $z \in \Omega(f)$ which is absurd because $y \notin h(\Omega(f))$.

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  • $\begingroup$ It is not necessarily true that $\Omega(g|_{\Omega(g)}) = \Omega(g)$. Similarly you don't know that $\Omega(f|_K) = \Omega(f)$. So, for example, if you assume $Y=\Omega(g)$ then it is not necessarily true that $\Omega(g) = Y$, and you cannot conclude that recurrent points are dense. $\endgroup$
    – koro
    Jun 2 '21 at 18:35

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