Semiconjugacy sends non wandering set onto non wandering set 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.
 A: 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 $y\in \Omega(g).$
A: 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 $y \notin h(\Omega(f))$.
