# 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.

• Just a guess, but, it would seem to hinge on obvious equality $h\circ f^n=g^n\circ h.$ May 26 '21 at 1:06
• @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.
– HFKy
May 26 '21 at 1:16
• Yeah, if true, you’ll definitely need compactness. May 26 '21 at 2:03

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 $$y\in \Omega(g).$$

• 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
– HFKy
May 26 '21 at 2:48
• 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.
– HFKy
May 26 '21 at 2:51
• At least with the $f$ I was thinking of, $X/\sim$ isn’t Hausdorff. Every point is in any neighborhood of the fixed point. May 26 '21 at 4:26
• 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 $x$ May 26 '21 at 4:30
• 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?
– HFKy
May 26 '21 at 12:31

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 $$y \notin h(\Omega(f))$$.

• 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.
– koro
Jun 2 '21 at 18:35