Let $M$ be a compact space and $f,g:M \to M$ whose non-wandering sets satisfy $\Omega(f)=\Omega(g)=M$. Can we have $\Omega(f \circ g)=M$?
Or more specifically, if $\Omega(f)=M$, can we have $\Omega(f^n)=M$ for any positive integral number $n$?
Any reference would be helpful. Thanks!
To the first question the answer is negative. There are two homeomorphisms of the circle with irrational rotation number such that their composition is Morse-Smale (in fact, you can multiply two $2\times 2$ matrices with complex eigenvalues to get one hyperbolic matrix, the action on the proyective space does the trick). This implies that $\Omega(f)=\Omega(g)=S^1$ but $\Omega(f\circ g)$ consits of two points.
To the last question, the answer is yes (and it is important that $\Omega(f)=M$), by that I mean $\Omega(f)=\Omega(f^m)=M$ for every $m\geq 1$.
The proof I know goes as follows:
Consider a basis $\{A_n\}$ of the topology of $M$ and $O_n=$ {$x\in A_n \ : \ f^j(x) \in A_n \ for \ some \ j \geq 1$ } $\cup \overline{A_n}^c$. The set $O_n$ is open (because $A_n$ is open) and dense since every point is non wandering (so, given an open set $U\subset A_n$ there exists $x\in U$ such that for $j\geq 1$ we have $f^j(x) \in U \subset A_n$). (Notice that if $\Omega(f)\neq M$ the set $O_n$ could fail to be dense in $\Omega(f)$).
If $x\in R=\bigcap O_n$ we get that $x$ is a limit point for $f$ (in fact, it will be recurrent since given a neighborhood $U$ of $x$, there is $x\in A_n \subset U$, and since $x\notin \overline{A_n}^c$ and $x\in O_n$ we get that it has a future iterate in $A_n$) and thus also for $f^m$. Notice that $x$ may a priori be not recurrent for $f^m$, but it will be a limit point.
Since the limit set is contained in the nonwandering, we obtain the result.
This aplies also to get that if $\Omega(f|_{\Omega(f)})=\Omega(f)$ then $\Omega(f)=\Omega(f^m)$.
I am wondering if there is an example $(f,X)$ and $n\ge2$ with $\Omega(f)\neq\Omega(f^n)$. It is interesting to know such examples.
There is an observation that for a homeo $f:X\to X$, if $x\in\omega(x,f)$, then $x\in\omega(x,f^n)$ for each $n\ge1$. The proof is:
Let $n\ge2$ be given. Note that $\omega(x,f)=\bigcup_{0\le k< n}\omega(f^kx,f^n)$. So there exists $k$ with $x\in\omega(f^kx,f^n)$.
If $k=0$ we are done.
Otherwise let $l=n-k\in[1,n-1]$. Then $f^lx\in\omega(x,f^n)$. We show inductively $f^{jl}x\in\omega(x,f^n)$ for each $j\ge1$. Since $\omega(x,f^n)$ is strictly $f^n$-invariant and $f^{nl}x\in\omega(x,f^n)$, we get $x\in\omega(x,f^n)$, too.
$f^{(j+1)l}x=f^l(f^{jl}x)\in f^l\omega(x,f^n)=\omega(f^lx,f^n)\subset\omega(x,f^n)$, where $\in$ is from induction hypothesis and $\subset$ is from the forward invariance of $\omega$-sets.
