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!



Is the composition of non-wandering maps still non-wandering?
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](http://en.wikipedia.org/wiki/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 $\Omega(f)=\Omega(f^m)$. 


