# Is the composition of non-wandering maps still non-wandering?

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!

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I think you mean to ask, "Must we have..." -- of course we "can" have the situation you describe. –  Vaughn Climenhaga Jul 1 '10 at 13:28
In your second paragraph, do you mean "for all positive integral numbers $n$" or "for some positive integral number $n$"? –  Mariano Suárez-Alvarez Jul 1 '10 at 17:50
@Climenhaga: Sorry, English is not my native language. I may have some mistakes when expressing the meaning. What you said is exactly that I want to ask. Thanks for correcting my mistake. @Su\'arez-alvarez: I mean "for all positive integral numbers". –  X.M. Du Jul 2 '10 at 9:47

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 $\Omega(f)=\Omega(f^m)$.

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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.

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