Attracting fixed point of $f$ if and only if repelling fixed point of $f^{-1}$

While working on some dynamical system problems, I came across an interesting exercise that has left me stumped for quite a few weeks now. I have a solution, but I don't think it holds enough water yet, so I was hoping to get some feedback by posting it here. First, some (perhaps strange) definitions (for simplicity we will assume that everything is in regards to locally compact metric spaces):

A fixed point $p$ of a continuous map $f$ is called topologically attracting if it has a neighbourhood $U$ in which the iterates $f^n$ are all defined on $U$ and the sequence $\{f^n|_U\}$ converges uniformly to the constant map $U \to \{p\}$.

A fixed point $p$ of a continuous map $f$ is called topologically repelling if it has a neighbourhood $U$ so that for all $x \neq p$ in $U$, there exists some $n \geq 1$ such that $f^n(x) \notin U.$ Hence, the only infinite orbit that completely lies in $U$ is the orbit of the fixed point itself.

The problem that I am attempting is as follows: Let $f: (X, d) \to (X, d)$ be a continuous map on a locally compact metric space with fixed point $p$. Suppose that $f$ maps a compact neighbourhood $K$ of $p$ homeomorphically onto a compact neighbourhood $K'$. Then the restriction $f: K \to K'$ is topologically repelling at $p$ if and only if its inverse $g: K' \to K$ is topologically attracting at $p$.

I believe I have the forward direction, but the backwards direction seems to escape me. This is what I have tried (feedback is much appreciated)!

($\impliedby$) Suppose that $g$ is topologically attracting at $p$, and suppose for a contradiction that $f$ is not topologically repelling. Because $g$ is topologically attracting, there exists a compact neighbourhood $N' \subseteq K'$ of $p$ such that the iterates $\{g^n\}$ converge uniformly on $N'$ to the constant map $N' \to \{p\}$. Since $f$ is not topologically repelling, then for every compact neighbourhood $N \subseteq K'$ of $p$, there exists $x_0 \neq p$ in $N$ such that for all $n \geq 1$, $f^n(x_0) \in N$. Take $N = N'$. Then after passing to a subsequence, we may assume that $f^n(x_0)$ converges to some limit $\hat{x}_0 \in N'$, for $N'$ is compact in a metric space.

Next, fix $n \in \mathbb{N}$. Then as the iterates $g^m$ uniformly converges, we have \begin{equation*} \lim_{m \to \infty} d(g^m(f^n(x_0)), p) = 0. \end{equation*} Moreover, we may interchange limits to obtain \begin{align*} 0 ={} & \lim_{n \to \infty} \lim_{m \to \infty} d(g^m(f^n(x_0)), p) \\ ={} & \lim_{m \to \infty} \lim_{n \to \infty} d(g^m(f^n(x_0)), p) \\ ={} & \lim_{m \to \infty} d(g^m(\hat{x}_0), p), \end{align*} where the last equality follows from continuity of $g^m$. Hence, for sufficiently large $m$, we have $g^m(\hat{x}_0) = p = g^m(p)$, which by injectivity implies that $\hat{x}_0 = p$. Thus, $f^n(x_0) \to p$ as $n \to \infty$. But since $p$ is a fixed point of $f$, we also have that $f^n(x_0) = f^n(p)$ for sufficiently large $n$. Therefore, $x_0 = p$ by injectivity of $f$, a contradiction.

(I have major misgivings about the last paragraph here...)

Solution 1:

I don't see an obvious reason for why you can interchange limits. Even if you can, seems like in your last steps you used the fact that $f^n(x_0)$ converging to a fixed point implies that it is eventually equal to this fixed point, which is not necessarily true.

But we can get $\lim f^n(x) = p$ in another way (no subsequences needed). Indeed, if that was not the case, there exists $\varepsilon >0$ and a subsequence $n_k$ such that $$ d(f^{n_k}(x),p) \geq \varepsilon, \ \forall k $$ but there for large enough $m$, $g^m(N') \subset B_{\varepsilon}(p)$, in particular for large $k$, $$ g^{n_k-n_1}(f^{n_k}(x)) = f^{n_1}(x) \in B_{\varepsilon}(p) $$ which is a contradiction, therefore $\lim f^n(x) = p$.

This means that $x$ is a point such that $$ \underset{n \rightarrow \infty}{\lim}g^n(x) = p = \underset{n \rightarrow -\infty}{\lim} g^n(x) $$ such point cannot exist if $p$ is an attractor for $g$, indeed, take $r = d(x,p)$, for large $m$ we must have $$ g^m(N') \subset B_{\frac{r}{2}}(p) $$ but $g^m(g^{-m}(x)) = x$, which is a contradiction.

If you wish for a direct proof without the use of contradictions, we may show directly that every point in $N'$ eventually leaves by iterating $f$. Indeed, given $x \in N'$, take $r = d(p,x)$ and $m$ large enough in order to have $ g^m(N') \subset B_{\frac{r}{2}}(p)$, then $f^m(x)$ cannot be in $N'$, otherwise $g^m(f^m(x)) = x \notin B_{\frac{r}{2}}(p)$.