Q:

$f^{n_i}(x)\to y$ implies $f^{-n_i}(y)\to x$?

Let $(X, d)$ be a compact metric space and $f:X\to X$ be a homeomorphism. If there exists a sequence $n_i$ such that $n_i\to\infty$ as $i\to\infty$ and $x, y\in X$ are such that $f^{n_i}(x)\to y$ as $i\to\infty$.Can we conclude that $f^{-n_i}(y)\to x$ as $i\to\infty$?

A:

No.
Consider $X=S^1\subset\mathbb C$ and $$f(z)=\frac{z+\frac12}{|z+\frac12|}$$ (that is we shift a bit to the right and project back to the circle).
Then $f^n(z)\to 1$ for all $z\in S^1$ except $-1$, but $f^{-n}(1)=1$ for all $n$.

