If $I = (I, ≤)$ and $J = (J,≤)$ are linearly ordered sets, does $f ≥ g$ awalys imply $f^{-1} ≤ g^{-1}$ for invertible $f, g \colon I → J$?
This is easy to prove if $f$ and $g$ are assumed to be monotone or at least piecewise monotone. I also think it’s true for finite sets $I$ and $J$, because $f ≥ g$ seems to imply $f = g$ in this case.
How about the general case?
Let $I$ be the set of odd numbers and $J$ — even numbers.
$$f(x)=\begin{cases} 2 & x=-1\\ 0 & x=1\\ x+1 & x\not\in \{-1,1\}, \end{cases}$$
$$g(x)=-f(-x)=\begin{cases} 0 & x=-1\\ -2 & x=1\\ x-1 & x\not\in \{-1,1\}. \end{cases}$$
$\forall x\in I\; f(x)\ge g(x)$ but $f^{-1}(0)> g^{-1}(0)$.