If $\bot$ is an operation on a set $X$ then we say that it is directly compatibile on the left with an order $\mathcal O$ on $X$ at any $x$ in $X$ if the inequality $$ x_1\prec_\mathcal Ox_2 $$ with $x_1$ and $x_2$ in $X$ implies the inequality $$ x\,\bot\,x_1\prec_\mathcal O x\,\bot\,x_2 $$ whereas we say that it is directly compatibile on the right if the first inequality implies the inequality $$ x_1\,\bot\,x\prec_\mathcal O x_2\,\bot\,x $$ Moreover, we say that $\bot$ is inversely compatibile on the left with $\mathcal O$ at $x$ if the inequality $$ x\,\bot\,x_1\prec_\mathcal O x\,\bot\,x_2 $$ implies the inequality $$ x_1\prec_\mathcal O x_2 $$ whereas we say that it is inversely compatibile on the right if the inequality $$ x_1\,\bot\,x\prec_\mathcal O x_2\,\bot\,x $$ implies the inequality $$ x_1\prec_\mathcal O x_2 $$ So, if $\bot$ is directly and inversely compatible on the left or on the right with $\mathcal O$ at $x$ then we say that it is simply compatible at $x$.
Finally, we say that $\bot$ is directly or inversely compatible on the left or on the right with $\mathcal O$ if it is directly or inversely compatibile on the left or on the right at each $x$ in $X$ so that we say simply that $\bot$ is directly or inversely compatible with $\mathcal O$ if it is directly or inversely compatibile with $\mathcal O$ at each $x$ in $X$.
So, I observed that if the inclusion $$ A\subseteq B $$ with $A$ and $B$ in $\mathscr P(X)$ holds then even the inclusion $$ A\cup X\subseteq B\cup X $$ holds so that by commutativity of $\cup$ I concluded that $\cup$ on $\mathscr P(X)$ is compatibile with $\subseteq$ at $X$ however it is not inversely compatible since if $B$ is empty and $A$ is $X$ then $A$ is not contained into $B$ if $X$ is not empty. So, I am searching now a counterexample showing that if $\bot$ is inversely compatibile (on the left or on the right) with $\mathcal O$ at any $x$ then it is not necessarily directly compatibile too: so, do this counterexample exists? perhaps inversly compatibile implies directly? Could someone help me, please?
N.B.
I point out that if $\mathcal O$ is totally then directly and inversely compatibility (on the left or on the right) at any $x$ are equivalent.
Here's another version.
In this case, $f$ is not surjective, but that is not one of the requirements, and this is a bit simpler.
Consider the poset (actually a lattice):
Define $f(0) = 0$, $f(1) = a_1$ and $f(a_i) = a_{i + 1}$ for all $i$.
If $x \neq 0$, then $f(0) < f(x)$, but $0 < x$ too;
if $x, y \neq 0$, then $f(x)\not< f(y)$.
Thus, $f(x) < f(y)$ implies that $x < y$, for all $x$ and $y$.
On the other hand, $a_1 < 1$ but $f(a_1) = a_2 \not< a_1 = f(1)$, and so $x < y$ doesn't imply $f(x) < f(y)$.
So if you define $x \bot y = f(x)$ then $\bot$ is inversely compatible but not directly compatible.