For $\mathcal R\subseteq X\times X^\prime$, $A\subseteq X$ and $B\subseteq X^\prime$ define:
$A\mathcal R=\{y\in X^\prime\mid \exists a\in A:(a,y)\in \mathcal R\}$
$\mathcal R B=\{x\in X\mid \exists b\in B:(x,b)\in \mathcal R\}$
Let $(X,\tau)$ and $(X^\prime,\tau^\prime)$ be finite topological spaces. Suppose that $\mathcal R $ is such that
$$A^\prime\subseteq X^\prime\implies\overline{\mathcal R A^\prime}\mathcal R
\subseteq \overline{A^\prime}\tag{1}$$
Does the corresponding then holds for $\mathcal R^{op}$, defined by $(y,x)\in\mathcal R^{op}\iff(x,y)\in\mathcal R$?
$$A\subseteq X\implies\overline{\mathcal R^{op} A}\mathcal R^{op} \subseteq \overline{A}\tag{2}$$
The only context I have so far, is that $(1)$ is a consistent extension of the definition of continuous functions on finite topological spaces. See A proof that continuity of $f:X\to Y$ is equivalent to $\overline{f^{-1}(M)}\subset f^{-1}(\overline{M})$.
