If $f:X\to Y$, and $A\subseteq X$, $B\subseteq Y$, then the equation $f[A] \cap B\subseteq f[A\cap f^{-1}[B]]$ holds. Indeed, let $y\in f[A]\cap B$, then $y=f(x)$ for some $x\in A$; since $f(x)\in B$ we also have $x \in f^{-1}[B]$ and hence $x\in A\cap f^{-1}[B]$, and $y\in f[A\cap f^{-1}[B]]$.
I would like to generalize this proof to bounded lattices. Here the forward and preimage maps $f$ and $f^{-1}$ are just monotone maps with a Galois connection $f(a)\le b\iff a\le f^{-1}(b)$. Is the theorem $f(a)\wedge b\le f(a\wedge f^{-1}(b))$ still true? If not, what extra property do sets bring to the table here?
The statement does not hold for arbitrary bounded lattices. If we construct the free lattice pair generated by a Galois connection like this, with one generator $b\in B$ and a Galois connection $f(a)\le b\iff a\le g(b)$, we obtain the structure:
$$\bot\le g\bot \le gb\le \top$$ $$\bot\le fgb\le b\wedge f\top\le b,\ f\top\le b\vee f\top\le \top$$
One can show using the Galois connection that $f(\bot)=f(g\bot)=\bot$, $g(\top)=g(f\top)=g(b\vee f\top)=\top$, and $g(fgb)=g(b\wedge f\top)=gb$, so these are all the points that are needed.
In this lattice already the claim does not hold: We have $f(\top)\wedge b>f(\top\wedge g(b))=fgb$. It is not clear to me if there are any axioms simpler than just $f(a)\wedge b\le f(a\wedge g(b))$ itself that imply it.