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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?

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  • $\begingroup$ That result is certainly true if $f$ preserves meets, that is $$f(x\wedge y)=f(x)\wedge f(y).$$ The converse is also true, provided $f$ is injective. If the converse were true in all cases, that would be a characterization. But this might not be helpful at all. $\endgroup$ – amrsa Sep 17 '18 at 20:03
  • $\begingroup$ @amrsa Unfortunately, that doesn't seem to be true even in the case of sets - it is not true that $f[A\cap B]=f[A]\cap f[B]$, so it wouldn't do so well as a generalization! $\endgroup$ – Mario Carneiro Sep 17 '18 at 23:04
  • $\begingroup$ Yes, I understand it's not helpful. I just meant that if $f$ preserves meets, then $$f(a\wedge f^{-1}(b))=f(a)\wedge f(f^{-1}(b))=f(a)\wedge b.$$ I realised the content of your comment a couple of days ago when I was about to answer the question... $\endgroup$ – amrsa Sep 18 '18 at 7:58
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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.

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