@Eldad Afik, that's a neat question! I wish this special case had occurred to me before, but even after trying to carefully read all the material I missed this one.

The formula for constructing \\(f\\) predicts that \\(f(b)=\bigwedge_Q \emptyset=\top_Q=3\\), but as you said that choice (and all other choices) fails to satisfy the Galois condition.

After walking through the proof much more carefully than I did before, I think I found the source of the problem. Although it appears that \\(g\\) "preserves all meets", this actually isn't the case. I believe the term "all meets" should be inspected in terms of the power set of \\(Q\\). For most subsets, i.e.

\[ \\{1\\}, \\{2\\}, \\{3\\}, \\{1,2\\}, \\{1,3\\}, \\{2,3\\}, \\{1,2,3\\}, \]

your map \\(g\\) sends each subset to \\(\\{a\\}\\), and the meet is preserved.

However, writing out the power set elements explicitly reveals that we left one out--the empty subset! "Preserve meets" means that we can transport the subset along \\(g\\) before or after taking the meet and get the same answer, \\(g(\bigwedge_Q S)=\bigwedge_P g_!(S)\\).

If \\(g\\) does "preserve meets" in the case of the empty set, we get

\[ g({\bigwedge}_Q \emptyset) = {\bigwedge}_P g_\!(\emptyset) = {\bigwedge}_P \emptyset \]

But the meet of \\(\emptyset\\) does not exist in \\(P\\)! (\\(P\\) has no top element.) So, if I'm looking at this in the correct way, the term "preserves all meets" doesn't just impose a condition on the map, but also a little universal condition on the codomain set: the codomain of the map must have a meet for the empty set. That was not at all obvious to me when the theorem just says that \\(P\\) is "any poset".

In your example, the codomain \\(P\\) has no top element, so no meet of \\(\emptyset\\), so it's impossible for *any* map into that poset to "preserve all meets".

The formula for constructing \\(f\\) predicts that \\(f(b)=\bigwedge_Q \emptyset=\top_Q=3\\), but as you said that choice (and all other choices) fails to satisfy the Galois condition.

After walking through the proof much more carefully than I did before, I think I found the source of the problem. Although it appears that \\(g\\) "preserves all meets", this actually isn't the case. I believe the term "all meets" should be inspected in terms of the power set of \\(Q\\). For most subsets, i.e.

\[ \\{1\\}, \\{2\\}, \\{3\\}, \\{1,2\\}, \\{1,3\\}, \\{2,3\\}, \\{1,2,3\\}, \]

your map \\(g\\) sends each subset to \\(\\{a\\}\\), and the meet is preserved.

However, writing out the power set elements explicitly reveals that we left one out--the empty subset! "Preserve meets" means that we can transport the subset along \\(g\\) before or after taking the meet and get the same answer, \\(g(\bigwedge_Q S)=\bigwedge_P g_!(S)\\).

If \\(g\\) does "preserve meets" in the case of the empty set, we get

\[ g({\bigwedge}_Q \emptyset) = {\bigwedge}_P g_\!(\emptyset) = {\bigwedge}_P \emptyset \]

But the meet of \\(\emptyset\\) does not exist in \\(P\\)! (\\(P\\) has no top element.) So, if I'm looking at this in the correct way, the term "preserves all meets" doesn't just impose a condition on the map, but also a little universal condition on the codomain set: the codomain of the map must have a meet for the empty set. That was not at all obvious to me when the theorem just says that \\(P\\) is "any poset".

In your example, the codomain \\(P\\) has no top element, so no meet of \\(\emptyset\\), so it's impossible for *any* map into that poset to "preserve all meets".