Many thanks for your detailed reply!
I have a questions about :
How do we compare the elements of the empty set to any other elements?
Is the above statement is true because all \\( q \in Q \\) satisfy \\( q \le q' \\) for any \\( q' \in \emptyset \\) ?
Using this logic I could also say that all \\( q \in Q \\) satisfy \\( q' \le q \\) for any \\( q' \in \emptyset \\) ,
which means all \\( q \in Q \\) are \\( q \equiv q' \\) for any \\( q' \in \emptyset \\), so using transitivity of equivalence,
we have that all \\( q \in Q \\) are equivalent, isn't it?
Another question arises when we take another preorder, thanks to [Grant Roy](https://forum.azimuthproject.org/profile/1768/Grant%20Roy),
as our \\( P \\), Q is still \\( 1 \to 2 \to 3 \\) as above, and \\( g : Q \to P \\):
\\[ g(3) = d \\]
\\[ g(2) = b \\]
\\[ g(1) = a \\]
\\( g\\) is monotone, meet preserving, and even the empty set should find a meet in our new \\( P\\), which is \\(e\\).
The construction for a left adjoint would assign
\\( f(e) = \bigwedge_Q \emptyset = 1 \\)
and the Galois connection is not satisfied again?!
I'd be happy for your insights.