Let $\mathfrak{A}$, $\mathfrak{B}$ be bounded posets.
The main question: Explicitly describe (and prove that it exists) the greatest Galois connection between $\mathfrak{A}$ and $\mathfrak{B}$ (as defined below).
Consider Galois connections between these posets. Order them by the formula $f\le g \Leftrightarrow f^\ast\le g^\ast$ (where $f^\ast$ is the lower adjoint for a Galois connection $f$).
I am to find the greatest of all Galois connections between $\mathfrak{A}$ and $\mathfrak{B}$. (This is my main question, how to describe this greatest connection.)
I conjecture that the greatest connection exists and its lower adjoint is defined by the formula:
$$f^\ast X = \left\{ \begin{array}{ll} \bot^{\mathfrak{B}} & \text{if } X = \bot^{\mathfrak{A}}\\ \top^{\mathfrak{B}} & \text{if } X \neq \bot^{\mathfrak{A}} \end{array} \right.$$
($\bot$ denotes the least element of a poset and $\top$ the greatest one).
Please help to prove (or disprove) that the above defined $f^\ast$ is really a lower adjoint and help to find the corresponding upper adjoint.
With help of @darijgrinberg:
Suppose $f^\ast$ has upper adjoint $f_\ast$. Then it is defined by the formula $f_{\ast} Y = \max \left\{ X \in \mathfrak{A} \mid f^{\ast} X \leq Y \right\} = \max \left\{ \begin{array}{ll} \mathfrak{A} & \text{if } Y = \top^{\mathfrak{B}}\\ \{ \bot^{\mathfrak{A}} \} & \text{if } Y \neq \top^{\mathfrak{B}} \end{array} \right. = \left\{ \begin{array}{ll} \top^{\mathfrak{A}} & \text{if } Y = \top^{\mathfrak{B}}\\ \bot^{\mathfrak{A}} & \text{if } Y \neq \top^{\mathfrak{B}} \end{array} \right.$
Then the routine verification that they are indeed adjoint:
$f^\ast X \sqsubseteq Y \Leftrightarrow X = \bot^{\mathfrak{A}} \land Y = \top^{\mathfrak{B}} \Leftrightarrow X \sqsubseteq f_\ast Y$.