Following picture is from Ordered Sets and Complete Lattices by Hilary A. Priestley. My question is how to prove part (ii)?
I know that when $F_R=F$ it follows that $G_R=G$. So all is needed is to show that $F_R(A) = F(A)$, for every $A \subseteq \mathcal{P}(G)$. Assume $m \in F_R(A)$. Then $gRm$ for some $g \in A$, which means $m \in F(\{g\})$. How to get to that $m \in F(A)$?
More formally
$$ \begin{eqnarray} && F_R(A) \\ && \text{(by the definition of $F_R$)} \\ &=& \{ y \in Y \mid \exists a \in A, a R y \} \\ && \text{(by the definition of $R$)} \\ &=& \{ y \in Y \mid \exists a \in A, y \in F(\{a\}) \} \\ && \text{(by set theoretic union)} \\ &=& \{ y \in Y \mid y \in \bigcup_{a \in A} F(\{a\}) \} \\ && \text{(since left adjoints preserve colimits)} \\ &=& \{ y \in Y \mid y \in F(A) \} \\ && \text{(set theory)} \\ &=& F(A) \\ \end{eqnarray} $$
For the "set theoretic union" part we used the fact that $\exists a \in A, y \in F(\{a\})$ iff $y\in\bigcup_{a \in A} F(\{a\})$.
For the "left adjoints preserve colimits" part we have the argument that $$\begin{eqnarray} && F(X_1) \cup F(X_2) \subseteq Y \\[3mm] &\iff& F(X_1) \subseteq \phantom{G(}Y\phantom{)} \quad\land\quad F(X_2) \subseteq \phantom{G(}Y\phantom{)} \\ &\iff& \phantom{F(}X_1\phantom{)} \subseteq G(Y) \quad\land\quad \phantom{F(}X_2\phantom{)} \subseteq G(Y) \\[3mm] &\iff& \phantom{F(}X_1 \cup X_2\phantom{)} \subseteq G(Y) \\ &\iff& F(X_1 \cup X_2) \subseteq \phantom{G(}Y \\ \end{eqnarray}$$
and put $Y = F(X_1) \cup F(X_2)$ and $Y = F(X_1 \cup X_2)$ to deduce the equality $F(X_1) \cup F(X_2) = F(X_1 \cup X_2)$.
The same holds for an infinite union.
$$ \begin{eqnarray} && G_R(A) \\ && \text{(by the definition of $G_R$)} \\ &=& \{ x \in X \mid \forall y \in Y, xRy \implies y \in B \} \\ && \text{(by the definition of $R$)} \\ &=& \{ x \in X \mid \forall y \in Y, y \in F(\{x\}) \implies y \in B \} \\ && \text{(the logical definition of a subset)} \\ &=& \{ x \in X \mid F(\{x\}) \subseteq \phantom{G(}B\phantom{)} \} \\ && \text{(applying the Galois connection)} \\ &=& \{ x \in X \mid \phantom{F(}\{x\}\phantom{)} \subseteq G(B) \} \\ &=& \{ x \in X \mid x \in G(B) \} \\ &=& G(B) \\ \end{eqnarray} $$
For clarity I will use $X$ (instead of $G$) and $Y$ (instead of $M$) for the sets, $F$ and $G$ for the mappings. I will write $xRy$ as shorthand for $(x,y) \in R$.
Let $(F,G)$ be a Galois connection between $\mathscr{P}(X)$ and $\mathscr{P}(Y)$. So $$ \begin{eqnarray} F : \mathscr{P}(X) \to \mathscr{P}(Y) \\ G : \mathscr{P}(Y) \to \mathscr{P}(X) \end{eqnarray}$$ satisfy $F(A) \le B$ iff $A \le G(B)$.
We well define $R \subseteq X \times Y$ by $$R := \{ (x,y) \in X \times Y \mid y \in F(\{x\}) \}$$
Now we have $$ \begin{eqnarray} F_R(A) &=& \{ y \in Y \mid \exists a \in A, aRy \} \\ G_R(B) &=& \{ x \in X \mid \forall y \in Y, xRy \implies y \in B \} \end{eqnarray}$$
And we want to show $F = F_R$ and $G = G_R$.
Now $$F_R(A) = \{ y \in Y \mid \exists a \in A, aRy \}$$ may be rewritten (by expanding the definition of $R$) as $$F_R(A) = \{ y \in Y \mid \exists a \in A, y \in F(\{a\}) \}$$ and you can see that this set is just every $y$ in $F(\{a\})$ for every $a \in A$, so it's just $F(A)$.
For $$G_R(B) = \{ x \in X \mid \forall y \in Y, y \in F(\{x\}) \implies y \in B \}$$ the inner condition is the same as saying that $F(\{x\}) \subseteq B$, apply the Galois connection to this to rewrite $G_R(B) = \{ x \in X \mid \{x\} \subseteq G(B) \}$ thus $G_R(B) = G(B)$.
So if $y \in f_R(A)$, then $y \in f({a})$ for some $a \in A$. Then clearly also $y \in \bigcup_{a \in A} f({a}) = f(\bigcup_{a \in A} {a})$. And because $A = \bigcup_{a \in A} {a}$, then also $y \in f(A)$. Not certain if that's completely accurate, but in that style. And then how to get from $y \in f(A)$ to $y \in f_R(A)$?
– onm Sep 02 '20 at 08:03