Galois connection from binary relations 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 $m \in F(A)$?

 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.

$$
\begin{eqnarray}
&& G_R(A) \\
&& \text{(by the definition of $G_R$)} \\
&=& \{ x \in X \mid \forall y \in Y, xRy \implies y \in B \} \\
&& \text{(by the definition of $R$)} \\
&=& \{ x \in X \mid \forall y \in Y, y \in F(\{x\}) \implies y \in B \} \\
&& \text{(the logical definition of a subset)} \\
&=& \{ x \in X \mid F(\{x\}) \subseteq \phantom{G(}B\phantom{)} \} \\
&& \text{(applying the Galois connection)} \\
&=& \{ x \in X \mid \phantom{F(}\{x\}\phantom{)} \subseteq G(B) \} \\
&=& \{ x \in X \mid x \in G(B) \} \\
&=& G(B) \\
\end{eqnarray}
$$
A: 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 $G_R(B) = G(B)$.
