Naturality condition for adjoint functors I am reading chapter 2 of Tom Leinster's Basic Category Theory. In section 2.1 he defines the naturality condition with two equations. For categories $\mathscr{A}$ and $\mathscr{B}$ and morphisms $p: A' \rightarrow A$, $f: A \rightarrow G(B)$ in $\mathscr{A}$ and $q: B \rightarrow B'$, $g: F(A) \rightarrow B$ in $\mathscr{B}$ if $F$ and $G$ are adjoint functors we have $\overline{q \circ g} = G(q) \circ \overline{g}$ and $\overline{f \circ p} = \overline{f} \circ F(p)$.
In section 2.2 he claims it follows from the naturality condition that for each $A \in \mathscr{A}$ we have a map $\eta_A : A \rightarrow GF(A)$ such that $\eta_A = \overline{1_{F(A)}}$. Dually, for each $B \in \mathscr{B}$ we have a map $\epsilon_B : FG(B) \rightarrow B$ such that $\epsilon_B = \overline{1_{G(B)}}$. Moreover, these define natural transformations.
I am struggling to see how this follows from the naturality conditions.
To construct a map $A \rightarrow GF(A)$ from the given conditions it seems like I would need to set $B = F(A)$, $A = A'$ and $p = 1_A$. This approach leads nowhere as all I can get is $f = \overline{\overline{f}}$.
 A: The naturality of Hom bijection can be expressed via following commutative diagram:
$\require{AMScd}$
$$\begin{CD}
id_{F(A)}\in Hom(F(A),F(A)) @>\text{Adjoint bijection}>> Hom(A,GF(A))\\
@V \text{postcompose}\;\; F(A\to A') VV @V \text{postcompose}\;\; GF(A\to A')VV\\
Hom(F(A),F(A')) @>\text{Adjoint bijection}>> Hom(A,GF(A'))\\
@A{\text{precompose} \;\; F(A\to A')}AA @A {\text{precompose} \;\; A\to A'}AA\\
id_{F(A')}\in Hom(F(A'),F(A')) @>\text{Adjoint bijection}>> Hom(A',GF(A'))
\end{CD}$$
Upward and downward directions each represent one part of axiom.
The thing here to notice is that under vertical maps, both $id_{F(A)}$ and $id_{F(A')}$ map to same morphism in $Hom(F(A),F(A'))$ (this is essentially by axiom on identity morphisms).
So their end-image in $Hom(A,GF(A'))$ must conincide when following arrows vertically then horizontally.
But by commutativity of diagram, then by definition of $\eta$ one must have 
$$
GF(A\to A')\circ\eta_A=\eta_{A'} \circ (A\to A')
$$ 
which is exactly the naturality  of $\eta$
