Proving that the counit of adjunction is a natural transformation Let $F:\mathcal A\to\mathcal B, G:\mathcal B\to \mathcal A$ be functors such that $F$ is left adjoint to $G$. I'm trying to prove that the counit of adjunction $\epsilon: FG\to 1_\mathscr{B}$ is a natural transformation.
So the task is to prove that the following square commutes:

i.e. $\overline {1_{G(B')}}\circ FG(g)=1_\mathscr{B}(g)\circ\overline {1_{G(B)}}$.
I tried writing down the conditions of $F$ and $G$ being adjoint. The first one is the commutativity of this square:

The LHS of the last equation is equal to the RHS of the equation we need to establish.
Then I tried to use the other requirement on the functors being adjoint, namely, the commutativity of this square:

(it's assumed that there's an arrow $f:A'\to A$ in the background).
I put $A=G(B'), A'=G(B)$ and got the highlighted equality from the commutativity of the diagram:

If I could set $f=id$, that would tell me $FG(g)=\overline {G(g)}$, and the proof would be finished. But I can't do that (I can't set $B'=B$, because I wouldn't be able to use the map $G(g)$ in the diagram).
So how should I proceed?
 A: There are lots of ways to show this, because there's actually a lot going on here. For example, $\epsilon_B$ is actually a universal morphism; that's one way you can show this. But a simple way is to use the relations that are given by the definition of an adjunction. To be clear, I’ll write $\varphi: \text{Hom}_{\mathcal{B}}(F(A), B) \to \text{Hom}_{\mathcal{A}}(A, G(B))$
so that $\epsilon_B = \varphi^{-1}(1_{G(B)})$
Now recall that for $k: B \to B’$ and $h: A’ \to A$, the diagrams below commute by naturality of $\varphi$.

Now for any $f: F(A) \to B$, the above diagrams give us the relations
$$
    \varphi(k \circ f) = G(k) \circ \varphi(f)
    \qquad 
    \varphi(f \circ F(h)) = \varphi(f) \circ h.
$$
To show that $\epsilon_B$ is natural, you’ll need to show that the diagram below commutes.

Observe that we have the relation
$$
\text{Hom}_{B}(F(G(B), B’) \cong \text{Hom}_{A}(G(B), G(B’))
$$
By just substituting $A = G(B)$. But, by our formulas given by the definition of an adjunction, we get that
$$
\varphi(\epsilon_{B’} \circ F(G(g)))
= 1_{G(B’)} \circ G(g) = G(g)
$$
while
$$
\varphi(g \circ \epsilon_B)
= 
G(g) \circ \varphi(\epsilon_B) = G(g) \circ 1_{G(B’)} = G(g).
$$
So we see that $\epsilon_{B’} \circ F(G(g))$ and $g \circ \epsilon_B$ are sent to the same element. But since $\phi$ is an isomorphism, this implies that $\epsilon_{B’} \circ F(G(g)) = g \circ \epsilon_B$. Hence the diagram must commutes, so you have naturality.
