On proof of showing uniqueness of adjoint equivalence functor 
Suppose that $C,D$ are categories, and fix a functor $F:C\to D$. Then if there is some $G$ such that $(F,G,e,\epsilon)$ is an adjoint equivalence, and if $G'$ is another such functor, $G$ and $G'$ are isomorphic by a unique isomorphism.

(Here adjoint equivalence means that $(F,G,e,\epsilon)$ is an adjoint with $e,\epsilon$ being natural isomorphisms)
Proof: Fix $F:C\to D$, and define the category $E_F$ by


*

*$Ob(E_F)$ consists of triplet $(G,e,\epsilon)$ such that $(F,G,e,\epsilon)$ is an adjoint equivalence.

*A morphism $(G_1,e_1,\epsilon_1)\to (G_2,e_2,\epsilon_2)$ is given by a natural transformation $f:G_1\rightarrow G_2$ such that $f\circ id_F$ compose $e_1$ coincides with $e_2$, and $\epsilon_2$ compose $id_F \circ f$ coincides with $\epsilon_1$ (this condition can be stated much more compactly by drawing commutative diagram, but I'm not sure how I could draw it here - Sorry about that)
Then the claim is that any two objects in this category is isomorphic, and this isomorphism is unique.
I have no problem understanding the part about objects being isomorphic, but I don't quite follow uniqueness proof.
This goes as follows:

By commutativity condition imposed on $f\in \text{Hom}((G_1,e_1,\epsilon_1),(G_2,e_2,\epsilon_2))$, given $x\in C$ and $y\in D$,
$$
f_{F(x)}\circ e_{1,x}=e_{2,x}\;\;\; \text{and} \;\;\; e_{2,y}\circ F(f_y)=\epsilon_{1,y}
$$
which can be put in the form
$$
f_{F(x)}=e_{2,x}\circ e^{-1}_{1,x}\;\;\;\;\;\text{and}\;\;\; F(f_y)=\epsilon^{-1}_{2,y}\circ \epsilon_{1,y}
$$
Applying $G_2$ to second equality yields $G_2F(f_y)=G_2(\epsilon_{2,y})^{-1}\circ G_2(\epsilon_{1,y})$ which from naturality of $e_2$ implies
$$
e_{2,G_2(y)}\circ f_y=G_2(\epsilon_{2,y})^{-1}\circ G_2(\epsilon_{1,y})\circ e_{2,G_1(y)}
$$
(Apply naturality diagram of $e_2$ to $G_1(y)\to G_2(y)$ by $f_y$)

Therefore $f_y=G_2(\epsilon_{1,y})\circ e_{2,G_1(y)}$ which uniquely determines $f$.

Question: How does "therefore" bit follows from previous line? 

The previous line is equivalent to
$$G_2(\epsilon_{2,y})\circ e_{2,G_2(y)}\circ f_y= G_2(\epsilon_{1,y})\circ e_{2,G_1(y)}.$$
The "therefore" follows from the "triangle identity" between the unit and counit for $F\dashv G_2$:
$$G_2(\epsilon_{2,y})\circ e_{2,G_2(y)}=id_{G_2(y)}.$$
