The right adjoint of a functor is unique up to a natural isomorphism I'm having trouble showing that the right adjoint of a functor $F \colon \mathcal{C} \to \mathcal{D}$ is unique up to isomorphism. (My definition of adjoint functors is the one using units and counits.)
Let $G_1$ and $G_2$ be two right adjoints of $F$, with units $e_1,e_2$ and counits $\epsilon_1,\epsilon_2$ respectively. We can construct natural transformations
$$f \colon G_1 \overset{e_2G_1}\longrightarrow G_2FG_1 \overset{G_2\epsilon_1}\longrightarrow G_2$$
and
$$g \colon G_2 \overset{e_1G_2}\longrightarrow G_1FG_2 \overset{G_1\epsilon_2}\longrightarrow G_1.$$
I'm not quite sure how to show that these are inverses of each other. For example, consider $g \circ f \colon:G_1\to G_1,$ which is the map
$$G_1\epsilon_2 \circ e_1G_2 \circ G_2\epsilon_1 \circ e_2G_1.$$ How do I show that this is the identity? I know that $G_i\epsilon_i \circ e_iG_i$ is the identity natural transformation on $G_i$, but how do I use that information here, since the subscripts in the composite are alternating?

Hint 1: Find the arrows in the following diagram which make its squares commutative, using naturality. 
$$\matrix{G_1 & \to & G_2FG_1 & \to & G_2\\
\downarrow && \downarrow && \downarrow \\
G_1FG_1 & \to & G_1FG_2FG_1 & \to & G_1FG_2 \\
&&\downarrow && \downarrow \\
&&G_1FG_1 & \to & G_1}$$

Hint 2: Alternatively, we can draw these natural transformations as diagonal morphisms $\swarrow$ in a square of functors, expressing $\alpha:UF\to GV$ as 
$$\matrix{&\overset F\longrightarrow&\\ {}_V\downarrow&\alpha&\downarrow_{\,U}\\
&\underset G\longrightarrow}$$
Then draw the units $\eta_i:=e_i$ as squares for $1\ \!1\to G_iF$ and the counits as squares for $FG_i\to 1\ \!1$, and consider the following arrangement:
$$\pmatrix{ \varepsilon_1 &\eta_2 \\ &\varepsilon_2&\eta_1} $$
Determine the functors on the borders, and show that, in general, pasting the squares horizontally commutes with pasting them vertically.
