Adjunctions and isomorphism. I'm stuck with the next exercise. I don't know how can I solve it.
Let $\mathcal{F}:\mathcal{C}\to\mathcal{D}$ and $\mathcal{E},\mathcal{G}:\mathcal{D}\to\mathcal{C}$ three adjoint functors $\mathcal{E}\dashv\mathcal{F}\dashv\mathcal{G}$. Let $(\alpha,\beta)$ the unit and counit of the adjunction $\mathcal{E}\dashv\mathcal{F}$ and $(\eta,\varepsilon)$ the unit and counit of the adjunction $\mathcal{F}\dashv \mathcal{G}$. Suposse that the next conditions (1) holds:

a) $\text{Id}_{\mathcal{EFE}}=\mathcal{E}_{\alpha}\circ\beta_{\varepsilon}$
b) $\text{Id}_{\mathcal{FEF}}=\alpha_{\mathcal{F}}\circ\mathcal{F}_{\beta}$

Prove that the next conditions holds

a) $\text{Id}_{\mathcal{FGF}}=\mathcal{F}_{\eta}\circ\varepsilon_{\mathcal{F}}$
b) $\text{Id}_{\mathcal{GFG}}=\eta_{\mathcal{G}}\circ \mathcal{G}_{\varepsilon}$

First, I proved that the condition a) is equivalent to the condition b), thus, I only need to prove one of a) or b). Next, I wrote the commutative diagrams related to my hypothesis:
By (1) we have the next commutative diagrams

And because $\mathcal{E}\dashv\mathcal{F}$ and $\mathcal{F}\dashv\mathcal{G}$ we have the next pair of commutative diagrams


Thus I need to prove that the next diagram is commutative:

But I don't know how. By combining the previous diagrams, I obtained the next commutative diagram:

But in fact that diagram does not help. I saw that if I prove the previous diagram, then in fact I will prove that $\mathcal{F}_{\eta}$ is a natural isomorphism, but, again, I'm stuck. Any hint? I really appreciate any help you provide me.
 A: This is one possible way to approach the problem.
By the definition of adjunction we have the following chain of natural isomorphisms 
$$\newcommand{\mc}{\mathcal} 
\newcommand{\id}{\text{Id}}
\begin{align*}
\mc D[\mc{FGF}(c),\mc{FGF}(c)] &\cong
\mc C[\mc{EFGF}(c),\mc{GF}(c)] \\ &\cong 
\mc D[\mc{FEFGF}(c),\mc{F}(c)] \\ &\cong 
\mc C[\mc{EFEFGF}(c),c]
\end{align*}\ .
$$
Since these are natural bijections we have that 
$$\mc F(\eta)\circ \epsilon_{\mc F}=\id_{\mc{FGF}}$$
holds only and only if their transformed morphisms in $\mc C[\mc{EFEFGF}(c),c]$.
From there an additional small step is required to complete the solution.
Here it follows this small step but I suggest you take a look at it only after having tried by yourself.
In the same way you can prove the other identity.
I hope this helps.

A computation shows that $\mc F(\eta_c) \circ \epsilon_{\mc F(c)}$ becomes $\beta_c \circ \mc{E}(\epsilon_{\mc F(c)})\circ \beta_{\mc{EFGF}(c)}$ while $\id_{\mc{FGF}(c)}$ becomes $\beta_c \circ \mc E(\epsilon_{\mc F(c)})\circ \mc{EF}(\beta_{\mc{GF}(c)})$. 
To conclude we can just observe that
$$
\begin{align*}
\mc{EF}(\beta_{\mc{GF}(c)}) &= \mc{EF}(\beta_{\mc{GF}(c)})\circ\mc E(\alpha_{\mc{FGF}(c)})\circ\beta_{\mc {EFGF}(c)}\\
&= \mc{E}(\mc F(\beta_{\mc{GF}(c)})\circ \alpha_{\mc{FGF}(c)})\circ \beta_{\mc{EFGF}(c)} \\
&= \beta_{\mc{EFGF}(c)}
\end{align*}$$
where the first equality follows from the hypothesis, the second by functoriality and the third by the triangle identities for adjunctions.
