Pair of adjoint functors and showing that two certain maps are inverses I'm currently self-studying "Categories and Sheaves"by Schapira and Kashiwara, and I've been stuck on problem 1.19 all day today, so I was hoping that someone could help me out.
Let $\mathcal{C}$, $\mathcal{C'}$ be categories and $L_v:\mathcal{C} \rightarrow \mathcal{C'}$, $R_v:\mathcal{C'} \rightarrow \mathcal{C}$ be functors such that $(L_v,R_v)$ is a pair of adjoint functors (v=1,2). Let $\epsilon_v: id_\mathcal{C} \rightarrow R_v L_v$ and 
$\eta_v: L_v \circ R_v  \rightarrow id_{\mathcal{C'}}$ be the adjunction morphisms. Prove that the two maps $\lambda$,$\mu$ :
$$Hom_{Fct(\mathcal{C},\mathcal{C'})}(L_1,L_2) \rightleftarrows^{\lambda}_{\mu} Hom_{Fct(\mathcal{C'},\mathcal{C})}(R_2,R_1)$$
given by :
$$\lambda(\varphi):R_2 \rightarrow^{\epsilon_1 \circ R_2} R_1 \circ L_1 \circ R_2 \rightarrow^{R_1 \circ \varphi \circ R_2} R_1 \circ L_2 \circ R_2 \rightarrow^{R_1 \circ \eta_2} R_1$$ for $\varphi \in Hom_{Fct(\mathcal{C},\mathcal{C'})}(L_1,L_2)$
$$\mu(\psi) : L_1 \rightarrow^{L_1 \circ \epsilon_2} L_1 \circ R_2 \circ L_2 \rightarrow^{L_1 \circ \psi \circ L_2} L_1 \circ R_1 \circ L_2 \rightarrow^{\eta_1 \circ L_2} L_2$$
for $\psi \in Hom_{Fct(\mathcal{C}',\mathcal{C})}(R_2,R_1)$ are inverse to eachother.
So far, I've been trying to play around with the zig-zag identity, but it doesn't seem to lead me anywhere. Any help will be greatly appreciated!
 A: Exercise $1.19$ of Categories and Sheaves, by Masaki Kashiwara and Pierre Schapira. Preview available at Google Books, and at Amazon. 
I'll try to write this answer so that it can be read without consulting either Kashiwara and Schapira's book (abbreviated by [KS]) or the question. 
The exercise asks to check two formulas. The two verifications are very similar, and I'll do only one of them. I'll try to prove the formula in question by rediscovering it. I hope the notation is self-explanatory, but if it isn't, thanks for letting me know.
For $i=1,2$ let 
$$
L_i:\mathcal C\to \mathcal C',\quad R_i:\mathcal C'\to \mathcal C
$$ 
be left and right adjoint functors, and let 
$$
\varepsilon_i:\text{id}_{\mathcal C}\to R_iL_i,\quad\eta_i:L_iR_i\to\text{id}_{\mathcal C'}
$$ 
be the adjunction morphisms. Let $X$ be in $\mathcal C$ and $X'$ be in $\mathcal C'$, and write 
$$
a_i:\mathcal C'(L_iX,X')\to \mathcal C(X,R_iX')
$$
for the adjunction isomorphism. Let 
$$
\varphi:L_1\to L_2
$$ be a morphism, and consider the diagram
$$
\begin{matrix}
&a_2&\\ 
\mathcal C'(L_2X,X')&\to&\mathcal C(X,R_2X')\\ &&&\\
\varphi(X)^*\downarrow&&\downarrow b\\ &&&\\
\mathcal C'(L_1X,X')&\to&\mathcal C(X,R_1X'),\\
&a_1&
\end{matrix}
$$
where $b$ is defined by the requirement that the diagram commutes. 
The goal is to find a morphism 
$$
\psi:R_2\to R_1
$$ 
such that 
$$
b=\psi(X')_*.
$$
Let 
$$
f:X\to R_2X'
$$ 
be a morphism. By the proof of Proposition [KS, $1.5.4$], we have
$$
a_2^{-1}f=\left[L_2X\xrightarrow{L_2f}L_2R_2X'\xrightarrow{\eta_2(X')}X'\right],
$$
by which I mean that $a_2^{-1}f$ is equal to the composition between the brackets. This gives
$$
\varphi(X)^*a_2^{-1}f=\left[L_1X\xrightarrow{\varphi(X)}L_2X\xrightarrow{L_2f}L_2R_2X'\xrightarrow{\eta_2(X')}X'\right].
$$
The functoriality of $\varphi$ yields 
$$
\varphi(X)^*a_2^{-1}f=\left[L_1X\xrightarrow{L_1f}L_1R_2X'\xrightarrow{\varphi(R_2X')}L_2R_2X'\xrightarrow{\eta_2(X')}X'\right].
$$
Using again the proof of Proposition [KS, $1.5.4$], we get 
$$
a_1\,\varphi(X)^*a_2^{-1}f=
$$
$$
\left[X\xrightarrow{\varepsilon_1(X)}R_1L_1X\xrightarrow{R_1L_1f}R_1L_2R_2X'\xrightarrow{R_1\varphi(R_2X')}R_1L_2R_2X'\xrightarrow{R_1\eta_2(X')}R_1X'\right],
$$
which, by functoriality of $\varepsilon_1$, is equal to 
$$
\left[X\xrightarrow{f}R_2X'\xrightarrow{\varepsilon_1(R_2X')}R_1L_1R_2X'\xrightarrow{R_1\varphi(R_2X')}R_1L_2R_2X'\xrightarrow{R_1\eta_2(X')}R_1X'\right].
$$
This implies the sought-for formula
$$
\psi(X')=\left[R_2X'\xrightarrow{\varepsilon_1(R_2X')}R_1L_1R_2X'\xrightarrow{R_1\varphi(R_2X')}R_1L_2R_2X'\xrightarrow{R_1\eta_2(X')}R_1X'\right].
$$
