Given the adjunction morphisms $\epsilon, \eta$, then $(\eta \circ L) \circ(L\circ \epsilon) = \text{id}_L$ is easy to check.
This is from page 29 of "categories and sheaves". I have a major confusion contusion regarding this and not sure where to even begin...
EDIT:
$\text{id}_L$ is the identity natural transformation for the functor $L : C \to C'$.
$\epsilon : \text{id}_{C'} \to R\circ L $ is constructed as $X \mapsto \psi(\text{id}_{L(X)})$ where $\psi : \text{Hom}_{C'}(L(X), L(X)) \simeq \text{Hom}_C(X, R \circ L(X))$.
$\eta : \text{id}_C \to R\circ L$ is constructed as?
$$ \require{AMScd} \begin{CD} \text{id}_C(X) @>{\text{id}_C(f)}>> \text{id}_C(Y)\\ @V\epsilon_XVV @VV\epsilon_YV \\ R\circ L(X) @>{R\circ L(f)}>> R \circ L(Y) \end{CD} $$
Then functor it over to $C'$ by composing the whole diagram with $L$:
$$ \require{AMScd} \begin{CD} L(X) @>{L(f)}>> L(Y)\\ @VL(\epsilon_X)VV @VVL(\epsilon_Y)V \\ L\circ R\circ L(X) @>{L\circ R\circ L(f)}>> L\circ R \circ L(Y) \end{CD} $$
from which we can use $\eta : L \circ R \to \text{id}_{C'}$ to yield:
$$ \require{AMScd} \begin{CD} L(X) @>{L(f)}>> L(Y)\\ @VL(\epsilon_X)VV @VVL(\epsilon_Y)V \\ L\circ R\circ L(X) @>{L\circ R\circ L(f)}>> L\circ R \circ L(Y) \\ @V\eta_{L(X)}VV @VV\eta_{L(Y)}V \\ L(X) @>{L(f)}>> L(Y) \end{CD} $$
The above visuallly exhibits that $(\eta \circ L ) \circ (L\circ \epsilon)$ is indeed a natural map of functors $L \to L$. Additionally it shows that for any $L(f) \in \text{Hom}_{C'}(L(X), L(Y))$ the natural map, so composed, takes us back to $L(f)$. The only map that can do that is unique, $\text{id}_L$.