I know there is a fair amount of similar questions to this one, but none of them deal directly with what I'm struggling with right now. For instance, in this one the right adjoint functor is constructed, while in my situation I already have the pair of functors. In this one the objective is to derive the unit-counit equations, (which is the same I'm trying to prove, essentially), but this question does not have any answers. This question deals with the converse, but that direction I was able to prove.
Goal
Given functors $L \colon C \to D, R \colon D \to C$, a family of morphisms $\eta_X \colon X \to RL(X), \epsilon_Y \colon LR(Y) \to Y$ such that $\forall f \colon L(X) \to Y. \exists ! g \colon X \to R(Y). f= \epsilon_Y \circ L(g) \land g=R(f) \circ \eta_X$ and vice-versa (for every $g$ there is an unique $f$), then $\eta: id_C \to RL, \epsilon: LR \to id_D$ are natural transformations and the natural-transformation-equalities hold.
First I need to prove that $\eta$ is a natural transformation. That is, given a morphism $f \colon X_1 \to X_2$ in $C$, we have $\eta_{X_2} \circ f = RL(f) \circ \eta_{X_1}$.
What I did
Define $g = \eta_{X_2} \circ f$. By hypothesis, there exists an unique $f' \colon L(X_1) \to L(X_2)$ such that $\epsilon_{L(X_2)} \circ L(g) = f'$ and $g= R(f') \circ \eta_{X_1}$. Hence $\eta_{X_2} \circ f = R(f') \circ \eta_{X_1}$. It is sufficient to prove $R(f')=RL(f)$. After some calculations, we have $R(f') = R(\epsilon_{L(X_2)} \circ L(\eta_{X_2})) \circ RL(f)$. How to prove that $R(\epsilon_{L(X_2)} \circ L(\eta_{X_2})) = id_{RL(X_2)}$?
Consider the morphism $id_{L(X_2)} \colon L(X_2) \to L(X_2)$. By hypothesis, there exists an unique $g \colon X_2 \to RL(X_2)$ such that $\epsilon_{L(X_2)} \circ L(g) = id_{L(X_2)}$ and $g=id_{RL(X_2)} \circ \eta_{X_2}$. But the only $g$ such that the second equality holds is $g=\eta_{X_2}$, hence $\epsilon_{L(X_2)} \circ L(\eta_{X_2}) = id_{L(X_2)}$.