Let $L: C \to D$ be left adjoint to the functor $R: D \to C.$
For an object $X$ of $C$, we have an isomorphism $Hom_D(L(X),L(X)) \xrightarrow{\sim} Hom_C(X,(R\circ L)X)$.
Show that there exists a unique morphism of functors $\eta : id_C \to R\circ L$ such that $\eta(X) $is the image of $id_{L(X)}$ under this isomorphism, for every $X\in C$.
Note:$\eta$ is called the unit of the adjunction
My try:
I've tried spelling out the left adjointness in the hypothesis and the morphism of functors of the thesis:
$\forall X \in C$ aand $Y \in D$: $\alpha_{X,Y}: Hom_D(LX,Y)\cong Hom_C(X,RY)$ s.t for all $f: X_1 \to X_2$: $$ \begin{matrix} \operatorname{Hom}_D(LX_2, Y) & \xrightarrow{\alpha_{X_2,Y}} & \operatorname{Hom}_C(X_2, RY) \\ \left\downarrow{\scriptstyle{h \mapsto h\circ L(f)}}\vphantom{\int}\right. & & \left\downarrow{\scriptstyle{g \mapsto g \circ f}}\vphantom{\int}\right.\\ \operatorname{Hom}_D(LX_1, Y)& \xrightarrow{\alpha_{X_1,Y}} & \operatorname{Hom}_C(X_1, RY) \end{matrix} $$
and for all $f: Y_1 \to Y_2$:
$$ \begin{matrix} \operatorname{Hom}_D(LX, Y_1) & \xrightarrow{\alpha_{X,Y_1}} & \operatorname{Hom}_C(X, RY_1) \\ \left\downarrow{\scriptstyle{h \mapsto f\circ h}}\vphantom{\int}\right. & & \left\downarrow{\scriptstyle{g \mapsto R(f) \circ g}}\vphantom{\int}\right.\\ \operatorname{Hom}_D(LX, Y_2)& \xrightarrow{\alpha_{X,Y_2}} & \operatorname{Hom}_C(X, RY_2) \end{matrix} $$
the diagrams commute.
Taking $Y=LX$, I have the diagrams corresponding to the stated isomorphism. I tried taking
$Y=LX_2$ in the first one so that I can take $id_{L(X_2)}$ in $Hom_D(LX_2,LX_2) $
and
$Y_1=LX_2, Y_2=LX_1$ in the second one
The thesis should be that $\eta$ is an isomorphism of functors, that is
for each $X \in C$, $\eta_X:\{id_CX \to (R \circ L) X\}_{X \in C}$ s.t
$$ \begin{matrix} id_C X & \xrightarrow{id_C(f)} & id_C Y \\ \left\downarrow{\scriptstyle{\eta_X}}\vphantom{\int}\right. & & \left\downarrow{\scriptstyle{\eta_Y}}\vphantom{\int}\right.\\ (R\circ L) X& \xrightarrow{(R\circ L)f} & (R\circ L) Y \end{matrix} $$
commutes.
and $\eta(X)=\alpha_{X,Y}(id_{L(X)}), X \in C$
but I am stuck here. I having trouble combining the diagrams above to prove the thesis. How should I do this?
Moreover I am denoting with $\eta_X$ the family of morphism that is part of the definition of morphism of functors since that how it was defined in my lecture, as an index of the $\eta$, so I am usure if in this problem $\eta(X)$ is the same denoted differently. Another notational difficulty is that $id_C$ should I be a functor, but the notation makes it look as the identity morphism in the category $C$.
$\require{AMScd}$I agree it can be confusing to extract properties of $\eta$ from this definition of adjunction ( :) even more so, but doably, in the enriched case). I've done it many times and still always forget how to do it at first glance. In the below I accidentally swapped the roles of $D,C$, please forgive this.
Given $f:X\to Y$, $RL(f)\eta_X$ is what you get if you plug in $1_{LX}$ in the top left corner and follow it around the right hand route in the square: $$\begin{CD}C(LX,LX)@>\cong>>D(X,RLX)\\@VC(1,L(f))VV@VVD(1,R(L(f))V\\C(LX,LY)@>>\cong>D(X,RLY)\end{CD}$$
So $RL(f)\eta_X$ is the transpose of $L(f)$. What's that? Well: $$\begin{CD}C(LY,LY)@>\cong>>D(Y,RLY)\\@VC(Lf,1)VV@VVD(f,1)V\\C(LX,LY)@>>\cong>D(X,RLY)\end{CD}$$If you plug in $1_{LY} $ in the top left you find that $L(f)$ transpose is just $D(f,1)(\eta_Y)=\eta_Y\circ f$. So $RL(f)\circ\eta_X=\eta_Y\circ f$ as desired!
Notice this "plugging in the identity" business is really about the Yoneda lemma. These natural transformations are equal iff. their values at the identity morphism are equal.