I am trying to replicate the solution to this question. But, in the process I got stuck. Here is my question:
Suppose $F: C\rightleftarrows D :G$ is an adjoin pair with unit $\eta :1_C\Rightarrow GF$ and counit $\epsilon : FG\Rightarrow 1_D$ respectively. For any other functor $H :D\to C$ with a natural transformation $\alpha:G\Rightarrow H,$ is it true that $$H(\epsilon_Y)\circ\alpha_{FG(Y)}\circ\eta_{G(Y)}\stackrel{?}{=}\alpha_Y$$ for all $Y\in D$ ?
I came up with the following (commutative) naturality squares $\require{AMScd}$ \begin{CD} G(Y) @>\alpha_Y>> H(Y)\\ @V \eta_{G(Y)} V V @VV \eta_{H(Y)} V\\ GHG(Y) @>GF(\alpha_Y)>> GFH(Y)\\ @V \alpha_{FG(Y)} V V @VV \alpha_{FH(Y)} V\\ HFG(Y) @>>HF(\alpha_Y)> HFH(Y) \end{CD} But does not seems helpful with the question that I am having.
Add:
If $Y$ is in the essential image of $F,$ using the triangle identities, the required identity reduces to $\alpha_{FGF(X)}\circ\eta_{GF(X)}\stackrel{?}{=}HF(\eta_X)\circ\alpha_{F(X)}$ for some $X\in C$ and, the commutative square $\require{AMScd}$ \begin{CD} GF(X) @>GF(\eta_X)>> GFGF(X)\\ @V \alpha_{F(X)} V V @VV \alpha_{FGF(X)} V\\ HF(X) @>>HF(\eta_X)> HFGF(X) \end{CD} confirmed it when $GF(\eta_X)=\eta_{GF(X)}$ (but I am not sure whether this true or not when $\eta_X$ is not an epimorphism). This highly suggest that the required identity can be correct.
You want to show that $$H(\epsilon_Y)\circ\alpha_{FG(Y)}\circ\eta_{G(Y)} = \alpha_Y$$ By naturality, i.e. since the square $$ \begin{CD} GFGY @>\alpha_{FGY}>> HFGY \\ @VG\epsilon_YVV @VVH\epsilon_YV \\ GY @>>\alpha_Y> HY \end{CD} $$ is commutative, this is $$ \alpha_Y \circ G\epsilon_Y\circ \eta_{GY} $$ and by one of the triangle identity in an adjunction, this is $$ \alpha_Y \circ \text{id}_{GY}.$$