Initial object, final object and unit and counit I'm wanting to try the following:
Let $F: C\longrightarrow{D}$ and $G:D\longrightarrow{C}$ functors.
If there exists a natural transformation $\eta:1_C\Longrightarrow {G\circ{F}}$ such that $\eta_c$ is an initial object in the category $(c\downarrow G)$, for all $c\in C$ then there exists a natural transformation $\epsilon:F\circ{G}\Longrightarrow{1_D}$ such that $\epsilon_d$ is a final object in the category $(F\downarrow d)$, for all $d\in D$.
These transformations are called the unit and counit respectively.
Indeed: Let $f:Fc\longrightarrow{d}$ be an object in the category $(F\downarrow d)$. For proving that there exists a unique $\beta:c\longrightarrow{Gd}$ such that $\epsilon_d\circ{F\beta=f}$.
I want to use the morphism $f$ on $G$ i.e. $Gf:(G\circ{F})c\longrightarrow{Gd}$ since it is a object in category $(c\downarrow G)$ then by the hypothesis there is a unique map $\alpha:F\widetilde{c}\longrightarrow{d}$ such that $G\alpha\circ{\eta_{\widetilde{c}}}=Gf$ where $\widetilde{c}=(G\circ{F})c$. I don't get the existence of $\beta$ with that idea.
I await a suggestion.
Thank you very much
 A: First one has to define $\epsilon_d$ for every $d$. For this, use the initiality of $\eta_{Gd}$: there is a unique map $\epsilon_d: FGd \to d$ making the triangle $1_{Gd}: Gd \xrightarrow{\eta_{Gd}} GFG d \xrightarrow{G\epsilon_d} Gd $ commute.
Now assume given $f: Fc \to d.$  We want to factor this map through $FGd \xrightarrow{\epsilon_d} d $ via a map $F \beta$ for some $\beta: c \to Gd.$ Where to get $\beta?$ Well, applying $G$ gives a map $GFc \to Gd$. Precomposing this with $\eta_c: c \to GF c$ gives a map with the right domain and codomain. Will $F \beta$ make the triangle $Fc \to FGd \to d$ commute? By the initiality of $\eta_{c},$ there is only one map $g$ such that $Gg \circ \eta_{c}$ is $\beta.$ This map is $f$ by the definition of $\beta$. But $\epsilon_d \circ F \beta$ also works as such a map: $G\epsilon_d \circ GF \beta \circ \eta_c = G\epsilon_d \circ \eta_{Gd} \circ \beta = \beta,$ the first equality by naturality and the second by the definition of $\epsilon_d.$ It follows that $\epsilon_d \circ F \beta = f,$ i.e. $f$ and $\beta$ are adjoint.
