Let $C,D$ be categories, $F:C\rightarrow D$ a functor that is adjoint at the left to a functor $G:D\rightarrow C$. Show that exists a functorial morphism $FG\rightarrow 1_D$ and $1_C\rightarrow GF$ (that is not necessarily a isomorphism).
I started doing this:
I wanna show that exists functorial morphisms like \begin{align*} \alpha: FG&\rightarrow 1_D \\ \beta: 1_C&\rightarrow GF \end{align*}
So, for each $X\in C$ I need to identifie some $\beta_X:X\rightarrow GFX$.
I know that $F$ is a functor that is adjoint at the left to a functor $G$, this means that for each $X\in C$ and $Y\in D$ exists a bijection $$\Phi_{X,Y}: Hom_C(X,GY)\rightarrow Hom_D(FX,Y).$$
As $F:C\rightarrow D$ then for $X\in C$ we have $FX\in D$ then exists a bijection $$\Phi_{X,FX}:Hom_C(X,GFX)\rightarrow Hom_D(FX,FX)$$ as $1_{FX}\in Hom_D(FX,FX)$ we have that exists $f_X \in Hom_C(X,GFX)$ such that $\Phi_{X,FX}(f_X)=1_{FX}$, so define for each $X\in C$ $$\beta_X=f_X$$
Now to show that this wil make a functorial morphism I need to show that the square is commutative with $u:X\rightarrow Y$
That's what I can't do.
Because $\Phi$ is a natural isomorphism, for any $f:X\to Y$, the following diagram commutes : $$\require{AMScd} \begin{CD} \operatorname{Hom}_{\mathcal C}(X,GFX) @>\Phi_{X,FX}>>\operatorname{Hom}_{\mathcal D}(FX,FX) \\ @V GFu \circ - VV @V Fu\circ - VV\\ \operatorname{Hom}_{\mathcal C}(X,GFY) @> \Phi_{X,FY} >> \operatorname{Hom}_{\mathcal D}(FX,FY) \\ @A -\circ u AA @A - \circ Fu AA\\ \operatorname{Hom}_{\mathcal C}(Y,GFY) @>\Phi_{Y,FY}>> \operatorname{Hom}_{\mathcal D}(FY,FY) \end{CD}$$
Now, if we follow the elements $\beta_X \in\operatorname{Hom}_{\mathcal C}(X,GFX) $ and $\beta_Y \in \operatorname{Hom}_{\mathcal C}(Y,GFY)$, we see they both arrive at $Fu \in \operatorname{Hom}_{\mathcal D}(FX,FY)$. Since $\Phi_{X,FY}$ is an isomorphism, we must have : $$GF u \circ \beta_X = \beta_Y \circ u$$
which is what we wanted.