How to use Yoneda's lemma in proving: $F$ fully-faithful iff unit is an isomorphism Let $F \dashv G$ be an adjunction with unit $e: Id \implies GF$.
It is well known that $e$ is an isomorphism if and only $F$ is fully-faithful. I've a proof of this fact that doesn't use Yoneda's lemma in any explicit way.
Here: Let $C,D$ be categories and $F:C\to D$ and $G:D\to C$ be adjoint functors. Then $F$ is fully faithful iff the unit is an isomorphism?
it is mentioned that using the Yoneda lemma can solve this question almost directly. Can you elaborate on how? 
For instance, even in the direction $e$ is an isomorphism implies $F$ is fully-faithful, the only way I see to use the Yoneda lemma doesn't shorten my direct proof.
For the other direction I'm actually not sure how to use explicitly.
By using Yoneda's lemma I mean use either:


*

*The Yoneda functor $Y$ is fully-faith

*There is an isomorphism between Fun$(Y(x), F)$ and $F(x)$ where $F$ is a presheaf
How can the Yoneda lemma be used to prove this claim easily?
 A: We have $$\mathsf{Hom}(A,B)\stackrel{\eta_B\circ-}{\to}\mathsf{Hom}(A,GFB)\stackrel{\varphi_{A,FB}}{\cong}\mathsf{Hom}(FA,FB)$$ natural in $A$ and $B$ with $\eta$ being the unit. This would immediately make fullness and faithfulness of $F$ be equivalent to $\eta$ being an isomorphism (via Yoneda) as long as we can show that $Ff=\varphi(\eta\circ f)$. Specifically, we'd have $$\mathsf{Hom}(A,B)\stackrel{F}{\cong}\mathsf{Hom}(FA,FB)\stackrel{\varphi^{-1}_{A,FB}}{\cong}\mathsf{Hom}(A,GFB)$$ is equal to $$\mathsf{Hom}(A,B)\stackrel{\eta_B\circ-}{\to}\mathsf{Hom}(A,GFB)$$ meaning it's an isomorphism and Yoneda (like all fully faithful functors) reflects isomorphisms. The other direction is obvious: $F$ would be a composition of isomorphisms. 
The necessary equality is immediate by naturality of $\varphi$ (in $A$ as above) which lets us reduce to $Fid=id=\varphi(\eta)$ and indeed $\eta$ is defined as (or provably equivalent to) as $\varphi^{-1}(id)$. (In the cases where $\eta$ is primitive, then the definition of $\varphi$ and its inverse will make the above equation fairly obvious.)
