Assume that $F: \mathcal{C} \rightarrow \mathcal{D}$ is left adjoint to $G: \mathcal{D} \rightarrow \mathcal{C}$ with counit $\varepsilon: F G \Rightarrow$ id $_{\mathcal{D}}$ and unit $\eta$ : id $_{\mathcal{C}} \Rightarrow G F$. Show that $$ \mathcal{D}\left(d_{1}, d_{2}\right) \stackrel{\varepsilon^{*}}{\rightarrow} \mathcal{D}\left(F G d_{1}, d_{2}\right) \stackrel{\text { adjunction }}{\cong} \mathcal{C}\left(G d_{1}, G d_{2}\right) $$ is the functor $G$. Conclude that
- $G$ is faithful if and only if each arrow $\varepsilon_{d}: F G d \rightarrow d$ is an epimorphism;
- $G$ is full if and only if each arrow $\varepsilon_{d}: F G d \rightarrow d$ is a split monomorphism;
- $G$ is full and faithful if and only if $\varepsilon$ is an isomorphism.
I know this question has been answered before: 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?
However, I'm confused about
Show that $$ \mathcal{D}\left(d_{1}, d_{2}\right) \stackrel{\varepsilon^{*}}{\rightarrow} \mathcal{D}\left(F G d_{1}, d_{2}\right) \stackrel{\text { adjunction }}{\cong} \mathcal{C}\left(G d_{1}, G d_{2}\right) $$ is the functor $G$.
I've never seen the notation $\varepsilon^*$ before which makes me hesitant to write out an explicit answer. Indeed, it seems trivial but I might be misunderstanding some subtlety. Any help is greatly appreciated!
I'm sorry people think the identity your presenting is trivial. It's really not. You can check out the proposition labeled (pre/post-composition with (co-)unit followed by adjunct is adjoint functor) here.
With regards to the subsequent implications check out Proposition 2.4 in nLab's page on adjoint functors.
Hope this answered any remaining questions.