# Relation between right adjoints and counits

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

1. $$G$$ is faithful if and only if each arrow $$\varepsilon_{d}: F G d \rightarrow d$$ is an epimorphism;
2. $$G$$ is full if and only if each arrow $$\varepsilon_{d}: F G d \rightarrow d$$ is a split monomorphism;
3. $$G$$ 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?

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!

• $f^*$ is the notation for precomposition with $f$: it sends a map $g$ to $g \circ f$. Some also call it "pulling back". Jan 20, 2022 at 20:46
• But it seems to be a weird pullback qhere only one component is pulled back- As far as I know, $\varepsilon$ is a natural iso between the identity functor of §D§ and the functor $FG$. Jan 20, 2022 at 21:22
• And it has a component at $d_1$, giving a morphism $F G d_1 \to d_1$. Writing subscripts all the time is laborious. Jan 20, 2022 at 22:25