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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?

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!

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  • $\begingroup$ $f^*$ is the notation for precomposition with $f$: it sends a map $g$ to $g \circ f$. Some also call it "pulling back". $\endgroup$ Jan 20, 2022 at 20:46
  • $\begingroup$ 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 $FG$. $\endgroup$ Jan 20, 2022 at 21:22
  • $\begingroup$ 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. $\endgroup$
    – Zhen Lin
    Jan 20, 2022 at 22:25

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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.

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