# Prove that adjunction restricts to an equivalence between full subcategories

Given problem 2.2.11 (a) from T. Leinster's "Basic Category Theory" (I modified the question a bit since it's difficult to draw an adjunction here, but the logic is the same):

Let a pair of functors $$F : \mathscr{A} \rightarrow \mathscr{B}$$ and $$G : \mathscr{B} \rightarrow \mathscr{A}$$ be adjunction such that $$F$$ is left adjoint to $$G$$, i.e. $$F \dashv G$$. Write $$\textbf{Fix}(GF)$$ for the full subcategory of $$\mathscr{A}$$ whose objects are those $$A \in \mathscr{A}$$ such that $$\eta_{A}$$ is an isomorphism, and dually $$\textbf{Fix}(FG) \subseteq \mathscr{B}$$. Prove that the adjunction $$(F, G, \eta, \epsilon)$$ restricts to an equivalence $$(F', G', \eta', \epsilon')$$ between $$\textbf{Fix}(GF)$$ and $$\textbf{Fix}(FG)$$.

I have difficulties understanding the question completely. Since we need to show that "adjunction restricts to an equivalence", do we first find two functors, say $$F'$$ and $$G'$$, and prove that they are both fully faithful and essentially surjective?

Isn't it sufficient to find one of the functors ($$F'$$ or $$G'$$) (s.t. they are full, faithful, and essentially surjective) to show an equivalence?

And if this is a case, why and how do I use the natural transformations $$\eta', \epsilon'$$ to show the equivalence?

Thanks!

You want to show that the restriction of $$F$$ to the fixpoints of $$GF$$ is fully faithful, and that it maps essentially surjectively on objects to the fixpoints of $$FG$$. Alternatively, you can prove that the restrictions of $$F$$, $$G$$, $$\eta,\varepsilon$$ to the respective fixpoints give an adjoint equivalence, that is, an adjunction with invertible unit and counit.
• @oneturkmen I mean $\mathbf{Fix}(GF)$ and so on. The idea is that those subcategories are effectively fixed by $GF$. – Kevin Carlson Mar 28 '19 at 17:00