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?



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.

  • $\begingroup$ What do you mean by "the fixpoints of GF" or "fixpoints"? $\endgroup$ – oneturkmen Mar 28 '19 at 8:33
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    $\begingroup$ @oneturkmen I mean $\mathbf{Fix}(GF)$ and so on. The idea is that those subcategories are effectively fixed by $GF$. $\endgroup$ – Kevin Arlin Mar 28 '19 at 17:00
  • $\begingroup$ It makes sense now. Thanks! $\endgroup$ – oneturkmen Mar 28 '19 at 17:06

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