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Let $F:C\to D$, $G:D\to C$ be functors such that $F\dashv G\dashv F$. I want to show that they are equivalencies of categories.We have the existence of $\eta:Id_C\to G\circ F$, $\epsilon:F\circ G\to Id_D$ for $F\dashv G$ and $\lambda:Id_D\to F\circ G$, $\mu:G\circ F\to Id_C$ for $G\dashv F$ which are natural transformation satisfying :

$\epsilon_{F(c)}\circ F(\eta_c)=Id_{F(c)}$, $G(\epsilon_d)\circ \eta_{G(d)}=Id_{G(d)}$

and $\mu_{G(d)}\circ G(\lambda_d)=Id_{G(d)}$, $F(\mu_c)\circ \lambda_{F(c)}=Id_{F(c)}$

To show they are equivalencies of categories I want to find natural isomorphisms : $\sigma:Id_C\to G\circ F$ and $\tau: F\circ G\to Id_D$. Natural candidates are $\eta,\epsilon$ with inverses given by $\mu,\lambda$ respectively. So it would be enough to show for example that $\eta_c\circ \mu_c=Id_{G\circ F(c)}$ and $\mu_c\circ \eta_c=Id_c$ but I can't seem to prove it. I don't understand how the diagrams we have enable us to get these relations.

Is it the way to go ?

raisinsec
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    It’s not true. There are counterexamples. (Look up “chains of adjoints”.) – Zhen Lin Mar 15 '23 at 13:18
  • @ZhenLin Ah interesting because it's an exercise, I'll check chains of adjoins then. Is the converse true ? That is if we have an equivalence, then we get the chain ? – raisinsec Mar 15 '23 at 13:35
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    Yes. Any equivalence is both a left adjoint and a right adjoint so you get an infinite alternating chain. – Zhen Lin Mar 15 '23 at 14:35
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    See also a counterexample here – Arnaud D. Mar 20 '23 at 08:34
  • @ZhenLin I was told that it was true if $F$ is fully faithful. However I still can't show it. My naives candidates still don't seem to work (how do I show $\eta_c\circ \mu_c=id_c$ even with bijectivity ?) so I tried using the different diagrams we have to create new natural transformations from the ones we have but it doesn't work. Do you have any hint ? – raisinsec Mar 22 '23 at 13:48
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    An adjunction where the right adjoint is fully faithful has the property that its counit is an isomorphism. Dually, an adjunction where the left adjoint is fully faithful has the property that its unit is an isomorphism. – Zhen Lin Mar 22 '23 at 14:00
  • @ZhenLin So here the only data we need to show the statement is $F\dashv G$, with your result we would get that both natural transformations are isomorphisms directly giving the result. $G\dashv F$ doesn't bring any new information ? – raisinsec Mar 22 '23 at 14:14
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    If you knew both $F$ and $G$ were fully faithful then it would suffice to know that one is adjoint to the other. But you only know $F$ is fully faithful. – Zhen Lin Mar 22 '23 at 15:41

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