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Suppose that $A$ and $B$ are equivalent categories. Then there exist functors $F: A \rightarrow B$ and $G: B \rightarrow A$ such that $F \circ G \cong 1$ and $G \circ F \cong 1$. I am trying to show that $G$ is a left and right adjoint of $F$ (Actually, the problem states that these functors have left and right adjoints, but I think that these are the correct adjoints).

I have been trying to work with the definition of adjunction that two functors are adjoint when $B(F(A), B) \cong A(A, G(B))$. I haven't been able to work out isomorphism is that would show that these two fuctors are adjoint. How should I proceed?

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  • $\begingroup$ I just want to emphasize a subtlety here : even if indeed $F$ and $G$ are adjoint, the natural isos $FG\simeq 1$ and $GF\simeq 1$ that show that there are equivalences need not be the unit and counit of the adjunction. The nlab explains that you can keep one of them, but the other one might need to be adjusted. $\endgroup$
    – Pece
    Nov 7 '17 at 15:19
  • $\begingroup$ Does this affect the answer given by @Long? His answer seems to avoid considerations of the unit and couunit. $\endgroup$
    – Topos
    Nov 7 '17 at 15:35
  • $\begingroup$ No it does not, the answer seems good to me. In fact, Long's answer inscribes in this remark as follow: it shows that $F$ is left adjoint to $G$ and keep the natural iso $1 \overset \simeq \to GF$ to be the unit; so it will be the counit that might not be exactly the isos $FG\overset \simeq \to 1$ that we have. (This is very understandable as this natural iso does not play any role in Long's proof.) $\endgroup$
    – Pece
    Nov 7 '17 at 16:06
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$(Fa,b) \simeq (GFa,Gb) \simeq (a,Gb)$ The first bijection is due to $G$ being a category equivalence and the second is via composing with the natural transformation $Id \to GF $.

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  • $\begingroup$ Does the first bijection follow from the fact that $G$ is full? $\endgroup$
    – Topos
    Nov 7 '17 at 15:02
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    $\begingroup$ @Topos $G$ being a functor only gives you a function $(Fa,b)\to (GFa,Gb)$. This will be a bijection if and only if $G$ is full and faithful. $\endgroup$
    – Arnaud D.
    Nov 7 '17 at 15:04
  • $\begingroup$ @Long How would I show that this isomorphism is natural in $A$ and $B$ though? Don't I also need to show that these functors are adjoint? $\endgroup$
    – Topos
    Nov 7 '17 at 22:43
  • $\begingroup$ @Topos by definition of naturality; it is quite straight forward. First show the left bijection is natural, then show it for the right. Adjointness is the consequence of the naturality of the bijection - by definition... $\endgroup$
    – Long
    Nov 7 '17 at 22:48
  • $\begingroup$ @Long I'm really new to category theory, so it's not as simple as you'd think for me. Could you show me how to start it off? $\endgroup$
    – Topos
    Nov 7 '17 at 23:21

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