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I know that this question has been already asked (Is an equivalence an adjunction?). However I am dealing now with the proof and I can't figure out how to write it properly.

This is my idea and the way I would like to follow: Let's consider $(F, G,\eta,\varepsilon)$ an equivalence of categories, so $F:\mathcal{A}\longrightarrow\mathcal{B}$ and $G:\mathcal{B}\longrightarrow\mathcal{A}$ functors, $\eta: 1_{\mathcal{A}}\longrightarrow G\circ F$ and $\varepsilon: F\circ G\longrightarrow 1_{\mathcal{B}}$ natural isomorphisms, that is $\eta_A: A\longrightarrow GF(A)$ and $\varepsilon_B: FG(B)\longrightarrow B$ isomorphisms in $\mathcal{A}$ and $\mathcal{B}$ respectively for every $A\in Ob(\mathcal{A})$ and $B\in Ob(\mathcal{B})$.

We need to show $F\dashv G$, which means that for all $A\in Ob(\mathcal{A})$ we need to prove that $\eta_A:A\longrightarrow GF(A)$ is an initial object in $(A\Rightarrow G)$. We recall that the comma category $(A\Rightarrow G)$ has got functions of the form $f:A\longrightarrow G(B)$ with $B\in Ob(\mathcal{B})$ as objects (more properly they are of the type $(B,f:A\longrightarrow G(B))$ ) and a map in $(A\Rightarrow G)$ between two objects $f:A\longrightarrow G(B)$ and $f':A\longrightarrow G(B')$ is given by a map $g \in \mathcal{B}(B,B')$ such that $f'=G(g)\circ f$.

Let $A\in Ob(\mathcal{A})$ (so $F(A)\in Ob(\mathcal{B})$), $\eta_A: A\longrightarrow G(F(A))\in Ob(A\Rightarrow G)$ and $f:A\longrightarrow G(B) \in Ob(A\Rightarrow G)$. We want now to prove that there exists a unique map $g:F(A)\longrightarrow B$ in $(A\Rightarrow G)$ such that $f=G(g)\circ \eta_A$.

Since $\eta$ is a natural isomorphism, in particular a natural transformation, between $1_{\mathcal{A}}$ and $G\circ F$, we infer that the following diagram commutes: $$ \begin{array}{ccc} A & \overset{f}{\longrightarrow} & A' \\ \scriptstyle{\eta_A}\downarrow\scriptstyle{\cong} & & \scriptstyle{\cong}\downarrow \scriptstyle{\eta_{A'}} \\ GF(A) & \overset{GF(f)}{\longrightarrow} & GF(A') \end{array} $$ for all $A,A'\in Ob(\mathcal{A})$, $f\in \mathcal{A}(A,A')$. Now we can choose $A'\equiv G(B)$ and define $g:=F(f)$, so we get: $$ \begin{array}{ccc} A & \overset{f}{\longrightarrow} & G(B) \\ \scriptstyle{\eta_A}\downarrow\scriptstyle{\cong} & & \scriptstyle{\cong}\downarrow \scriptstyle{\eta_{G(B)}} \\ GF(A) & \overset{GF(f)}{\longrightarrow} & GF(G(B)) \end{array} $$ and we would find: $f=\eta_{G(B)}^{-1}\circ G(g)\circ \eta_A$. But there is the $\eta_{G(B)}^{-1}$ and I don't want it there...how can I send it away? (if this is possible, otherwise I need to change approach to the proof)

Thanks very much for any help!

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    $\begingroup$ Not every equivalence is an adjunction: you may have to change the unit or the counit. $\endgroup$ – Kevin Arlin Jan 13 '17 at 22:35
  • $\begingroup$ Yes, this I know. However what is the best way to change it? I thought about something like $A\longrightarrow GF(GF(A))$, but after a while this haven't sounded sensical to me anymore. $\endgroup$ – any_one Jan 14 '17 at 13:04
  • $\begingroup$ Just for the reference, the notation used in this question comes from Leinster's book (section 2.3). $\endgroup$ – user634426 Jun 22 '20 at 1:01
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Hint: an equivalence is fully faithful, so that

$$G: Hom(FG(B), B) \to Hom(GFG(B), G(B)) $$

is surjective.

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  • $\begingroup$ Hi David, was my suggestion to use botox inappropriate? $\endgroup$ – Andrea Marino Mar 27 at 21:47

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