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Let $C$ and $D$ be categories and $F:C\to D$, $G:D\to C$ two functors.

$F$ is left-adjoint to $G$, if there are natural transformations $\eta:id_C\to GF$ and $\epsilon:FG\to id_D$ such that \begin{eqnarray} F&\xrightarrow{F\eta}&FGF&\xrightarrow{\epsilon F}F\\ G&\xrightarrow{\eta G}&GFG&\xrightarrow{G\epsilon}G \end{eqnarray} are the identity (!) transformations.

$F$ is an equivalence of categories (with inverse $G$) if there are natural isomorphisms $\eta:id_C\to GF$ and $\epsilon:FG\to id_D$ without any further properties.

Is $F$ left-adjoint to $G$, if $F$ is an equivalence of categories (with inverse $G$)?
If not, suppose that $F$ is an equivalence of categories with inverse $G':D\to C$ and suppose further that $F$ is left-adjoint to $G$. Does it follow that there is a natural isomorphism $G\to G'$ or is there even an identity $G=G'$?

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  • $\begingroup$ The answer to the first question is "yes", $F$ is both left-adjoint and right-adjoint to $G$. The answer to the second question is "yes, there exists a natural isomorphism $G\to G'$, but it is not necessarily an identity". $\endgroup$
    – Oskar
    Dec 6, 2013 at 14:47

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Every equivalence can be "improved" to an adjoint equivalence, but you may have to change the unit or the counit: see e.g. here for a proof. Essentially the same thing is proved as Theorem 4.2.3 in the Homotopy Type Theory book.

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  • $\begingroup$ What is the Homotopy Type Theory book? $\endgroup$ Dec 6, 2013 at 16:35
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    $\begingroup$ @MartinBrandenburg: it can be characterized as either the unique book with title "Homotopy Type Theory" or as the unique book on the subject of Homotopy Type Theory. $\endgroup$ Dec 6, 2013 at 20:55
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    $\begingroup$ homotopytypetheory.org/book $\endgroup$
    – Berci
    Dec 8, 2013 at 0:42
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    $\begingroup$ Adjoints are unique up to unique isomorphism. $\endgroup$
    – Zhen Lin
    Dec 8, 2013 at 17:53
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    $\begingroup$ Suppose we have already shown that the natural iso $\eta : 1_{\mathbf{C}} \to GF$ witnessing the equivalence is the unit of the desired adjunction. Why can't the counit just be the other iso $\epsilon : FG \to 1_{\mathbf{D}}$ of the equivalence in general? It satisfies the UMP the counit since for each $g: FC \to D$, we have $\epsilon_D \circ (\epsilon_D^{-1} \circ g) = g$ and $\epsilon_D^{-1} \circ g = Ff$ for some unique $f : C \to GD$, since we can show that $F$ is fully faithful. What is wrong with this argument? $\endgroup$ Jan 28, 2016 at 7:35

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