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If functors $F: \mathcal C \leftrightarrows \mathcal D: G$ furnish us with an equivalence of categories, is it necessarily the case that $F$ is left adjoint to $G$?

This question has been discussed on StackExchange before, but the accepted answer felt unsatisfying. The answer links this nLab page which discusses the need to modify the unit or counit to upgrade an equivalence to an adjoint equivalence. I'm having trouble understanding why the units and counits take such a central place in the discussion, however.

In fact, we can relax the hypotheses to ask the following more general(?) question.

Suppose that $F: \mathcal C \to \mathcal D$ is a fully faithful functor and assume the existence of another functor $G: \mathcal D \to \mathcal C$ such that there is a natural isomorphism $\epsilon: FG \cong \mathrm{id}_\mathcal{D}$. Is it true that $F$ is a left adjoint to $G$?

Assuming that the definition of an adjunction $F \vdash G$ is simply $\mathrm{Hom}(FC, D) \cong \mathrm{Hom}(C, GD)$ for all $C \in \mathcal C, D \in \mathcal D$, the answer appears to be yes via the following reasoning. $$ \mathrm{Hom}(C, GD) \cong \mathrm{Hom}(FC, FGD)\cong \mathrm{Hom}(FC, D) $$ where the first iso follows from the faithfulness of $F$, and the second from $\epsilon: FG \cong \mathrm{id}_\mathcal{D}$. If I had to guess, the second iso is where issues arise.

Posetologist
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1 Answers1

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An adjunction is a pair of functors together with a choice of natural isomorphism $\mathrm{Hom}(C,GD)\cong \mathrm{Hom}(FC,D)$, a choice which is equivalent to choosing the unit and counit. Similarly, to give an equivalence is to give $F,G,$ and also the unit and counit. It's possible to choose the unit and counit of the equivalence such that they don't satisfy the triangle identities, which gives an equivalence that's not an adjunction. It's true, though, that if $F$ is an equivalence inverse to $G,$ then $F$ is a left adjoint of $G$ (for an appropriate choice of unit and counit.)

Kevin Carlson
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