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The following may be a basic question.

Let $\mathscr{C}, \mathscr{D}$ be categories. Assume that $F: \mathscr{C}\to \mathscr{D}$ and $G: \mathscr{D}\to \mathscr{C}$ define an equivalence of categories, i.e. there are natural isomorphisms $$\epsilon: FG \to \operatorname{id}_\mathscr{D}, \quad \eta: \operatorname{id}_{\mathscr{C}} \to GF.$$

Are the identities $$F(\eta_U)= \epsilon_{FU}^{-1}, \quad G(\epsilon_X) = \eta_{GX}^{-1}$$ automatically satisfied?

I think we can choose $\epsilon, \eta$ to be natural isomorphisms satisfying these identities, but I'm not sure if they are automatic. After trying a little bit, I am starting to believe that these identities are not automatically satisfied. Can someone comfirm this hunch?

Andromeda
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    Those identities in fact do not hold in general; they are equivalent to the triangular identities, since $\epsilon,\eta$ are isos, and when they are satisfied one talks of adjoint equivalence. – Ezio Greggio Jun 25 '22 at 21:33
  • @EzioGreggio Thanks! But it is true that if there is an equivalence, there is also an adjoint equivalence? (I think it is, but just doublechecking). – Andromeda Jun 25 '22 at 21:34
  • https://math.stackexchange.com/q/2096446/815585 – FShrike Jun 25 '22 at 21:35
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    It can be shown that every equivalence yields an adjunction but the unit and counit need not be the same – FShrike Jun 25 '22 at 21:35
  • This is because they need not satisfy the triangle identities. You modify the unit/counit appropriately so that it works out – FShrike Jun 25 '22 at 21:36
  • @FShrike Yes, but do the unit and the counit of the adjoint equivalence remain isomorphisms? – Andromeda Jun 25 '22 at 21:36
  • See Ezio’s comment – FShrike Jun 25 '22 at 21:38
  • @Andromeda yes, since $\eta$ and $\epsilon$ have the universal property of resp. the unit and the counit of an adjunction, you can recover an adjoint equivalence for each one. Maybe you would find this useful: https://math.stackexchange.com/questions/4411590/clarify-on-adjoint-equivalences – Ezio Greggio Jun 25 '22 at 21:39
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    @EzioGreggio Thanks! Perhaps you like to summarise your comments to my question in an answer? I'll gladly accept :) – Andromeda Jun 25 '22 at 21:41
  • This post also gives a nice overview: https://math.stackexchange.com/questions/3359074/given-an-equivalence-f-g-eta-epsilon-one-has-f-dashv-g?rq=1 We can change one of the natural isomorphisms by another isomorphism to obtain an adjoint equivalence. – Andromeda Jun 25 '22 at 21:44

1 Answers1

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Let $F:\sf C\to \sf D$ be an equivalence, coming with a quasi-inverse $ G:\sf D\to C$ and natural isos $\eta:\mathrm{id}_\mathsf C\Rightarrow GF$, $\epsilon:GF\Rightarrow\mathrm{id}_\mathsf D$.

Suppose to have an arrow $f:c\to Gd$, for $c\in\mathsf C,d\in \mathsf D$. Since $\eta_c$ is an iso, you get an arrow $f\circ \eta_c^{-1}:GFc\to Gd$. As $G$ is fully faithful, there is a unique arrow $g:Fc\to d$ such that $Gg=f\circ \eta_c^{-1}$; i.e. there is a unique $g:Fc\to d$ such that $Gg\circ \eta_c=f$, proving $F\dashv G$. Call $\xi$ the counit of this adjunction: it is a natural iso, because, for every $d\in \sf D$, one has $\mathrm {id}_{Gd}=G\xi_d\circ \eta_{Gd}$, meaning that $G\xi_d$ is an iso, and so that $\xi_d$ is an iso, for a fully faithful functor reflects isos. (But even if the unit is not an iso, you can prove that if $F\dashv G$ and $G$ is fully faithful, the counit is already an iso).

Hence you can recover an adjoint equivalence $F\dashv G$ with unit $\eta$, and the same is true for the counit, by duality.

Ezio Greggio
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