If functors $F,G$ constitute an equivalence, then they are adjoint in both directions

Suppose functors $F:\mathscr{A} \to \mathscr{B}$ and $G: \mathscr{B} \to \mathscr{A}$ constitute an equivalence of categories, in that there exist natural isomorphisms $$\eta : 1_{\mathscr{A}} \to GF,$$ $$\gamma:FG \to 1_{\mathscr{B}}.$$ I read a proof that, if necessary, $\gamma$ may be exchanged for another natural isomorphism $$\varepsilon:FG \to 1_{\mathscr{B}}$$ so that the pair $\eta,\varepsilon$ satisfy the triangle identities necessary to be the unit and co-unit for an adjunction $$F \dashv G.$$

Explicitly, the author takes as $\varepsilon$ the composite $$FG \xrightarrow{FG \gamma^{-1}} FGFG \xrightarrow{(F\eta G)^{-1}} FG \xrightarrow{\quad \gamma \quad} 1_{\mathscr{B}}.$$

Now, I actually follow the proof - I just don't understand his motivation for defining $\varepsilon$ that way.

For instance, with $\varepsilon$ so defined, the first triangle identity boils down to the commutativity of the following diagram (where $A \in \mathrm{ob}(\mathscr{A})$): commutativity that follows from the (proven) equalities like $FG \gamma^{-1} = \gamma^{-1}FG$ and $GF \eta = \eta GF$, and the observation that the top square of the diagram is a particular naturality square of $\gamma^{-1}$.

Like I said, once he declared what $\varepsilon$ was going to be, I understand every claim he makes. My question is, what motivation led him to that choice of $\varepsilon$? I don't see what's intuitive about its construction - its composite expression still seems arbitrary.

Is there obvious motivation that I am missing? I am familiar with the basics of whiskering and horizontal composition of natural transformations. BTW here is a link to the proof: http://www.logicmatters.net/resources/pdfs/GentleIntro.pdf (page 262)

I do not know about any intuitive and direct argument that allows to prove that the $\epsilon$ defined in the question is the counit of the adjoint equivalence, nevertheless there is a little indirect argument that makes the said $\epsilon$ naturally arise.

The (wanna be) unit $\eta \colon 1_{\mathcal A} \to GF$ of the equivalence induces a family of morphisms between the presheaves $$\mathcal B[Fx , -] \stackrel{\varphi_{x}}{\longrightarrow} \mathcal A[x,G-]$$ $$\alpha \in \mathcal B[Fx,y] \mapsto G\alpha \circ \eta_x \in \mathcal A[x,Gy]$$ which are natural in $y$.

With a little effort one can prove that also the morphism $$\mathcal A[x,G-] \stackrel{\psi_x}\longrightarrow \mathcal B[Fx,-]$$ $$\beta \in \mathcal A[x,Gy] \mapsto \gamma_y \circ F\beta \in \mathcal B[Fx,y]$$ is a natural transformation which composed with $\varphi_x$ gives us a natural transformation $$\mathcal B[Fx,-] \longrightarrow \mathcal B[Fx,-]$$ $$\alpha \in \mathcal B[Fx,y] \mapsto \gamma_y \circ FG\alpha \circ F\eta_x=\alpha \circ \gamma_{Fx} \circ F\eta_x\ .$$

This natural isomorphism is the natural one induced by right composition with the morphism $\gamma_{Fx} \circ F\eta_x$, which is an isomorphism, being the composite of two isomorphisms.

So we ended up having that $\psi_x \circ \varphi_x = \mathcal B[\gamma_{Fx}\circ F\eta_x,-]$ hence we have the equality $$\mathcal B[F\eta_x^{-1}\circ\gamma_{Fx}^{-1},-]\circ \psi_x \circ \varphi_x = 1_{\mathcal B[Fx,-]}$$ which proves that $$\mathcal B[Fx,-] \stackrel{\varphi_x}\longrightarrow \mathcal A[x,G-]$$ is a natural isomorphism and hence by one of the characterizations of adjunctions this proves that the families of functions $$\varphi_{x,y} \colon \mathcal B[Fx,y] \longrightarrow \mathcal A[x,Gy]$$ $$\alpha \in \mathcal B[Fx,y] \mapsto G\alpha \circ \eta_x$$ give us a natural transformation between the two functors $\mathcal B[F-,-]$ and $\mathcal A[-,G-]$, i.e. an adjunction whose unit is $\eta$ and the counit $\epsilon$ is obtained by yoneda as the morphism $$\epsilon_y = \varphi_{x,y}^{-1}(1_{Gy})=\mathcal B[F\eta_x^{-1}\circ \gamma_{Fx}^{-1},y]\circ\psi_x(1_{Gy})=\gamma_y \circ F\eta_x^{-1} \circ \gamma_{Fx}^{-1}$$ that is the wished morphism.

Hope this helps.

• Thank you for taking the time to respond. I'm going over your answer now
– joeb
Jun 26 '17 at 21:55