# Category Theory in Context, Proposition 4.4.5: Promoting equivalence to adjunction

Begin with an equivalence $$F: C \to D$$, $$G: D \to C$$ along with natural isomorphisms_ to the identities, so $$\eta: 1_C \simeq GF$$ and $$\epsilon: FG \simeq 1_D$$. The claim is that we can replace $$\epsilon$$ by $$\epsilon'$$ such that the resulting ($$\eta$$, $$\epsilon'$$) pair obey the triangle identities.

Begin by defining $$\gamma \equiv G \overset{\eta G}{\rightarrow} GFG \overset {G \epsilon}{\rightarrow} G$$. This need not be identity, and will not be if $$(\eta, \epsilon)$$ is not an adjunction. So define $$\epsilon' \equiv \epsilon \circ F\gamma^{-1}$$. The claim is that by "the naturality of the $$\eta$$ the diagram below commutes, which proves one triangle identity $$G \epsilon' \circ \eta G = 1_G$$ ": 1. Which part of the diagram is $$G \epsilon'$$?
2. Which part of the diagram is $$\eta G$$?
3. Why does the left triangle commute?
4. Does the right triangle commute because of the definition of $$\gamma$$?
5. Does the commutativity of the whole diagram follow from the commutativity of the left and right triangles?

My answers to the questions are these, which I feel are probably wrong:

1. $$G\epsilon' = G\epsilon \circ GF \gamma^{-1}$$. So the right-hand side of the top row is $$G \epsilon'$$
2. $$\eta G$$ is the left-hand side of the top row.
3. Computing the bottom leg, we get: $$\eta G \circ \gamma^{-1} = \eta G \circ (G \epsilon \circ \eta G)^{-1} = \eta G \circ (\eta G)^{-1} \circ (G \epsilon)^{-1} = (G\epsilon)^{-1}$$. Computing the top leg, we get $$GF\gamma^{-1} \circ \eta G = GF((G\epsilon)^{-1} \circ (\eta G)^{-1}) \circ \eta G$$. I'm stuck here, I'm not sure how to reduce this.
4. By definition, $$\gamma = G \epsilon \circ \eta G$$, from which it follows that the right triangle commutes.
##### Full Snippet, Prosition 4.4.5, Promoting equivalence to adjoint equivalence 1. The second and third horizontal natural transformations are $$G\epsilon \circ GF \gamma^{-1} = G(\epsilon \circ F\gamma^{-1}) = G \epsilon'$$.
2. I don't quite follow this question - there are two arrows labelled $$\eta G$$.
3. That's naturality of $$\eta G : G \implies GFG$$ applied to $$\gamma^{-1}$$.