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 now is that the map $\epsilon'_F \circ F \eta$ is idempotent.
- This appears like a typo to me. Should it not be $\epsilon' F \circ F \eta$ is idempotent, from the triangle identity? (notice that the $F$ is not a subscript of $\epsilon'$).
Next, the claim is that an idempotent isomorphism is an identity.
- Does this follow because $s^2 = s$ where $s$ is an iso implies that $s^2 \circ s^{-1} = s \circ s^{-1}$, or $s = id$?
Finally, the claim is that $\epsilon' F \circ F \eta$ is idempotent follows by chasing the diagram:
What is $\eta_{GF}, \epsilon'_F, \epsilon'_{FGF}$? I imagine that it really means $\eta GF$, $\epsilon' F$, and $\epsilon' FGF$?
How does one conveniently show that the diagram really does commute?
And finally, I imagine that there must be an easier way to prove these theorems without chasing commutative diagrams? Do string diagrams help? Some other formalism?


(1) and (3): For a natural transformation $\alpha$ and a (composable) functor $T$ we have by definition $$(\alpha T)_x:=\alpha_{T(x)}$$ this explains why the notations $\alpha T$ and $\alpha_T$ mean the same thing.
(2): Yes, correct.
(4): Each square commutes because of a naturality condition of one of the natural transformations. And the lower triangle commutes because of definition of $\epsilon'$.
(5): Well, yes, one can do a proof using string diagrams or general bicategorical arguments, though I wouldn't say they are 'simpler' or 'easier' (at least to me).