TITLE: Borceux. Handbook of Categorical Algebra I. Proposition 3.4.2.
QUESTION [5 upvotes]: I'm trying to understand proposition 3.' HoCA (vol. I).
Proposition 3.4.2 Consider a functor $F\colon \mathcal A \to \mathcal B$ with both a left adjoint functor $G$ and a right adjoint functor $H$. If functor is full and faithful, si is the other adjoint functor.
There are some minor typos in the proof, but they're not a big deal.
Borceux calls $\epsilon \colon GF \Rightarrow 1_{\mathcal A}$, $\eta\colon 1_{\mathcal B} \Rightarrow FG$ and $\alpha \colon FH \Rightarrow 1_{\mathcal B}$, $\beta\colon 1_{\mathcal A} \Rightarrow HF$ the counits and units of the adjuctions.
By duality, it is enough to prove, wlog, that if H is fully faithful then the same holds for G. Further, this statement is equivalent (by proposition 3.4.1) to say that $\eta$ is iso whenever $\alpha$ is.
The claim is that
$\alpha \circ (1_F *\epsilon * 1_H) \circ (1_{FG}*\alpha^{-1}) = \eta^{-1}$.
By the naturality of $eta$,
$\alpha \circ (1_F *\epsilon * 1_H) \circ (1_{FG}*\alpha^{-1}) \circ \eta = 
\alpha \circ (1_F *\epsilon * 1_H) \circ (\eta*1_{FH})\circ \alpha^{-1}.$
The trangular equality, i.e., $(1_F *\epsilon) \circ (\eta*1_{F}) = 1_{\mathcal F}$ (thus, in particular $(1_F *\epsilon *1_H) \circ (\eta*1_{FH}) = 1_{\mathcal FH}$), gives us
$alpha \circ \alpha^{-1} = 1_{\mathcal B}$.
On the other hand, first note that the triangular equality 
$(\alpha *1_F)\circ (1_F * \beta) = 1_F$
implies that $1_F* \beta = (\alpha *1_F)^{-1}\quad(\heartsuit)$.
Now consider
$\star = \eta\circ \alpha \circ (1_F *\epsilon * 1_H) \circ (1_{FG}*\alpha^{-1})$
$\star = (\text{by naturality of }\alpha) = (\alpha*1_{FG})\circ (1_{FH}*\eta) \circ (1_F *\epsilon * 1_H) \circ (1_{FG}*\alpha^{-1})$
$\star = (\text{by naturality of }1_F *\epsilon * 1_H) = (\alpha*1_{FG})\circ (1_F *\epsilon * 1_{HFG})\circ (1_{FGFH}*\eta) \circ (1_{FG}*\alpha^{-1})$
$\star = (\text{by naturality of }1_{FG}*\alpha^{-1}) = (\alpha*1_{FG})\circ (1_F *\epsilon * 1_{HFG})\circ (1_{FG}*\alpha^{-1}*1_{FG}) \circ (1_{FG} *\eta)$
Using $(\heartsuit)$, $\star$ is thus equal to
$(\alpha *1_{FG}) \circ (1_F*\epsilon *1_{HFG}) \circ (1_{FGF}*\beta *1_G) \circ (1_{FG}*\eta)$.
(Borceux writes
$(1_{FG}*\alpha) \circ (1_F*\epsilon *1_{HFG}) \circ (1_{FGF}*\beta *1_G) \circ (1_{FG}*\eta)$,
but I think there's a small typo).
The next (and last) equality is
$(\alpha *1_{FG}) \circ (1_F*\epsilon *1_{HFG}) \circ (1_{FGF}*\beta *1_G) \circ (1_{FG}*\eta) = (1_F*1_G)\circ (1_F*1_G)$.
Where does this equality come from?
Aside question. Proving all these last equalities I've realized that, if $A$ is a category, $F, G\colon A \to A$ are endofunctors, $\alpha\colon F\to 1_A$, $\beta\colon 1_A \to G$ are natural transformations, then
$\beta \circ \alpha = (\alpha *1_G)\circ (1_F *\beta)$.
I suppose that, maybe using the interchange law for 2-cells and doing some checks, lemma general enough to prove immediately the equalities in the last part of the proof.

REPLY [4 votes]: I will use the same notation. By naturality,
$$(1_F * \epsilon * 1_{H F G}) \circ (1_{F G F} * \beta * 1_G) = (1_F * \beta * 1_G) \circ (1_F * \epsilon * 1_G)$$
and so
$$(\alpha * 1_{F G}) \circ (1_F * \epsilon * 1_{H F G}) \circ (1_{F G F} * \beta * 1_G) \circ (1_{F G} * \eta)$$
is equal to
$$(\alpha * 1_{F G}) \circ  (1_F * \beta * 1_G) \circ (1_F * \epsilon * 1_G) \circ (1_{F G} * \eta)$$
which by the triangle identities is just $1_{F G}$.