Proof.
We will prove that $(1) \Rightarrow (2)$; the remaining assertions follow immediately from Proposition 6.1.2.9. Fix an object $D \in \operatorname{\mathcal{D}}$. Applying assertion $(1)$ in the case $C = G(D)$, we deduce that there is a unique morphism $\epsilon _{D}: (F \circ G)(D) \rightarrow D$ for which the composition
\[ G(D) \xrightarrow {\eta _{G(D)}} (G \circ F \circ G)(D) \xrightarrow { G( \epsilon _ D)} G(D) \]
is the identity morphism from $G(D)$ to itself.
We first claim that the construction $D \mapsto \epsilon _ D$ is a natural transformation of functors from $F \circ G$ to $\operatorname{id}_{\operatorname{\mathcal{D}}}$. Let $h: D \rightarrow D'$ be a morphism in the category $\operatorname{\mathcal{D}}$; we wish to show that the diagram
\[ \xymatrix@R =50pt@C=50pt{ (F \circ G)(D) \ar [d]^-{ (F \circ G)(h) } \ar [r]^-{ \epsilon _ D } & D \ar [d]^{h} \\ (F \circ G)(D') \ar [r]^-{ \epsilon _{D'} } & D' } \]
commutes. Consider the diagram
\[ \xymatrix@R =50pt@C=50pt{ G(D) \ar [r]^-{ \eta _{G(D)} } \ar [d]^{ G(h) } & (G \circ F \circ G)(D) \ar [r]^-{ G(\epsilon _ D)} \ar [d]^{G(F(G(h))) } & G(D) \ar [d]^{ G(h) } \ar [d] \\ G(D') \ar [r]^-{ \eta _{G(D')} } & (G \circ F \circ G)(D') \ar [r]^-{ G(\epsilon _{D'}) } & G(D') } \]
in the category $\operatorname{\mathcal{C}}$. It follows from the definitions of $\epsilon _ D$ and $\epsilon _{D'}$ that both horizontal compositions are equal to the identity, so the outer rectangle commutes. Since $\eta $ is a natural transformation, the left square commutes. It follows that the compositions $G(h) \circ G(\epsilon _ D) \circ \eta _{G(D)}$ and $G(\epsilon _{D'}) \circ G(F(G(h))) \circ \eta _{G(D)}$ are the same: that is, the morphisms
\[ h \circ \epsilon _ D, \epsilon _{D'} \circ F(G(h)) \in \operatorname{Hom}_{\operatorname{\mathcal{D}}}( (F \circ G)(D), D') \]
have the same right adjunct. Invoking assumption $(1)$, we deduce that $h \circ \epsilon _{D} = \epsilon _{D'} \circ F(G(h))$, as desired.
It follows immediately from the construction that the pair of natural transformations $(\eta , \epsilon )$ satisfies condition $(Z2)$ of Definition 6.1.0.2. To complete the proof, it will suffice to show that it also satisfies condition $(Z1)$. Let $C$ be an object of $\operatorname{\mathcal{C}}$; we wish to show that the composite map
\[ F(C) \xrightarrow { F( \eta _ C )} (F \circ G \circ F)(C) \xrightarrow { \epsilon _{ F(C) } } F(C) \]
is equal to the identity map $\operatorname{id}_{F(C)}$. Note that the right adjunct of $\epsilon _{F(C)} \circ F(\eta _ C)$ is the composite map
\[ C \xrightarrow {\eta _ C} (G \circ F)(C) \xrightarrow { (G \circ F)(\eta _ C) } (G \circ F \circ G \circ F)(C) \xrightarrow { G( \epsilon _{F(C)})} (G \circ F)(C). \]
By virtue of the fact that $(\eta , \epsilon )$ satisfies $(Z2)$, this composition is equal to $\eta _ C$, which is also the right adjunct of the identity map $\operatorname{id}_{F(C)}$. Invoking assumption $(1)$, we conclude that $\epsilon _{F(C)} \circ F(\eta _ C) = \operatorname{id}_{ F(C)}$, as desired.
$\square$