[proofplan] We prove both implications directly from the adjunction data. First, if $G$ is fully faithful, we use fullness to lift the unit component $\eta_{Gd}:Gd\to GFGd$ to a morphism $\delta_d:d\to FGd$, and faithfulness then lets us check after applying $G$ that $\delta_d$ is inverse to $\varepsilon_d$. Conversely, if every counit component is an isomorphism, we identify the map on hom-sets induced by $G$ with an explicit composite of adjunction bijections and postcomposition by $\varepsilon_d^{-1}$; this gives bijectivity on every hom-set, hence full faithfulness. [/proofplan] [step:Lift the unit through full faithfulness and construct a candidate inverse to the counit] Assume first that $G$ is fully faithful. Fix an object $d\in \mathcal D$. Since $G$ is full, the map \begin{align*} G_{d,FGd}:\operatorname{Hom}_{\mathcal D}(d,FGd) &\to \operatorname{Hom}_{\mathcal C}(Gd,GFGd) \\ \alpha &\mapsto G\alpha \end{align*} is surjective. Therefore there exists a morphism $\delta_d:d\to FGd$ in $\mathcal D$ such that \begin{align*} G\delta_d=\eta_{Gd}. \end{align*} We will prove that $\delta_d$ is a two-sided inverse to $\varepsilon_d$. [guided] Assume $G$ is fully faithful, and fix $d\in \mathcal D$. To prove that $\varepsilon_d:FGd\to d$ is an isomorphism, we must construct a morphism going in the opposite direction. The only natural morphism with the right shape after applying $G$ is the unit component \begin{align*} \eta_{Gd}:Gd\to GFGd. \end{align*} Fullness of $G$ says that every morphism between objects in the image of $G$ comes from a morphism in $\mathcal D$. More precisely, the function \begin{align*} G_{d,FGd}:\operatorname{Hom}_{\mathcal D}(d,FGd) &\to \operatorname{Hom}_{\mathcal C}(Gd,GFGd) \\ \alpha &\mapsto G\alpha \end{align*} is surjective. Applying this surjectivity to $\eta_{Gd}$ gives a morphism $\delta_d:d\to FGd$ such that \begin{align*} G\delta_d=\eta_{Gd}. \end{align*} The strategy is now to show that $\delta_d$ is inverse to $\varepsilon_d$. Because $G$ is faithful, it is enough to prove the required equalities after applying $G$. [/guided] [/step] [step:Show the lifted morphism is a right inverse to the counit] By the triangle identity for the adjunction $F\dashv G$, applied to the object $d\in\mathcal D$, we have \begin{align*} G\varepsilon_d\circ \eta_{Gd}=\operatorname{id}_{Gd}. \end{align*} Using $G\delta_d=\eta_{Gd}$, this gives \begin{align*} G(\varepsilon_d\circ \delta_d) =G\varepsilon_d\circ G\delta_d =G\varepsilon_d\circ \eta_{Gd} =\operatorname{id}_{Gd} =G(\operatorname{id}_d). \end{align*} Since $G$ is faithful, it follows that \begin{align*} \varepsilon_d\circ \delta_d=\operatorname{id}_d. \end{align*} [guided] We first prove that $\delta_d$ is a right inverse to $\varepsilon_d$, meaning $\varepsilon_d\circ \delta_d=\operatorname{id}_d$. The second triangle identity for the adjunction $F\dashv G$ states that, for every object $d\in\mathcal D$, \begin{align*} G\varepsilon_d\circ \eta_{Gd}=\operatorname{id}_{Gd}. \end{align*} Our construction of $\delta_d$ gave $G\delta_d=\eta_{Gd}$. Substituting this into the triangle identity yields \begin{align*} G(\varepsilon_d\circ \delta_d) =G\varepsilon_d\circ G\delta_d =G\varepsilon_d\circ \eta_{Gd} =\operatorname{id}_{Gd} =G(\operatorname{id}_d). \end{align*} Faithfulness of $G$ means that the map on each hom-set is injective. Since $\varepsilon_d\circ \delta_d$ and $\operatorname{id}_d$ are morphisms $d\to d$ whose images under $G$ are equal, faithfulness gives \begin{align*} \varepsilon_d\circ \delta_d=\operatorname{id}_d. \end{align*} [/guided] [/step] [step:Show the lifted morphism is a left inverse to the counit] It remains to prove $\delta_d\circ \varepsilon_d=\operatorname{id}_{FGd}$. Use the adjunction bijection \begin{align*} \Phi_{Gd,FGd}:\operatorname{Hom}_{\mathcal D}(FGd,FGd)&\to \operatorname{Hom}_{\mathcal C}(Gd,GFGd) \\ h&\mapsto Gh\circ \eta_{Gd}. \end{align*} This map is injective because it is a bijection. For $h=\delta_d\circ\varepsilon_d$, functoriality, the definition of $\delta_d$, and the triangle identity give \begin{align*} \Phi_{Gd,FGd}(\delta_d\circ\varepsilon_d) &=G(\delta_d\circ\varepsilon_d)\circ \eta_{Gd} \\ &=G\delta_d\circ G\varepsilon_d\circ \eta_{Gd} \\ &=\eta_{Gd}\circ G\varepsilon_d\circ \eta_{Gd} \\ &=\eta_{Gd}\circ \operatorname{id}_{Gd} \\ &=\eta_{Gd}. \end{align*} For the identity morphism on $FGd$, \begin{align*} \Phi_{Gd,FGd}(\operatorname{id}_{FGd}) =G(\operatorname{id}_{FGd})\circ \eta_{Gd} =\operatorname{id}_{GFGd}\circ \eta_{Gd} =\eta_{Gd}. \end{align*} Injectivity of $\Phi_{Gd,FGd}$ therefore gives \begin{align*} \delta_d\circ \varepsilon_d=\operatorname{id}_{FGd}. \end{align*} Thus $\varepsilon_d$ is an isomorphism with inverse $\delta_d$. [guided] We now prove the other inverse identity, $\delta_d\circ \varepsilon_d=\operatorname{id}_{FGd}$. We must avoid cancelling $\varepsilon_d$ on the left, since the previous step only proved that $\varepsilon_d$ has a right inverse. Instead we use the hom-set bijection supplied by the adjunction. For the objects $Gd\in\mathcal C$ and $FGd\in\mathcal D$, the adjunction gives a bijection \begin{align*} \Phi_{Gd,FGd}:\operatorname{Hom}_{\mathcal D}(FGd,FGd)&\to \operatorname{Hom}_{\mathcal C}(Gd,GFGd) \\ h&\mapsto Gh\circ \eta_{Gd}. \end{align*} Because $\Phi_{Gd,FGd}$ is a bijection, it is injective. Therefore it is enough to show that $\delta_d\circ\varepsilon_d$ and $\operatorname{id}_{FGd}$ have the same image under $\Phi_{Gd,FGd}$. First compute the image of $\delta_d\circ\varepsilon_d$. Functoriality of $G$ gives $G(\delta_d\circ\varepsilon_d)=G\delta_d\circ G\varepsilon_d$, and the construction of $\delta_d$ gives $G\delta_d=\eta_{Gd}$. Hence \begin{align*} \Phi_{Gd,FGd}(\delta_d\circ\varepsilon_d) &=G(\delta_d\circ\varepsilon_d)\circ \eta_{Gd} \\ &=G\delta_d\circ G\varepsilon_d\circ \eta_{Gd} \\ &=\eta_{Gd}\circ G\varepsilon_d\circ \eta_{Gd}. \end{align*} The triangle identity for the adjunction at the object $d\in\mathcal D$ states that \begin{align*} G\varepsilon_d\circ \eta_{Gd}=\operatorname{id}_{Gd}. \end{align*} Substituting this identity into the preceding calculation gives \begin{align*} \Phi_{Gd,FGd}(\delta_d\circ\varepsilon_d) =\eta_{Gd}\circ \operatorname{id}_{Gd} =\eta_{Gd}. \end{align*} Now compute the image of the identity morphism on $FGd$. Since $G$ preserves identity morphisms, \begin{align*} \Phi_{Gd,FGd}(\operatorname{id}_{FGd}) =G(\operatorname{id}_{FGd})\circ \eta_{Gd} =\operatorname{id}_{GFGd}\circ \eta_{Gd} =\eta_{Gd}. \end{align*} Thus \begin{align*} \Phi_{Gd,FGd}(\delta_d\circ\varepsilon_d) =\Phi_{Gd,FGd}(\operatorname{id}_{FGd}). \end{align*} Injectivity of the adjunction bijection $\Phi_{Gd,FGd}$ gives \begin{align*} \delta_d\circ \varepsilon_d=\operatorname{id}_{FGd}. \end{align*} Together with the already proved identity $\varepsilon_d\circ\delta_d=\operatorname{id}_d$, this shows that $\delta_d$ is a two-sided inverse to $\varepsilon_d$. Therefore $\varepsilon_d$ is an isomorphism. [/guided] [/step] [step:Use invertible counits to prove faithfulness of $G$] Conversely, assume that $\varepsilon_d:FGd\to d$ is an isomorphism for every object $d\in\mathcal D$. Let $d,d'\in\mathcal D$ be objects. Suppose $f,g:d\to d'$ are morphisms in $\mathcal D$ such that $Gf=Gg$. Applying $F$ gives \begin{align*} FGf=FGg. \end{align*} By naturality of the counit, we have \begin{align*} f\circ \varepsilon_d=\varepsilon_{d'}\circ FGf, \qquad g\circ \varepsilon_d=\varepsilon_{d'}\circ FGg. \end{align*} Hence \begin{align*} f\circ \varepsilon_d=g\circ \varepsilon_d. \end{align*} Since $\varepsilon_d$ is an isomorphism, it is an epimorphism, so $f=g$. Therefore $G$ is faithful. [guided] Assume now that every counit component $\varepsilon_d:FGd\to d$ is an isomorphism. We first prove faithfulness. Fix objects $d,d'\in\mathcal D$, and let $f,g:d\to d'$ be morphisms with \begin{align*} Gf=Gg. \end{align*} Applying the functor $F$ gives \begin{align*} FGf=FGg. \end{align*} Naturality of the counit $\varepsilon:FG\Rightarrow \operatorname{id}_{\mathcal D}$ says that for every morphism $h:x\to y$ in $\mathcal D$, \begin{align*} h\circ \varepsilon_x=\varepsilon_y\circ FGh. \end{align*} Using this with $h=f$ and then with $h=g$, we get \begin{align*} f\circ \varepsilon_d=\varepsilon_{d'}\circ FGf, \qquad g\circ \varepsilon_d=\varepsilon_{d'}\circ FGg. \end{align*} Since $FGf=FGg$, the right-hand sides agree, so \begin{align*} f\circ \varepsilon_d=g\circ \varepsilon_d. \end{align*} Because $\varepsilon_d$ is an isomorphism, we may compose on the right with its inverse $\varepsilon_d^{-1}:d\to FGd$. This yields \begin{align*} f =f\circ \varepsilon_d\circ \varepsilon_d^{-1} =g\circ \varepsilon_d\circ \varepsilon_d^{-1} =g. \end{align*} Thus the map induced by $G$ on $\operatorname{Hom}_{\mathcal D}(d,d')$ is injective, and $G$ is faithful. [/guided] [/step] [step:Use invertible counits to prove fullness of $G$] Let $d,d'\in\mathcal D$ be objects, and let \begin{align*} u:Gd\to Gd' \end{align*} be a morphism in $\mathcal C$. Define a morphism $\bar u:d\to d'$ in $\mathcal D$ by \begin{align*} \bar u:=\varepsilon_{d'}\circ F u\circ \varepsilon_d^{-1}. \end{align*} We prove that $G\bar u=u$. Naturality of $\eta$ applied to $u:Gd\to Gd'$ gives \begin{align*} GFu\circ \eta_{Gd}=\eta_{Gd'}\circ u. \end{align*} The triangle identity gives \begin{align*} G\varepsilon_d\circ \eta_{Gd}=\operatorname{id}_{Gd}, \qquad G\varepsilon_{d'}\circ \eta_{Gd'}=\operatorname{id}_{Gd'}. \end{align*} Since $\varepsilon_d$ is an isomorphism, applying $G$ shows that $G\varepsilon_d$ is an isomorphism with inverse $G(\varepsilon_d^{-1})$. From \begin{align*} G\varepsilon_d\circ \eta_{Gd}=\operatorname{id}_{Gd} \end{align*} we obtain \begin{align*} G(\varepsilon_d^{-1})=\eta_{Gd}. \end{align*} Therefore \begin{align*} G\bar u &=G\varepsilon_{d'}\circ GFu\circ G(\varepsilon_d^{-1}) \\ &=G\varepsilon_{d'}\circ GFu\circ \eta_{Gd} \\ &=G\varepsilon_{d'}\circ \eta_{Gd'}\circ u \\ &=\operatorname{id}_{Gd'}\circ u \\ &=u. \end{align*} Thus every morphism $u:Gd\to Gd'$ lies in the image of the hom-map induced by $G$, so $G$ is full. [guided] We now prove fullness. Fix objects $d,d'\in\mathcal D$, and let \begin{align*} u:Gd\to Gd' \end{align*} be any morphism in $\mathcal C$. To show fullness, we must find a morphism $\bar u:d\to d'$ in $\mathcal D$ such that $G\bar u=u$. Because the counit components are isomorphisms, the objects $d$ and $d'$ are isomorphic to $FGd$ and $FGd'$. This suggests transporting $u$ across $F$ and then using the counit isomorphisms. Define \begin{align*} \bar u:=\varepsilon_{d'}\circ F u\circ \varepsilon_d^{-1}:d\to d'. \end{align*} The types are correct: $\varepsilon_d^{-1}:d\to FGd$, then $Fu:FGd\to FGd'$, and then $\varepsilon_{d'}:FGd'\to d'$. We compute $G\bar u$. Naturality of the unit $\eta:\operatorname{id}_{\mathcal C}\Rightarrow GF$ applied to $u:Gd\to Gd'$ gives \begin{align*} GFu\circ \eta_{Gd}=\eta_{Gd'}\circ u. \end{align*} The triangle identity gives \begin{align*} G\varepsilon_d\circ \eta_{Gd}=\operatorname{id}_{Gd}, \qquad G\varepsilon_{d'}\circ \eta_{Gd'}=\operatorname{id}_{Gd'}. \end{align*} Since $\varepsilon_d$ is an isomorphism, $G\varepsilon_d$ is an isomorphism with inverse $G(\varepsilon_d^{-1})$. The equality \begin{align*} G\varepsilon_d\circ \eta_{Gd}=\operatorname{id}_{Gd} \end{align*} therefore identifies $\eta_{Gd}$ as the inverse of $G\varepsilon_d$, so \begin{align*} G(\varepsilon_d^{-1})=\eta_{Gd}. \end{align*} Now apply $G$ to the definition of $\bar u$ and substitute these identities: \begin{align*} G\bar u &=G\varepsilon_{d'}\circ GFu\circ G(\varepsilon_d^{-1}) \\ &=G\varepsilon_{d'}\circ GFu\circ \eta_{Gd} \\ &=G\varepsilon_{d'}\circ \eta_{Gd'}\circ u \\ &=\operatorname{id}_{Gd'}\circ u \\ &=u. \end{align*} Thus every morphism $u:Gd\to Gd'$ is the image under $G$ of some morphism $\bar u:d\to d'$, so the hom-map induced by $G$ is surjective. Hence $G$ is full. [/guided] [/step] [step:Conclude the equivalence] The first three steps show that full faithfulness of $G$ implies that every counit component $\varepsilon_d$ is an isomorphism. The last two steps show that if every counit component is an isomorphism, then $G$ is both faithful and full. Therefore $G$ is fully faithful if and only if $\varepsilon_d:FGd\to d$ is an isomorphism for every object $d\in\mathcal D$. [/step]