[proofplan] We compare the hom-set map induced by $F$ with the adjunction bijection. Under the adjunction, a morphism $Ff: Fc \to Fc'$ corresponds to the composite $\eta_{c'} \circ f: c \to GFc'$, so if every unit component is an isomorphism, the induced hom-set map is a bijection. Conversely, if $F$ is fully faithful, fullness lifts the counit component $\varepsilon_{Fc}: FGFc \to Fc$ to a morphism $\mu_c: GFc \to c$, and faithfulness plus the triangle identity forces $\mu_c$ to be a two-sided inverse to $\eta_c$. [/proofplan] [step:Identify the adjunction bijection on morphisms of the form $Ff$] For objects $c,c' \in \mathcal C$, define the hom-set map induced by $F$ by \begin{align*} F_{c,c'}: \operatorname{Hom}_{\mathcal C}(c,c') &\to \operatorname{Hom}_{\mathcal D}(Fc,Fc') \\ f &\mapsto Ff. \end{align*} The adjunction gives a bijection \begin{align*} \Phi_{c,Fc'}: \operatorname{Hom}_{\mathcal D}(Fc,Fc') &\to \operatorname{Hom}_{\mathcal C}(c,GFc') \\ \alpha &\mapsto G\alpha \circ \eta_c. \end{align*} For $f: c \to c'$ in $\mathcal C$, naturality of $\eta: \operatorname{id}_{\mathcal C} \Rightarrow GF$ gives \begin{align*} GFf \circ \eta_c = \eta_{c'} \circ f. \end{align*} Therefore \begin{align*} \Phi_{c,Fc'}(F_{c,c'}(f)) = \Phi_{c,Fc'}(Ff) = G(Ff) \circ \eta_c = \eta_{c'} \circ f. \end{align*} [guided] Fix objects $c,c' \in \mathcal C$. To prove that $F$ is fully faithful, we must understand the map on hom-sets \begin{align*} F_{c,c'}: \operatorname{Hom}_{\mathcal C}(c,c') &\to \operatorname{Hom}_{\mathcal D}(Fc,Fc') \\ f &\mapsto Ff. \end{align*} The adjunction $F \dashv G$ provides, for every object $d \in \mathcal D$, a bijection \begin{align*} \Phi_{c,d}: \operatorname{Hom}_{\mathcal D}(Fc,d) &\to \operatorname{Hom}_{\mathcal C}(c,Gd) \\ \alpha &\mapsto G\alpha \circ \eta_c. \end{align*} We specialize this to $d = Fc'$, obtaining \begin{align*} \Phi_{c,Fc'}: \operatorname{Hom}_{\mathcal D}(Fc,Fc') &\to \operatorname{Hom}_{\mathcal C}(c,GFc'). \end{align*} Now let $f: c \to c'$ be a morphism in $\mathcal C$. Applying $\Phi_{c,Fc'}$ to $Ff: Fc \to Fc'$ gives \begin{align*} \Phi_{c,Fc'}(Ff) = G(Ff) \circ \eta_c. \end{align*} Naturality of the unit $\eta: \operatorname{id}_{\mathcal C} \Rightarrow GF$ at the morphism $f: c \to c'$ says exactly that the square with sides $f$, $GFf$, $\eta_c$, and $\eta_{c'}$ commutes: \begin{align*} GFf \circ \eta_c = \eta_{c'} \circ f. \end{align*} Hence \begin{align*} \Phi_{c,Fc'}(F_{c,c'}(f)) = \Phi_{c,Fc'}(Ff) = G(Ff) \circ \eta_c = \eta_{c'} \circ f. \end{align*} Thus, after applying the adjunction bijection, the map induced by $F$ becomes the map that sends $f: c \to c'$ to $\eta_{c'} \circ f: c \to GFc'$. [/guided] [/step] [step:Deduce full faithfulness when every unit component is an isomorphism] Assume that $\eta_x: x \to GFx$ is an isomorphism for every object $x \in \mathcal C$. For fixed objects $c,c' \in \mathcal C$, define \begin{align*} H_{c,c'}: \operatorname{Hom}_{\mathcal C}(c,c') &\to \operatorname{Hom}_{\mathcal C}(c,GFc') \\ f &\mapsto \eta_{c'} \circ f. \end{align*} Since $\eta_{c'}$ is an isomorphism, $H_{c,c'}$ is a bijection, with inverse \begin{align*} K_{c,c'}: \operatorname{Hom}_{\mathcal C}(c,GFc') &\to \operatorname{Hom}_{\mathcal C}(c,c') \\ h &\mapsto \eta_{c'}^{-1} \circ h. \end{align*} The previous step proves \begin{align*} \Phi_{c,Fc'} \circ F_{c,c'} = H_{c,c'}. \end{align*} Since $\Phi_{c,Fc'}$ and $H_{c,c'}$ are bijections, $F_{c,c'}$ is a bijection. This holds for all $c,c' \in \mathcal C$, so $F$ is fully faithful. [/step] [step:Construct a candidate inverse to the unit from fullness] Assume now that $F$ is fully faithful. Fix an object $c \in \mathcal C$. Since $F$ is full, the map \begin{align*} F_{GFc,c}: \operatorname{Hom}_{\mathcal C}(GFc,c) &\to \operatorname{Hom}_{\mathcal D}(FGFc,Fc) \\ m &\mapsto Fm \end{align*} is surjective. The counit component \begin{align*} \varepsilon_{Fc}: FGFc \to Fc \end{align*} is a morphism in $\mathcal D$, so there exists a morphism \begin{align*} \mu_c: GFc \to c \end{align*} in $\mathcal C$ such that \begin{align*} F\mu_c = \varepsilon_{Fc}. \end{align*} [/step] [step:Use faithfulness and the triangle identity to prove $\mu_c \circ \eta_c = \operatorname{id}_c$] The first triangle identity for the adjunction says \begin{align*} \varepsilon_{Fc} \circ F\eta_c = \operatorname{id}_{Fc}. \end{align*} Using $F\mu_c = \varepsilon_{Fc}$, functoriality of $F$, and preservation of identity morphisms, we compute \begin{align*} F(\mu_c \circ \eta_c) = F\mu_c \circ F\eta_c = \varepsilon_{Fc} \circ F\eta_c = \operatorname{id}_{Fc} = F(\operatorname{id}_c). \end{align*} Since $F$ is faithful, the map $F_{c,c}$ is injective. Therefore \begin{align*} \mu_c \circ \eta_c = \operatorname{id}_c. \end{align*} [/step] [step:Use faithfulness and naturality of the unit to prove $\eta_c \circ \mu_c = \operatorname{id}_{GFc}$] Naturality of $\eta$ at the morphism $\mu_c: GFc \to c$ gives \begin{align*} GF\mu_c \circ \eta_{GFc} = \eta_c \circ \mu_c. \end{align*} Since $F\mu_c = \varepsilon_{Fc}$, this becomes \begin{align*} G\varepsilon_{Fc} \circ \eta_{GFc} = \eta_c \circ \mu_c. \end{align*} The second triangle identity for the adjunction, applied to the object $Fc \in \mathcal D$, says \begin{align*} G\varepsilon_{Fc} \circ \eta_{GFc} = \operatorname{id}_{GFc}. \end{align*} Hence \begin{align*} \eta_c \circ \mu_c = \operatorname{id}_{GFc}. \end{align*} Together with $\mu_c \circ \eta_c = \operatorname{id}_c$, this proves that $\eta_c$ is an isomorphism with inverse $\mu_c$. [guided] We already constructed a morphism $\mu_c: GFc \to c$ satisfying $F\mu_c = \varepsilon_{Fc}$. The previous step proved that $\mu_c$ is a left inverse to $\eta_c$: \begin{align*} \mu_c \circ \eta_c = \operatorname{id}_c. \end{align*} It remains to prove that it is also a right inverse. Naturality of the unit $\eta: \operatorname{id}_{\mathcal C} \Rightarrow GF$ applies to every morphism in $\mathcal C$, so we apply it to the morphism \begin{align*} \mu_c: GFc \to c. \end{align*} The naturality square gives \begin{align*} GF\mu_c \circ \eta_{GFc} = \eta_c \circ \mu_c. \end{align*} Because $\mu_c$ was chosen so that $F\mu_c = \varepsilon_{Fc}$, applying $G$ gives \begin{align*} GF\mu_c = G\varepsilon_{Fc}. \end{align*} Therefore \begin{align*} \eta_c \circ \mu_c = G\varepsilon_{Fc} \circ \eta_{GFc}. \end{align*} Now use the second triangle identity for the adjunction. Applied to the object $Fc \in \mathcal D$, it states that \begin{align*} G\varepsilon_{Fc} \circ \eta_{GFc} = \operatorname{id}_{GFc}. \end{align*} Substituting this into the previous equation yields \begin{align*} \eta_c \circ \mu_c = \operatorname{id}_{GFc}. \end{align*} Thus $\mu_c$ is both a left inverse and a right inverse to $\eta_c$. Consequently $\eta_c: c \to GFc$ is an isomorphism in $\mathcal C$, with inverse $\mu_c: GFc \to c$. [/guided] [/step] [step:Conclude the equivalence] We have proved that if every unit component $\eta_c$ is an isomorphism, then each induced hom-set map \begin{align*} F_{c,c'}: \operatorname{Hom}_{\mathcal C}(c,c') \to \operatorname{Hom}_{\mathcal D}(Fc,Fc') \end{align*} is a bijection, so $F$ is fully faithful. Conversely, if $F$ is fully faithful, then for every object $c \in \mathcal C$ the component $\eta_c: c \to GFc$ has inverse $\mu_c: GFc \to c$, so $\eta_c$ is an isomorphism. This proves the desired if and only if. [/step]