[proofplan] We construct the adjunction directly from the unit $\eta$ and counit $\varepsilon$. The map $\Phi_{c,d}$ sends a morphism $f: F c \to d$ to its adjunct $G(f)\circ \eta_c$, and the proposed inverse $\Psi_{c,d}$ sends $g: c \to Gd$ to $\varepsilon_d \circ F(g)$. The two triangle identities, together with naturality of $\eta$ and $\varepsilon$, show that these two assignments are inverse to each other. Finally, functoriality and naturality of $\eta$ verify naturality of the resulting hom-set bijection in both variables. [/proofplan] [step:Define the two transpose maps between the hom-sets] Fix objects $c \in \mathcal C$ and $d \in \mathcal D$. Since $\mathcal C$ and $\mathcal D$ are locally small, the collections $\operatorname{Hom}_{\mathcal D}(F c,d)$ and $\operatorname{Hom}_{\mathcal C}(c,Gd)$ are sets. Define \begin{align*} \Phi_{c,d}: \operatorname{Hom}_{\mathcal D}(F c,d) &\to \operatorname{Hom}_{\mathcal C}(c,Gd) \\ f &\mapsto G(f)\circ \eta_c, \end{align*} where $\eta_c: c \to G F c$ is the component of $\eta$ at $c$, and $G(f): G F c \to Gd$ is the image of $f:Fc \to d$ under $G$. Define \begin{align*} \Psi_{c,d}: \operatorname{Hom}_{\mathcal C}(c,Gd) &\to \operatorname{Hom}_{\mathcal D}(Fc,d) \\ g &\mapsto \varepsilon_d \circ F(g), \end{align*} where $F(g): Fc \to F G d$ is the image of $g:c \to Gd$ under $F$, and $\varepsilon_d: F G d \to d$ is the component of $\varepsilon$ at $d$. Both composites have the declared domains and codomains, so $\Phi_{c,d}$ and $\Psi_{c,d}$ are well-defined maps of sets. [/step] [step:Show that applying the transpose maps to $f:Fc \to d$ gives back $f$] Let $f:Fc \to d$ be a morphism in $\mathcal D$. Then \begin{align*} \Psi_{c,d}(\Phi_{c,d}(f)) &= \Psi_{c,d}\bigl(G(f)\circ \eta_c\bigr) \\ &= \varepsilon_d \circ F\bigl(G(f)\circ \eta_c\bigr) \\ &= \varepsilon_d \circ F(G(f)) \circ F(\eta_c). \end{align*} By naturality of $\varepsilon:FG \Rightarrow \operatorname{id}_{\mathcal D}$ applied to the morphism $f:Fc \to d$, we have \begin{align*} f \circ \varepsilon_{Fc} = \varepsilon_d \circ F(G(f)). \end{align*} Substituting this equality gives \begin{align*} \Psi_{c,d}(\Phi_{c,d}(f)) &= f \circ \varepsilon_{Fc} \circ F(\eta_c). \end{align*} By the triangle identity at $c$, \begin{align*} \varepsilon_{Fc}\circ F(\eta_c)=\operatorname{id}_{Fc}. \end{align*} Therefore \begin{align*} \Psi_{c,d}(\Phi_{c,d}(f)) &= f \circ \operatorname{id}_{Fc} \\ &= f. \end{align*} [guided] We start with a morphism $f:Fc \to d$ in $\mathcal D$ and follow it through the two proposed transpose operations. First $\Phi_{c,d}$ sends $f$ to \begin{align*} \Phi_{c,d}(f)=G(f)\circ \eta_c:c\to Gd. \end{align*} Applying $\Psi_{c,d}$ to this morphism gives \begin{align*} \Psi_{c,d}(\Phi_{c,d}(f)) &= \varepsilon_d \circ F\bigl(G(f)\circ \eta_c\bigr). \end{align*} Since $F$ is a functor, it preserves composition, so \begin{align*} F\bigl(G(f)\circ \eta_c\bigr) =F(G(f))\circ F(\eta_c). \end{align*} Thus \begin{align*} \Psi_{c,d}(\Phi_{c,d}(f)) &= \varepsilon_d \circ F(G(f)) \circ F(\eta_c). \end{align*} The reason this expression simplifies is naturality of the counit. The natural transformation $\varepsilon:FG\Rightarrow \operatorname{id}_{\mathcal D}$ says that for every morphism $h:x\to y$ in $\mathcal D$, \begin{align*} h\circ \varepsilon_x=\varepsilon_y\circ F(G(h)). \end{align*} We apply this with $x=Fc$, $y=d$, and $h=f:Fc\to d$. This gives \begin{align*} f\circ \varepsilon_{Fc}=\varepsilon_d\circ F(G(f)). \end{align*} Substituting into the previous expression yields \begin{align*} \Psi_{c,d}(\Phi_{c,d}(f)) &= f\circ \varepsilon_{Fc}\circ F(\eta_c). \end{align*} Now the first triangle identity applies exactly to the remaining middle composite: \begin{align*} \varepsilon_{Fc}\circ F(\eta_c)=\operatorname{id}_{Fc}. \end{align*} Therefore \begin{align*} \Psi_{c,d}(\Phi_{c,d}(f)) &= f\circ \operatorname{id}_{Fc} \\ &= f. \end{align*} So $\Psi_{c,d}\circ \Phi_{c,d}$ is the identity map on $\operatorname{Hom}_{\mathcal D}(Fc,d)$. [/guided] [/step] [step:Show that applying the transpose maps to $g:c \to Gd$ gives back $g$] Let $g:c \to Gd$ be a morphism in $\mathcal C$. Then \begin{align*} \Phi_{c,d}(\Psi_{c,d}(g)) &= \Phi_{c,d}\bigl(\varepsilon_d \circ F(g)\bigr) \\ &= G\bigl(\varepsilon_d \circ F(g)\bigr)\circ \eta_c \\ &= G(\varepsilon_d)\circ G(F(g))\circ \eta_c. \end{align*} By naturality of $\eta:\operatorname{id}_{\mathcal C}\Rightarrow GF$ applied to the morphism $g:c\to Gd$, we have \begin{align*} G(F(g))\circ \eta_c=\eta_{Gd}\circ g. \end{align*} Substituting this equality gives \begin{align*} \Phi_{c,d}(\Psi_{c,d}(g)) &=G(\varepsilon_d)\circ \eta_{Gd}\circ g. \end{align*} By the triangle identity at $d$, \begin{align*} G(\varepsilon_d)\circ \eta_{Gd}=\operatorname{id}_{Gd}. \end{align*} Therefore \begin{align*} \Phi_{c,d}(\Psi_{c,d}(g)) &=\operatorname{id}_{Gd}\circ g \\ &=g. \end{align*} [guided] Now we start with a morphism $g:c\to Gd$ in $\mathcal C$ and apply the two operations in the opposite order. The map $\Psi_{c,d}$ sends $g$ to \begin{align*} \Psi_{c,d}(g)=\varepsilon_d\circ F(g):Fc\to d. \end{align*} Applying $\Phi_{c,d}$ gives \begin{align*} \Phi_{c,d}(\Psi_{c,d}(g)) &=G\bigl(\varepsilon_d\circ F(g)\bigr)\circ \eta_c. \end{align*} Since $G$ is a functor, it preserves composition, so \begin{align*} G\bigl(\varepsilon_d\circ F(g)\bigr) =G(\varepsilon_d)\circ G(F(g)). \end{align*} Hence \begin{align*} \Phi_{c,d}(\Psi_{c,d}(g)) &=G(\varepsilon_d)\circ G(F(g))\circ \eta_c. \end{align*} The expression $G(F(g))\circ \eta_c$ is simplified using naturality of the unit. The natural transformation $\eta:\operatorname{id}_{\mathcal C}\Rightarrow GF$ says that for every morphism $a:x\to y$ in $\mathcal C$, \begin{align*} G(F(a))\circ \eta_x=\eta_y\circ a. \end{align*} We apply this with $x=c$, $y=Gd$, and $a=g:c\to Gd$. This gives \begin{align*} G(F(g))\circ \eta_c=\eta_{Gd}\circ g. \end{align*} Substituting, we obtain \begin{align*} \Phi_{c,d}(\Psi_{c,d}(g)) &=G(\varepsilon_d)\circ \eta_{Gd}\circ g. \end{align*} Now the second triangle identity applies: \begin{align*} G(\varepsilon_d)\circ \eta_{Gd}=\operatorname{id}_{Gd}. \end{align*} Therefore \begin{align*} \Phi_{c,d}(\Psi_{c,d}(g)) &=\operatorname{id}_{Gd}\circ g \\ &=g. \end{align*} So $\Phi_{c,d}\circ \Psi_{c,d}$ is the identity map on $\operatorname{Hom}_{\mathcal C}(c,Gd)$. [/guided] [/step] [step:Conclude that the transpose maps are bijections] The previous two steps show that \begin{align*} \Psi_{c,d}\circ \Phi_{c,d} &=\operatorname{id}_{\operatorname{Hom}_{\mathcal D}(Fc,d)}, \\ \Phi_{c,d}\circ \Psi_{c,d} &=\operatorname{id}_{\operatorname{Hom}_{\mathcal C}(c,Gd)}. \end{align*} Hence $\Phi_{c,d}$ is a bijection with inverse $\Psi_{c,d}$ for every $c\in\mathcal C$ and $d\in\mathcal D$. [/step] [step:Verify naturality of the hom-set bijection in both variables] Let $u:c'\to c$ be a morphism in $\mathcal C$, let $v:d\to d'$ be a morphism in $\mathcal D$, and let $f:Fc\to d$ be a morphism in $\mathcal D$. The induced morphism in $\operatorname{Hom}_{\mathcal D}(Fc',d')$ is \begin{align*} v\circ f\circ F(u):Fc'\to d'. \end{align*} We compute \begin{align*} \Phi_{c',d'}(v\circ f\circ F(u)) &=G(v\circ f\circ F(u))\circ \eta_{c'} \\ &=G(v)\circ G(f)\circ G(F(u))\circ \eta_{c'}. \end{align*} By naturality of $\eta$ applied to $u:c'\to c$, \begin{align*} G(F(u))\circ \eta_{c'}=\eta_c\circ u. \end{align*} Therefore \begin{align*} \Phi_{c',d'}(v\circ f\circ F(u)) &=G(v)\circ G(f)\circ \eta_c\circ u \\ &=G(v)\circ \Phi_{c,d}(f)\circ u. \end{align*} This is precisely the naturality condition for the family of bijections $\Phi_{c,d}$, contravariantly in $c$ and covariantly in $d$. [guided] The hom-set bijection must be compatible with changing both variables. Let $u:c'\to c$ be a morphism of $\mathcal C$ and let $v:d\to d'$ be a morphism of $\mathcal D$. A morphism $f:Fc\to d$ is transported on the $\mathcal D$-side by precomposition with $F(u):Fc'\to Fc$ and postcomposition with $v:d\to d'$, giving \begin{align*} v\circ f\circ F(u):Fc'\to d'. \end{align*} We apply $\Phi_{c',d'}$ to this transported morphism: \begin{align*} \Phi_{c',d'}(v\circ f\circ F(u)) &=G(v\circ f\circ F(u))\circ \eta_{c'}. \end{align*} Since $G$ is a functor, it preserves composition, so \begin{align*} G(v\circ f\circ F(u)) =G(v)\circ G(f)\circ G(F(u)). \end{align*} Thus \begin{align*} \Phi_{c',d'}(v\circ f\circ F(u)) &=G(v)\circ G(f)\circ G(F(u))\circ \eta_{c'}. \end{align*} The remaining expression involving $u$ is controlled by naturality of the unit. Applying naturality of $\eta:\operatorname{id}_{\mathcal C}\Rightarrow GF$ to the morphism $u:c'\to c$ gives \begin{align*} G(F(u))\circ \eta_{c'}=\eta_c\circ u. \end{align*} Substituting this identity gives \begin{align*} \Phi_{c',d'}(v\circ f\circ F(u)) &=G(v)\circ G(f)\circ \eta_c\circ u \\ &=G(v)\circ \Phi_{c,d}(f)\circ u. \end{align*} This equality says exactly that transposing after changing $f$ by $u$ and $v$ gives the same result as first transposing $f$ and then changing the resulting morphism $c\to Gd$ by precomposition with $u$ and postcomposition with $G(v)$. Hence the bijections are natural in both variables. [/guided] [/step] [step:Identify the resulting adjunction] For every $c\in\mathcal C$ and $d\in\mathcal D$, we have constructed a natural bijection \begin{align*} \operatorname{Hom}_{\mathcal D}(Fc,d)\cong \operatorname{Hom}_{\mathcal C}(c,Gd). \end{align*} By definition, this natural family of bijections is an adjunction with $F$ left adjoint to $G$. Therefore $F\dashv G$, with unit $\eta$ and counit $\varepsilon$ as stated. [/step]