[proofplan] We prove the equivalence by constructing the data in each formulation explicitly. Starting from natural hom-set bijections, we transpose identity morphisms to obtain the unit and counit, and naturality of the bijections gives the formulas for transposition and the triangle identities. Starting from a unit and counit satisfying the triangle identities, we define the hom-set bijections by the two displayed formulas and verify directly that the formulas are natural and inverse. Finally, the universal-arrow formulation is exactly the representing property of the same bijections at each object $c$, with functorial variation encoded by naturality in $c$. [/proofplan]

[step:Transpose identity morphisms to construct the unit and counit]Assume first that for each $c \in \mathcal C$ and $d \in \mathcal D$ we are given a bijection \begin{align*} \Phi_{c,d}: \mathcal D(Fc,d) \longrightarrow \mathcal C(c,Gd), \end{align*} natural in both variables. Define a family of morphisms $\eta_c:c \to G(Fc)$ by \begin{align*} \eta_c := \Phi_{c,Fc}(\operatorname{id}_{Fc}), \end{align*} and define a family of morphisms $\varepsilon_d:F(Gd)\to d$ by \begin{align*} \varepsilon_d := \Phi_{Gd,d}^{-1}(\operatorname{id}_{Gd}). \end{align*} We verify that $\eta$ is natural. Let $u:c' \to c$ be a morphism in $\mathcal C$. Naturality of $\Phi$ in the $\mathcal C$-variable gives the commutative square comparing precomposition by $u$ on the right with precomposition by $F(u)$ on the left. Applying this square to $\operatorname{id}_{Fc}$ gives \begin{align*} G(Fu)\circ \eta_{c'} = \eta_c \circ u. \end{align*} Thus $\eta:\operatorname{id}_{\mathcal C}\Rightarrow GF$ is natural. We verify that $\varepsilon$ is natural. Let $v:d \to d'$ be a morphism in $\mathcal D$. Naturality of $\Phi$ in the $\mathcal D$-variable, applied to $\varepsilon_d:F(Gd)\to d$, gives \begin{align*} \Phi_{Gd,d'}(v\circ \varepsilon_d)=G(v)\circ \Phi_{Gd,d}(\varepsilon_d)=G(v). \end{align*} Naturality in the $\mathcal C$-variable, applied to the morphism $G(v):Gd\to Gd'$, gives \begin{align*} \Phi_{Gd,d'}(\varepsilon_{d'}\circ F(Gv)) = \Phi_{Gd',d'}(\varepsilon_{d'})\circ G(v) = G(v). \end{align*} Since $\Phi_{Gd,d'}$ is injective, we obtain \begin{align*} v\circ \varepsilon_d=\varepsilon_{d'}\circ F(Gv). \end{align*} Therefore $\varepsilon:FG\Rightarrow \operatorname{id}_{\mathcal D}$ is natural.[/step]

[guided]The identities are obtained by transposing the two most canonical morphisms available: identity morphisms. For each $c \in \mathcal C$, the identity $\operatorname{id}_{Fc}:Fc\to Fc$ is an element of $\mathcal D(Fc,Fc)$, so its image under \begin{align*} \Phi_{c,Fc}: \mathcal D(Fc,Fc)\longrightarrow \mathcal C(c,G(Fc)) \end{align*} is a morphism $c\to G(Fc)$. This is the component $\eta_c$ of the unit: \begin{align*} \eta_c := \Phi_{c,Fc}(\operatorname{id}_{Fc}). \end{align*} Similarly, for each $d \in \mathcal D$, the identity $\operatorname{id}_{Gd}:Gd\to Gd$ is an element of $\mathcal C(Gd,Gd)$. Since $\Phi_{Gd,d}$ is a bijection, it has an inverse \begin{align*} \Phi_{Gd,d}^{-1}: \mathcal C(Gd,Gd)\longrightarrow \mathcal D(F(Gd),d), \end{align*} and we define \begin{align*} \varepsilon_d := \Phi_{Gd,d}^{-1}(\operatorname{id}_{Gd}). \end{align*} Now we check that these components assemble into natural transformations. Let $u:c'\to c$ be a morphism in $\mathcal C$. Naturality of $\Phi$ in $c$ says that transposing after precomposing by $F(u)$ is the same as precomposing the transpose by $u$. Applying this to $\operatorname{id}_{Fc}$ gives \begin{align*} \Phi_{c',Fc}(\operatorname{id}_{Fc}\circ F(u)) = \Phi_{c,Fc}(\operatorname{id}_{Fc})\circ u. \end{align*} The left side is $\Phi_{c',Fc}(F(u))$, while the right side is $\eta_c\circ u$. Naturality in the $\mathcal D$-variable, applied to $F(u):Fc'\to Fc$, also gives \begin{align*} \Phi_{c',Fc}(F(u))=G(Fu)\circ \Phi_{c',Fc'}(\operatorname{id}_{Fc'}) =G(Fu)\circ \eta_{c'}. \end{align*} Hence \begin{align*} G(Fu)\circ \eta_{c'}=\eta_c\circ u, \end{align*} which is exactly naturality of $\eta:\operatorname{id}_{\mathcal C}\Rightarrow GF$. For $\varepsilon$, let $v:d\to d'$ be a morphism in $\mathcal D$. Naturality in $d$ gives \begin{align*} \Phi_{Gd,d'}(v\circ \varepsilon_d) = G(v)\circ \Phi_{Gd,d}(\varepsilon_d) = G(v)\circ \operatorname{id}_{Gd} = G(v). \end{align*} On the other hand, naturality in $c$ applied to $G(v):Gd\to Gd'$ gives \begin{align*} \Phi_{Gd,d'}(\varepsilon_{d'}\circ F(Gv)) = \Phi_{Gd',d'}(\varepsilon_{d'})\circ G(v) = \operatorname{id}_{Gd'}\circ G(v) = G(v). \end{align*} Both $v\circ \varepsilon_d$ and $\varepsilon_{d'}\circ F(Gv)$ have the same transpose under the bijection $\Phi_{Gd,d'}$. Since this map is injective, \begin{align*} v\circ \varepsilon_d=\varepsilon_{d'}\circ F(Gv). \end{align*} Thus $\varepsilon$ is natural.[/guided]

[step:Derive the transposition formulas and the triangle identities]For every morphism $f:Fc\to d$ in $\mathcal D$, naturality of $\Phi$ in $d$ gives \begin{align*} \Phi_{c,d}(f)=G(f)\circ \eta_c. \end{align*} For every morphism $g:c\to Gd$ in $\mathcal C$, naturality of $\Phi$ in $c$ gives \begin{align*} \Phi_{c,d}^{-1}(g)=\varepsilon_d\circ F(g). \end{align*} Applying the inverse formula to $g=\eta_c:c\to G(Fc)$ gives \begin{align*} \Phi_{c,Fc}^{-1}(\eta_c)=\varepsilon_{Fc}\circ F(\eta_c). \end{align*} But $\eta_c=\Phi_{c,Fc}(\operatorname{id}_{Fc})$, so injectivity of $\Phi_{c,Fc}$ implies \begin{align*} \varepsilon_{Fc}\circ F(\eta_c)=\operatorname{id}_{Fc}. \end{align*} Applying the forward formula to $f=\varepsilon_d:F(Gd)\to d$ gives \begin{align*} \Phi_{Gd,d}(\varepsilon_d)=G(\varepsilon_d)\circ \eta_{Gd}. \end{align*} By definition of $\varepsilon_d$, the left side is $\operatorname{id}_{Gd}$, hence \begin{align*} G(\varepsilon_d)\circ \eta_{Gd}=\operatorname{id}_{Gd}. \end{align*} Thus the triangle identities hold.[/step]

[guided]The naturality of the hom-set bijection determines how every morphism transposes. Let $f:Fc\to d$ be a morphism in $\mathcal D$. Since $f$ is a morphism from $Fc$ to $d$, postcomposition by $f$ sends $\operatorname{id}_{Fc}$ to $f$. Naturality of $\Phi$ in the $\mathcal D$-variable says that this postcomposition corresponds, after transposition, to postcomposition by $G(f)$ in $\mathcal C$. Therefore \begin{align*} \Phi_{c,d}(f) &= \Phi_{c,d}(f\circ \operatorname{id}_{Fc})\\ &= G(f)\circ \Phi_{c,Fc}(\operatorname{id}_{Fc})\\ &= G(f)\circ \eta_c. \end{align*} Now let $g:c\to Gd$ be a morphism in $\mathcal C$. We claim that its inverse transpose is $\varepsilon_d\circ F(g)$. The morphism $F(g):F c\to F(Gd)$ is defined because $F:\mathcal C\to\mathcal D$ is a functor and $g:c\to Gd$ is a morphism in $\mathcal C$. Therefore the composite \begin{align*} \varepsilon_d\circ F(g):Fc\to d \end{align*} is a morphism in $\mathcal D$. Naturality of $\Phi$ in the $\mathcal C$-variable gives \begin{align*} \Phi_{c,d}(\varepsilon_d\circ F(g)) = \Phi_{Gd,d}(\varepsilon_d)\circ g = \operatorname{id}_{Gd}\circ g = g. \end{align*} Since $\Phi_{c,d}$ is bijective, this proves \begin{align*} \Phi_{c,d}^{-1}(g)=\varepsilon_d\circ F(g). \end{align*} The triangle identities are now forced by applying these formulas to the unit and counit themselves. First take $g=\eta_c:c\to G(Fc)$. The inverse formula gives \begin{align*} \Phi_{c,Fc}^{-1}(\eta_c)=\varepsilon_{Fc}\circ F(\eta_c). \end{align*} But $\eta_c$ was defined as $\Phi_{c,Fc}(\operatorname{id}_{Fc})$, so the unique morphism whose transpose is $\eta_c$ is $\operatorname{id}_{Fc}$. Hence \begin{align*} \varepsilon_{Fc}\circ F(\eta_c)=\operatorname{id}_{Fc}. \end{align*} Second take $f=\varepsilon_d:F(Gd)\to d$. The forward formula gives \begin{align*} \Phi_{Gd,d}(\varepsilon_d)=G(\varepsilon_d)\circ \eta_{Gd}. \end{align*} By the definition of $\varepsilon_d$ as the inverse transpose of $\operatorname{id}_{Gd}$, the left side is $\operatorname{id}_{Gd}$. Therefore \begin{align*} G(\varepsilon_d)\circ \eta_{Gd}=\operatorname{id}_{Gd}. \end{align*} These are precisely the two triangle identities.[/guided]

[step:Use the triangle identities to recover the hom-set bijections]Conversely, assume we are given natural transformations \begin{align*} \eta:\operatorname{id}_{\mathcal C}\Rightarrow GF, \qquad \varepsilon:FG\Rightarrow \operatorname{id}_{\mathcal D}, \end{align*} satisfying the triangle identities. For each $c \in \mathcal C$ and $d \in \mathcal D$, define maps \begin{align*} \Phi_{c,d}: \mathcal D(Fc,d) &\longrightarrow \mathcal C(c,Gd),\\ f &\longmapsto G(f)\circ \eta_c, \end{align*} and \begin{align*} \Psi_{c,d}: \mathcal C(c,Gd) &\longrightarrow \mathcal D(Fc,d),\\ g &\longmapsto \varepsilon_d\circ F(g). \end{align*} For $f:Fc\to d$, using functoriality of $F$ and $G$, naturality of $\varepsilon$, and the first triangle identity, we compute \begin{align*} \Psi_{c,d}(\Phi_{c,d}(f)) &= \varepsilon_d\circ F(G(f)\circ \eta_c)\\ &= \varepsilon_d\circ F(Gf)\circ F(\eta_c)\\ &= f\circ \varepsilon_{Fc}\circ F(\eta_c)\\ &= f\circ \operatorname{id}_{Fc}\\ &= f. \end{align*} For $g:c\to Gd$, using functoriality of $G$, naturality of $\eta$, and the second triangle identity, we compute \begin{align*} \Phi_{c,d}(\Psi_{c,d}(g)) &= G(\varepsilon_d\circ F(g))\circ \eta_c\\ &= G(\varepsilon_d)\circ G(Fg)\circ \eta_c\\ &= G(\varepsilon_d)\circ \eta_{Gd}\circ g\\ &= \operatorname{id}_{Gd}\circ g\\ &= g. \end{align*} Thus $\Phi_{c,d}$ and $\Psi_{c,d}$ are inverse bijections. Naturality of $\Phi$ in $c$ follows from naturality of $\eta$, and naturality of $\Phi$ in $d$ follows from functoriality of $G$. Hence the unit-counit data determine a natural family of hom-set bijections.[/step]

[guided]We now start with the unit and counit and reconstruct the hom-set bijection. For objects $c\in\mathcal C$ and $d\in\mathcal D$, define \begin{align*} \Phi_{c,d}: \mathcal D(Fc,d) &\longrightarrow \mathcal C(c,Gd),\\ f &\longmapsto G(f)\circ \eta_c. \end{align*} This sends a morphism $f:Fc\to d$ to the composite \begin{align*} c \xrightarrow{\eta_c} G(Fc) \xrightarrow{G(f)} Gd. \end{align*} Define a candidate inverse by \begin{align*} \Psi_{c,d}: \mathcal C(c,Gd) &\longrightarrow \mathcal D(Fc,d),\\ g &\longmapsto \varepsilon_d\circ F(g). \end{align*} This sends a morphism $g:c\to Gd$ to the composite \begin{align*} Fc \xrightarrow{F(g)} F(Gd) \xrightarrow{\varepsilon_d} d. \end{align*} We verify that these two maps are inverse. Let $f:Fc\to d$. Then \begin{align*} \Psi_{c,d}(\Phi_{c,d}(f)) &= \varepsilon_d\circ F(G(f)\circ \eta_c)\\ &= \varepsilon_d\circ F(Gf)\circ F(\eta_c), \end{align*} where the second equality uses functoriality of $F$. Naturality of $\varepsilon:FG\Rightarrow \operatorname{id}_{\mathcal D}$ for the morphism $f:Fc\to d$ says \begin{align*} f\circ \varepsilon_{Fc}=\varepsilon_d\circ F(Gf). \end{align*} Substituting this into the previous expression gives \begin{align*} \Psi_{c,d}(\Phi_{c,d}(f)) &= f\circ \varepsilon_{Fc}\circ F(\eta_c). \end{align*} The first triangle identity says \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*} Now let $g:c\to Gd$. Then \begin{align*} \Phi_{c,d}(\Psi_{c,d}(g)) &= G(\varepsilon_d\circ F(g))\circ \eta_c\\ &= G(\varepsilon_d)\circ G(Fg)\circ \eta_c, \end{align*} where the second equality uses functoriality of $G$. Naturality of $\eta:\operatorname{id}_{\mathcal C}\Rightarrow GF$ for the morphism $g:c\to Gd$ says \begin{align*} G(Fg)\circ \eta_c=\eta_{Gd}\circ g. \end{align*} Thus \begin{align*} \Phi_{c,d}(\Psi_{c,d}(g)) &= G(\varepsilon_d)\circ \eta_{Gd}\circ g. \end{align*} The second triangle identity says \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*} Hence $\Phi_{c,d}$ is a bijection with inverse $\Psi_{c,d}$. It remains to check naturality. Let $u:c'\to c$ be a morphism in $\mathcal C$ and $f:Fc\to d$ a morphism in $\mathcal D$. Naturality of $\eta$ gives \begin{align*} G(Fu)\circ \eta_{c'}=\eta_c\circ u. \end{align*} Therefore \begin{align*} \Phi_{c',d}(f\circ F(u)) &= G(f\circ F(u))\circ \eta_{c'}\\ &= G(f)\circ G(Fu)\circ \eta_{c'}\\ &= G(f)\circ \eta_c\circ u\\ &= \Phi_{c,d}(f)\circ u. \end{align*} This is naturality in $c$. Let $v:d\to d'$ be a morphism in $\mathcal D$. For $f:Fc\to d$, \begin{align*} \Phi_{c,d'}(v\circ f) &= G(v\circ f)\circ \eta_c\\ &= G(v)\circ G(f)\circ \eta_c\\ &= G(v)\circ \Phi_{c,d}(f). \end{align*} This is naturality in $d$. Thus the unit-counit data recover a natural family of hom-set bijections.[/guided]

[step:Interpret the hom-set bijections as universal arrows]Assume a natural family of bijections $\Phi_{c,d}$ is given. For each $c \in \mathcal C$, define \begin{align*} \eta_c:c\to G(Fc) \end{align*} by \begin{align*} \eta_c=\Phi_{c,Fc}(\operatorname{id}_{Fc}). \end{align*} For every object $d\in\mathcal D$ and every morphism $g:c\to Gd$ in $\mathcal C$, the bijection $\Phi_{c,d}$ gives a unique morphism \begin{align*} \bar g:=\Phi_{c,d}^{-1}(g):Fc\to d \end{align*} such that \begin{align*} G(\bar g)\circ \eta_c=g. \end{align*} Therefore $(Fc,\eta_c)$ is a universal arrow from $c$ to $G$. Conversely, suppose that for every $c\in\mathcal C$ we are given a universal arrow \begin{align*} \eta_c:c\to G(Fc), \end{align*} and that the representing objects vary functorially as the values of $F:\mathcal C\to\mathcal D$. The universal property gives, for every $d\in\mathcal D$, a bijection \begin{align*} \Phi_{c,d}: \mathcal D(Fc,d) &\longrightarrow \mathcal C(c,Gd),\\ f &\longmapsto G(f)\circ \eta_c. \end{align*} The assumed functorial variation in $c$ gives naturality in $c$, while functoriality of $G$ gives naturality in $d$. Hence the universal-arrow data determine the hom-set adjunction data.[/step]

[guided]A universal arrow from $c$ to $G$ consists of an object of $\mathcal D$, here denoted $Fc$, and a morphism \begin{align*} \eta_c:c\to G(Fc), \end{align*} such that every morphism $g:c\to Gd$ factors uniquely through $\eta_c$ by applying $G$ to a unique morphism $Fc\to d$. Starting from the hom-set bijections, we define \begin{align*} \eta_c=\Phi_{c,Fc}(\operatorname{id}_{Fc}). \end{align*} Let $d\in\mathcal D$ and let $g:c\to Gd$ be any morphism in $\mathcal C$. Since \begin{align*} \Phi_{c,d}: \mathcal D(Fc,d)\longrightarrow \mathcal C(c,Gd) \end{align*} is a bijection, there is a unique morphism \begin{align*} \bar g:=\Phi_{c,d}^{-1}(g):Fc\to d \end{align*} with $\Phi_{c,d}(\bar g)=g$. By the transposition formula already proved, \begin{align*} \Phi_{c,d}(\bar g)=G(\bar g)\circ \eta_c. \end{align*} Therefore \begin{align*} G(\bar g)\circ \eta_c=g. \end{align*} The uniqueness of $\bar g$ is exactly the injectivity of $\Phi_{c,d}$. Hence $(Fc,\eta_c)$ is a universal arrow from $c$ to $G$. Conversely, assume that for every $c\in\mathcal C$ there is a universal arrow \begin{align*} \eta_c:c\to G(Fc), \end{align*} and that the objects $Fc$ are the values of a functor $F:\mathcal C\to\mathcal D$. The universal property says precisely that, for every $d\in\mathcal D$, composition with $\eta_c$ after applying $G$ gives a bijection \begin{align*} \Phi_{c,d}: \mathcal D(Fc,d) &\longrightarrow \mathcal C(c,Gd),\\ f &\longmapsto G(f)\circ \eta_c. \end{align*} Functoriality in $d$ follows because for $v:d\to d'$, \begin{align*} \Phi_{c,d'}(v\circ f) = G(v\circ f)\circ \eta_c = G(v)\circ G(f)\circ \eta_c = G(v)\circ \Phi_{c,d}(f). \end{align*} The hypothesis that the universal objects vary functorially as $F$ supplies the corresponding naturality in $c$. Thus the universal-arrow formulation recovers the same natural hom-set bijections.[/guided]

[step:Conclude that the constructions are mutually inverse] Starting with hom-set bijections, the constructed unit and counit give back the same bijections because the formulas \begin{align*} \Phi_{c,d}(f)=G(f)\circ \eta_c, \qquad \Phi_{c,d}^{-1}(g)=\varepsilon_d\circ F(g) \end{align*} were derived from naturality and the definitions of $\eta$ and $\varepsilon$. Starting with unit-counit data, the hom-set bijections constructed from them transpose identity morphisms back to the original unit and counit: \begin{align*} \Phi_{c,Fc}(\operatorname{id}_{Fc}) &= G(\operatorname{id}_{Fc})\circ \eta_c = \eta_c,\\ \Phi_{Gd,d}^{-1}(\operatorname{id}_{Gd}) &= \varepsilon_d\circ F(\operatorname{id}_{Gd}) = \varepsilon_d. \end{align*} The universal-arrow formulation is the same data viewed objectwise through the same bijections. Hence the three descriptions are equivalent. [/step]