[proofplan]
We define the unit by transposing the identity morphism of $F c$, and define the counit by inverse-transposing the identity morphism of $G d$. Naturality of the hom-set bijections then proves that these componentwise definitions assemble into natural transformations. Finally, each triangle identity is proved by applying the relevant transposition map and observing that both sides have the same transpose; bijectivity of the hom-set correspondence then forces equality.
[/proofplan]
[step:Define the unit and counit by transposing identity morphisms]
For each object $c \in \mathcal C$, define the morphism
\begin{align*}
\eta_c:c \to G F c
\end{align*}
in $\mathcal C$ by
\begin{align*}
\eta_c := \Phi_{c,Fc}(\operatorname{id}_{F c}).
\end{align*}
For each object $d \in \mathcal D$, define the morphism
\begin{align*}
\varepsilon_d:F G d \to d
\end{align*}
in $\mathcal D$ by
\begin{align*}
\varepsilon_d := \Phi_{Gd,d}^{-1}(\operatorname{id}_{G d}).
\end{align*}
Equivalently,
\begin{align*}
\Phi_{Gd,d}(\varepsilon_d)=\operatorname{id}_{Gd}.
\end{align*}
These definitions have the required source and target because
\begin{align*}
\Phi_{c,Fc}: \operatorname{Hom}_{\mathcal D}(F c,F c) \to \operatorname{Hom}_{\mathcal C}(c,GF c)
\end{align*}
and
\begin{align*}
\Phi_{Gd,d}: \operatorname{Hom}_{\mathcal D}(F G d,d) \to \operatorname{Hom}_{\mathcal C}(G d,G d)
\end{align*}
are bijections.
[/step]
[step:Prove that the unit is natural]
Let $u:c \to c'$ be a morphism in $\mathcal C$. We must prove
\begin{align*}
G F u \circ \eta_c = \eta_{c'} \circ u.
\end{align*}
By naturality of $\Phi$ in the $\mathcal D$-variable, applied to the morphism $F u:F c \to F c'$ and the morphism $\operatorname{id}_{F c}:F c \to F c$, we have
\begin{align*}
\Phi_{c,Fc'}(F u \circ \operatorname{id}_{F c})
=
G(Fu)\circ \Phi_{c,Fc}(\operatorname{id}_{F c}).
\end{align*}
Using the definition of $\eta_c$, this becomes
\begin{align*}
\Phi_{c,Fc'}(F u)=G F u \circ \eta_c.
\end{align*}
By naturality of $\Phi$ in the $\mathcal C$-variable, applied to the morphism $u:c \to c'$ and the morphism $\operatorname{id}_{F c'}:F c' \to F c'$, we have
\begin{align*}
\Phi_{c,Fc'}(\operatorname{id}_{F c'} \circ F u)
=
\Phi_{c',Fc'}(\operatorname{id}_{F c'}) \circ u.
\end{align*}
Using the definition of $\eta_{c'}$, this becomes
\begin{align*}
\Phi_{c,Fc'}(F u)=\eta_{c'} \circ u.
\end{align*}
Comparing the two displayed identities gives
\begin{align*}
G F u \circ \eta_c = \eta_{c'} \circ u.
\end{align*}
Hence the components $(\eta_c)_{c \in \mathcal C}$ define a natural transformation
\begin{align*}
\eta:\operatorname{id}_{\mathcal C}\Rightarrow G F.
\end{align*}
[guided]
Fix a morphism $u:c \to c'$ in $\mathcal C$. Naturality of $\eta$ means that the square with top arrow $u:c \to c'$ and bottom arrow $GFu:GFc \to GFc'$ commutes, so we must prove
\begin{align*}
G F u \circ \eta_c = \eta_{c'} \circ u.
\end{align*}
The point is that both sides are the same morphism obtained by transposing $F u:F c \to F c'$ in two different ways. First use naturality of $\Phi$ in the $\mathcal D$-variable. We apply it to the morphism $F u:F c \to F c'$ in $\mathcal D$ and to $\operatorname{id}_{F c}:F c \to F c$. The naturality formula gives
\begin{align*}
\Phi_{c,Fc'}(F u \circ \operatorname{id}_{F c})
=
G(Fu)\circ \Phi_{c,Fc}(\operatorname{id}_{F c}).
\end{align*}
Since $F u \circ \operatorname{id}_{F c}=F u$ and $\Phi_{c,Fc}(\operatorname{id}_{F c})=\eta_c$ by definition, this simplifies to
\begin{align*}
\Phi_{c,Fc'}(F u)=G F u \circ \eta_c.
\end{align*}
Now use naturality of $\Phi$ in the $\mathcal C$-variable. We apply it to $u:c \to c'$ and to $\operatorname{id}_{F c'}:F c' \to F c'$. The naturality formula gives
\begin{align*}
\Phi_{c,Fc'}(\operatorname{id}_{F c'} \circ F u)
=
\Phi_{c',Fc'}(\operatorname{id}_{F c'}) \circ u.
\end{align*}
Since $\operatorname{id}_{F c'} \circ F u=F u$ and $\Phi_{c',Fc'}(\operatorname{id}_{F c'})=\eta_{c'}$, this becomes
\begin{align*}
\Phi_{c,Fc'}(F u)=\eta_{c'} \circ u.
\end{align*}
Both $G F u \circ \eta_c$ and $\eta_{c'} \circ u$ are therefore equal to the same morphism $\Phi_{c,Fc'}(F u):c \to G F c'$. Hence
\begin{align*}
G F u \circ \eta_c = \eta_{c'} \circ u.
\end{align*}
Because $u:c \to c'$ was arbitrary, the family $(\eta_c)_{c \in \mathcal C}$ is a natural transformation
\begin{align*}
\eta:\operatorname{id}_{\mathcal C}\Rightarrow G F.
\end{align*}
[/guided]
[/step]
[step:Prove that the counit is natural]
Let $v:d \to d'$ be a morphism in $\mathcal D$. We must prove
\begin{align*}
v \circ \varepsilon_d = \varepsilon_{d'} \circ F G v.
\end{align*}
Both sides are morphisms $F G d \to d'$ in $\mathcal D$. Since
\begin{align*}
\Phi_{Gd,d'}:\operatorname{Hom}_{\mathcal D}(F G d,d') \to \operatorname{Hom}_{\mathcal C}(Gd,Gd')
\end{align*}
is a bijection, it suffices to show that their images under $\Phi_{Gd,d'}$ are equal.
By naturality of $\Phi$ in the $\mathcal D$-variable, applied to $v:d \to d'$ and $\varepsilon_d:F G d \to d$, we obtain
\begin{align*}
\Phi_{Gd,d'}(v \circ \varepsilon_d)
=
Gv \circ \Phi_{Gd,d}(\varepsilon_d)
=
Gv \circ \operatorname{id}_{Gd}
=
Gv.
\end{align*}
By naturality of $\Phi$ in the $\mathcal C$-variable, applied to $Gv:Gd \to Gd'$ and $\varepsilon_{d'}:F G d' \to d'$, we obtain
\begin{align*}
\Phi_{Gd,d'}(\varepsilon_{d'} \circ F G v)
=
\Phi_{Gd',d'}(\varepsilon_{d'}) \circ Gv
=
\operatorname{id}_{Gd'} \circ Gv
=
Gv.
\end{align*}
Thus the two morphisms have the same image under the bijection $\Phi_{Gd,d'}$, and therefore
\begin{align*}
v \circ \varepsilon_d = \varepsilon_{d'} \circ F G v.
\end{align*}
Hence the components $(\varepsilon_d)_{d \in \mathcal D}$ define a natural transformation
\begin{align*}
\varepsilon:F G\Rightarrow \operatorname{id}_{\mathcal D}.
\end{align*}
[guided]
Fix a morphism $v:d \to d'$ in $\mathcal D$. Naturality of the counit asks for commutativity of the square with top arrow $FGv:FGd \to FGd'$ and bottom arrow $v:d \to d'$, namely
\begin{align*}
v \circ \varepsilon_d = \varepsilon_{d'} \circ F G v.
\end{align*}
Both sides are morphisms $F G d \to d'$ in $\mathcal D$. To prove equality, we apply the bijection
\begin{align*}
\Phi_{Gd,d'}:\operatorname{Hom}_{\mathcal D}(F G d,d') \to \operatorname{Hom}_{\mathcal C}(Gd,Gd').
\end{align*}
Because this map is injective, equality of images under $\Phi_{Gd,d'}$ implies equality of the original morphisms.
First compute the image of $v \circ \varepsilon_d$. Naturality of $\Phi$ in the $\mathcal D$-variable applies to $v:d \to d'$ and $\varepsilon_d:F G d \to d$, giving
\begin{align*}
\Phi_{Gd,d'}(v \circ \varepsilon_d)
=
Gv \circ \Phi_{Gd,d}(\varepsilon_d).
\end{align*}
By the definition of $\varepsilon_d$, we have $\Phi_{Gd,d}(\varepsilon_d)=\operatorname{id}_{Gd}$. Therefore
\begin{align*}
\Phi_{Gd,d'}(v \circ \varepsilon_d)
=
Gv \circ \operatorname{id}_{Gd}
=
Gv.
\end{align*}
Now compute the image of $\varepsilon_{d'} \circ F G v$. Naturality of $\Phi$ in the $\mathcal C$-variable applies to the morphism $Gv:Gd \to Gd'$ and the morphism $\varepsilon_{d'}:F G d' \to d'$, giving
\begin{align*}
\Phi_{Gd,d'}(\varepsilon_{d'} \circ F G v)
=
\Phi_{Gd',d'}(\varepsilon_{d'}) \circ Gv.
\end{align*}
By the definition of $\varepsilon_{d'}$, we have $\Phi_{Gd',d'}(\varepsilon_{d'})=\operatorname{id}_{Gd'}$. Hence
\begin{align*}
\Phi_{Gd,d'}(\varepsilon_{d'} \circ F G v)
=
\operatorname{id}_{Gd'} \circ Gv
=
Gv.
\end{align*}
Both morphisms $v \circ \varepsilon_d$ and $\varepsilon_{d'} \circ F G v$ have image $Gv$ under the bijection $\Phi_{Gd,d'}$. Since $\Phi_{Gd,d'}$ is injective, the morphisms are equal:
\begin{align*}
v \circ \varepsilon_d = \varepsilon_{d'} \circ F G v.
\end{align*}
Because $v:d \to d'$ was arbitrary, the family $(\varepsilon_d)_{d \in \mathcal D}$ defines a natural transformation
\begin{align*}
\varepsilon:F G\Rightarrow \operatorname{id}_{\mathcal D}.
\end{align*}
[/guided]
[/step]
[step:Verify the first triangle identity by comparing transposes]
Let $c \in \mathcal C$ be an object. We prove
\begin{align*}
\varepsilon_{F c} \circ F(\eta_c)=\operatorname{id}_{F c}.
\end{align*}
Both sides are morphisms $F c \to F c$ in $\mathcal D$. Since
\begin{align*}
\Phi_{c,Fc}:\operatorname{Hom}_{\mathcal D}(F c,F c) \to \operatorname{Hom}_{\mathcal C}(c,GF c)
\end{align*}
is a bijection, it suffices to compare their images under $\Phi_{c,Fc}$.
By naturality of $\Phi$ in the $\mathcal C$-variable, applied to $\eta_c:c \to G F c$ and $\varepsilon_{F c}:F G F c \to F c$, we have
\begin{align*}
\Phi_{c,Fc}(\varepsilon_{F c} \circ F(\eta_c))
=
\Phi_{GFc,Fc}(\varepsilon_{F c}) \circ \eta_c.
\end{align*}
By the definition of $\varepsilon_{F c}$,
\begin{align*}
\Phi_{GFc,Fc}(\varepsilon_{F c})=\operatorname{id}_{GFc}.
\end{align*}
Therefore
\begin{align*}
\Phi_{c,Fc}(\varepsilon_{F c} \circ F(\eta_c))
=
\operatorname{id}_{GFc}\circ \eta_c
=
\eta_c.
\end{align*}
On the other hand, by the definition of $\eta_c$,
\begin{align*}
\Phi_{c,Fc}(\operatorname{id}_{F c})=\eta_c.
\end{align*}
Thus the two morphisms have the same image under the bijection $\Phi_{c,Fc}$, and hence
\begin{align*}
\varepsilon_{F c} \circ F(\eta_c)=\operatorname{id}_{F c}.
\end{align*}
[guided]
Fix an object $c \in \mathcal C$. The first triangle identity is an equality in $\mathcal D$:
\begin{align*}
\varepsilon_{F c} \circ F(\eta_c)=\operatorname{id}_{F c}.
\end{align*}
Both sides are morphisms $F c \to F c$. The hom-set bijection
\begin{align*}
\Phi_{c,Fc}:\operatorname{Hom}_{\mathcal D}(F c,F c) \to \operatorname{Hom}_{\mathcal C}(c,GF c)
\end{align*}
is injective, so it is enough to show that the two sides have the same transpose under $\Phi_{c,Fc}$.
We first compute the transpose of $\varepsilon_{F c} \circ F(\eta_c)$. Naturality of $\Phi$ in the $\mathcal C$-variable applies to the morphism $\eta_c:c \to G F c$ and the morphism $\varepsilon_{F c}:F G F c \to F c$. This gives
\begin{align*}
\Phi_{c,Fc}(\varepsilon_{F c} \circ F(\eta_c))
=
\Phi_{GFc,Fc}(\varepsilon_{F c}) \circ \eta_c.
\end{align*}
The component $\varepsilon_{F c}$ was defined as the inverse transpose of $\operatorname{id}_{G F c}$, so
\begin{align*}
\Phi_{GFc,Fc}(\varepsilon_{F c})=\operatorname{id}_{G F c}.
\end{align*}
Substitution yields
\begin{align*}
\Phi_{c,Fc}(\varepsilon_{F c} \circ F(\eta_c))
=
\operatorname{id}_{GFc}\circ \eta_c
=
\eta_c.
\end{align*}
The transpose of the identity morphism $\operatorname{id}_{F c}:F c \to F c$ is also $\eta_c$, by the definition of the unit:
\begin{align*}
\Phi_{c,Fc}(\operatorname{id}_{F c})=\eta_c.
\end{align*}
Thus
\begin{align*}
\Phi_{c,Fc}(\varepsilon_{F c} \circ F(\eta_c))
=
\Phi_{c,Fc}(\operatorname{id}_{F c}).
\end{align*}
Since $\Phi_{c,Fc}$ is injective, the original morphisms are equal:
\begin{align*}
\varepsilon_{F c} \circ F(\eta_c)=\operatorname{id}_{F c}.
\end{align*}
[/guided]
[/step]
[step:Verify the second triangle identity by transposing the counit]
Let $d \in \mathcal D$ be an object. We prove
\begin{align*}
G(\varepsilon_d)\circ \eta_{Gd}=\operatorname{id}_{Gd}.
\end{align*}
By naturality of $\Phi$ in the $\mathcal D$-variable, applied to $\varepsilon_d:F G d \to d$ and $\operatorname{id}_{F G d}:F G d \to F G d$, we have
\begin{align*}
\Phi_{Gd,d}(\varepsilon_d \circ \operatorname{id}_{F G d})
=
G(\varepsilon_d)\circ \Phi_{Gd,FGd}(\operatorname{id}_{F G d}).
\end{align*}
Using the definitions of $\varepsilon_d$ and $\eta_{Gd}$, this becomes
\begin{align*}
\operatorname{id}_{Gd}
=
G(\varepsilon_d)\circ \eta_{Gd}.
\end{align*}
Therefore
\begin{align*}
G(\varepsilon_d)\circ \eta_{Gd}=\operatorname{id}_{Gd}.
\end{align*}
[guided]
Fix an object $d \in \mathcal D$. The second triangle identity is an equality in $\mathcal C$:
\begin{align*}
G(\varepsilon_d)\circ \eta_{Gd}=\operatorname{id}_{Gd}.
\end{align*}
Here $\eta_{Gd}:Gd \to G F G d$ is the unit component at the object $Gd \in \mathcal C$, and $G(\varepsilon_d):G F G d \to G d$ is the image under $G$ of the counit component $\varepsilon_d:F G d \to d$.
We compute the transpose of $\varepsilon_d$ in a way that exposes exactly this composite. Naturality of $\Phi$ in the $\mathcal D$-variable applies to the morphism $\varepsilon_d:F G d \to d$ and the identity morphism $\operatorname{id}_{F G d}:F G d \to F G d$. It gives
\begin{align*}
\Phi_{Gd,d}(\varepsilon_d \circ \operatorname{id}_{F G d})
=
G(\varepsilon_d)\circ \Phi_{Gd,FGd}(\operatorname{id}_{F G d}).
\end{align*}
The left-hand side simplifies because $\varepsilon_d \circ \operatorname{id}_{F G d}=\varepsilon_d$, and by definition of the counit,
\begin{align*}
\Phi_{Gd,d}(\varepsilon_d)=\operatorname{id}_{Gd}.
\end{align*}
The factor on the right-hand side is the unit at $Gd$, because
\begin{align*}
\eta_{Gd}=\Phi_{Gd,FGd}(\operatorname{id}_{F G d}).
\end{align*}
Therefore the displayed naturality identity becomes
\begin{align*}
\operatorname{id}_{Gd}
=
G(\varepsilon_d)\circ \eta_{Gd}.
\end{align*}
This is exactly the second triangle identity.
[/guided]
[/step]
[step:Conclude that the hom-set adjunction yields the required adjunction data]
The families
\begin{align*}
(\eta_c)_{c \in \mathcal C},
\qquad
(\varepsilon_d)_{d \in \mathcal D}
\end{align*}
have been shown to define natural transformations
\begin{align*}
\eta:\operatorname{id}_{\mathcal C}\Rightarrow G F,
\qquad
\varepsilon:F G\Rightarrow \operatorname{id}_{\mathcal D}.
\end{align*}
The preceding two steps prove that for every $c \in \mathcal C$ and every $d \in \mathcal D$,
\begin{align*}
\varepsilon_{F c}\circ F(\eta_c)&=\operatorname{id}_{F c},\\
G(\varepsilon_d)\circ \eta_{Gd}&=\operatorname{id}_{Gd}.
\end{align*}
Thus the unit and counit determined by the hom-set adjunction satisfy the triangle identities.
[/step]
