Proof of Theorem 8.2.4.6.
Let $\lambda : \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}_{-}^{\operatorname{op}} \times \operatorname{\mathcal{C}}_{+}$ be a coupling which is corepresented by a functor $F: \operatorname{\mathcal{C}}_{-} \rightarrow \operatorname{\mathcal{C}}_{+}$, and let $\mu : \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{D}}_{-}^{\operatorname{op}} \times \operatorname{\mathcal{D}}_{+}$ be a coupling which is represented by a functor $G: \operatorname{\mathcal{D}}_{+} \rightarrow \operatorname{\mathcal{D}}_{-}$. It follows from Theorem 8.2.2.11 that the coupling
\[ \Phi : \operatorname{Fun}_{\pm }(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) \rightarrow \operatorname{Fun}( \operatorname{\mathcal{C}}_{-}, \operatorname{\mathcal{D}}_{-} )^{\operatorname{op}} \times \operatorname{Fun}( \operatorname{\mathcal{C}}_{+}, \operatorname{\mathcal{D}}_{+} ) \]
of Remark 8.2.2.3 is representable, and that an object $(T_{-}, T, T_+) \in \operatorname{Fun}_{\pm }(\operatorname{\mathcal{C}},\operatorname{\mathcal{D}})$ is universal if and only if the functor $T$ carries couniversal objects of $\operatorname{\mathcal{C}}$ to universal objects of $\operatorname{\mathcal{D}}$. We will complete the proof by showing that the coupling $\Phi $ is representable by the functor
\[ H: \operatorname{Fun}( \operatorname{\mathcal{C}}_{+}, \operatorname{\mathcal{D}}_{+} ) \rightarrow \operatorname{Fun}(\operatorname{\mathcal{C}}_{-}, \operatorname{\mathcal{D}}_{-}) \quad \quad T_{+} \mapsto G \circ T_{+} \circ F. \]
Choose a morphism of couplings
8.47
\begin{equation} \begin{gathered}\label{equation:internal-hom-coupling1} \xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{ \widetilde{F} } \ar [d]^{\lambda } & \operatorname{Tw}( \operatorname{\mathcal{C}}_{+} ) \ar [d] \\ \operatorname{\mathcal{C}}_{-}^{\operatorname{op}} \times \operatorname{\mathcal{C}}_{+} \ar [r]^-{ F^{\operatorname{op}} \times \operatorname{id}} & \operatorname{\mathcal{C}}_{+}^{\operatorname{op}} \times \operatorname{\mathcal{C}}_{+} } \end{gathered} \end{equation}
which exhibits $\lambda $ as corepresented by $F$, and a morphism of couplings
8.48
\begin{equation} \begin{gathered}\label{equation:internal-hom-coupling2} \xymatrix@R =50pt@C=50pt{ \operatorname{Tw}(\operatorname{\mathcal{D}}_{+}) \ar [r]^-{ \widetilde{G} } \ar [d] & \operatorname{\mathcal{D}}\ar [d]^{\mu } \\ \operatorname{\mathcal{D}}_{+}^{\operatorname{op}} \times \operatorname{\mathcal{D}}_{+} \ar [r]^-{G^{\operatorname{op}} \times \operatorname{id}} & \operatorname{\mathcal{D}}_{-}^{\operatorname{op}} \times \operatorname{\mathcal{D}}_{+} } \end{gathered} \end{equation}
which exhibits $\mu $ as represented by $G$.
Let $E: \operatorname{Tw}( \operatorname{Fun}( \operatorname{\mathcal{C}}_{+}, \operatorname{\mathcal{D}}_{+} ) ) \rightarrow \operatorname{Fun}_{\pm }( \operatorname{Tw}(\operatorname{\mathcal{C}}_{+}), \operatorname{Tw}(\operatorname{\mathcal{D}}_{+} ) )$ be the comparison map of Example 8.2.4.8. Precomposition with (8.47) and postcomposition with (8.48) determines a functor $E': \operatorname{Fun}_{\pm }( \operatorname{Tw}(\operatorname{\mathcal{C}}_{+}), \operatorname{Tw}(\operatorname{\mathcal{D}}_{+} ) ) \rightarrow \operatorname{Fun}_{\pm }(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}})$ for which the diagram
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Tw}( \operatorname{Fun}( \operatorname{\mathcal{C}}_{+}, \operatorname{\mathcal{D}}_{+} ) ) \ar [r]^-{E' \circ E} \ar [d] & \operatorname{Fun}_{\pm }(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) \ar [d] \\ \operatorname{Fun}(\operatorname{\mathcal{C}}_{+}, \operatorname{\mathcal{D}}_{+} )^{\operatorname{op}} \times \operatorname{Fun}(\operatorname{\mathcal{C}}_{+}, \operatorname{\mathcal{D}}_{+}) \ar [r]^-{ H^{\operatorname{op}} \times \operatorname{id}} & \operatorname{Fun}(\operatorname{\mathcal{C}}_{-}, \operatorname{\mathcal{D}}_{-} )^{\operatorname{op}} \times \operatorname{Fun}(\operatorname{\mathcal{C}}_{+}, \operatorname{\mathcal{D}}_{+}) } \]
is commutative. We will complete the proof by showing that this diagram exhibits the coupling $\Phi $ as represented by $H$.
Fix a functor $T_{+}: \operatorname{\mathcal{C}}_{+} \rightarrow \operatorname{\mathcal{D}}_{+}$; we wish to show that the composite functor
\[ \operatorname{Tw}( \operatorname{Fun}( \operatorname{\mathcal{C}}_{+}, \operatorname{\mathcal{D}}_{+} ) ) \xrightarrow {E} \operatorname{Fun}_{\pm }( \operatorname{Tw}(\operatorname{\mathcal{C}}_{+}), \operatorname{Tw}(\operatorname{\mathcal{D}}_{+} ) ) \xrightarrow {E'} \operatorname{Fun}_{\pm }(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}}) \]
carries $\operatorname{id}_{ T_{+} }$ to a universal object of $\operatorname{Fun}_{\pm }(\operatorname{\mathcal{C}}, \operatorname{\mathcal{D}})$. Unwinding the definitions, we see that the image of $\operatorname{id}_{ T_{+} }$ is given by the triple $(G \circ T_{+} \circ F, \widetilde{G} \circ \operatorname{Tw}( T_{+} ) \circ \widetilde{F}, T_{+} ) \in \operatorname{Fun}_{\pm }( \operatorname{\mathcal{C}}, \operatorname{\mathcal{D}})$. Using the criterion of Theorem 8.2.2.11, we are reduced to showing that the composite functor
\[ \operatorname{\mathcal{C}}\xrightarrow { \widetilde{F} } \operatorname{Tw}( \operatorname{\mathcal{C}}_{+} ) \xrightarrow { \operatorname{Tw}( T_{+} ) } \operatorname{Tw}( \operatorname{\mathcal{D}}_{+} ) \xrightarrow { \widetilde{G} } \operatorname{\mathcal{D}} \]
carries every couniversal object $X \in \operatorname{\mathcal{C}}$ to a universal object of $\operatorname{\mathcal{D}}$. Proposition 8.2.3.7 guarantees that $\widetilde{F}(X) \in \operatorname{Tw}( \operatorname{\mathcal{C}}_{+} )$ corresponds to an isomorphism in $\operatorname{\mathcal{C}}_{+}$, so its image under $\operatorname{Tw}( T_{+} )$ corresponds to an isomorphism in $\operatorname{\mathcal{D}}_{+}$; the desired result now follows from our hypothesis that the functor $\widetilde{G}$ carries isomorphisms in $\operatorname{\mathcal{D}}_{+}$ to universal objects of $\operatorname{\mathcal{D}}$.
$\square$