the functor ${{\mathscr{C}\mathrm{at}}}_\infty^\delta\to {{\mathscr{C}\mathrm{at}}}_\infty$ is continuous.

### Pairings of $\infty$-categories

We recall the theory of *pairings of $\infty$-categories*; essentially all this material is taken from [@Lurie:ZsNVU6Ru], though we provide proofs for certain details that will be important to us when discussing adjunctions. Recall that the $\infty$-category of pairings $${\operatorname}{CPair}\subseteq {\operatorname}{Fun}(\Lambda_0^2,{{\mathscr{C}\mathrm{at}}}_\infty)$$ is the full subcategory consisting of diagrams $$\begin{tikzpicture}
 \node  (m-1-1)  at (0,1) {${\mathcal}{M}$};
 \node  (m-1-2)  at (1,0) {${\mathcal}{D}^{op}$};
 \node  (m-2-1)  at (-1,0) {${\mathcal}{C}$};
\draw[->] (m-1-1) -- node {$\lambda_D$} (m-1-2);
\draw[->] (m-1-1) -- node [swap] {$\lambda_C$} (m-2-1);
\end{tikzpicture}$$ such that $\lambda:{\mathcal}{M}\xrightarrow{\lambda_C\times\lambda_D} {\mathcal}{C}\times{\mathcal}{D}^{op}$ is equivalent to a right fibration.

Given such a right fibration, $\lambda$ is classified by a functor (cf. [@Lurie:2009un § 2.2.1]) $${\mathcal}{D}\times{\mathcal}{C}^{op}\to {\mathscr{S}}$$ to the $\infty$-category, ${\mathscr{S}}$, of spaces; or equivalently a functor $$\label{eq:lD'}\lambda':{\mathcal}{D}\to {\operatorname}{Fun}({\mathcal}{C}^{op},{\mathscr{S}})=:{\mathscr{P}}({\mathcal}{C})$$ to the $\infty$-category of presheaves over ${\mathcal}{C}$. Here $\lambda'$ takes each vertex $d\in {\mathcal}{D}$ to the right fibration $$\label{eq:Pred}{\mathcal}{M}\times_{{\mathcal}{D}^{op}}\{d\}\to {\mathcal}{C}.$$

As in [@Lurie:ZsNVU6Ru], we call an object of $m\in{\mathcal}{M}$ *right universal* if it is a terminal object of ${\mathcal}{M}\times_{{\mathcal}{D}^{op}}\{\lambda_D(m)\}$ and we call a right fibration $$\label{eq:RFibrPair}\lambda:{\mathcal}{M}\to {\mathcal}{C}\times{\mathcal}{D}^{op}$$ a *right representable pairing*, if for each $d\in{\mathcal}{D}^{op}$, there exists a right universal object in the fibre ${\mathcal}{M}\times_{{\mathcal}{D}^{op}}\{d\}$ over $d$. In this case, for each $d\in{\mathcal}{D}$, the 