I have recently been reading on 2-fibrations. It is well-known (from Hermida) that the codomain functor $cod \colon \textbf{Fib} \to \textbf{Cat}$ taking each fibration to its base category from the 2-category of fibrations to the category of categories, is a 2-fibration.
[Question] When we replace $\textbf{Fib}$ with $\textbf{BiFib}$ the 2-category of bifibrations, does the functor $cod \colon \textbf{BiFib} \to \textbf{Cat}$ become a 2-bifibration?
It seems like it would be to me: We know that pullbacks in $\textbf{Cat}$ enjoy the 2-dimentional universal property and they yield 2-cartesian 1-cells for $cod \colon \textbf{Fib} \to \textbf{Cat}$. Since a morphism of bifibrations should have both pullback and pushforward. Also we know that for $\textbf{Fib}(\mathcal{A})$(2-category of fibrations over $\mathcal{A}$), given two fibrations $\mathcal{p : E \to A}$ and $\mathcal{q : D \to A}$ then a fibred functor between them is a functor $F \colon \mathcal{E} \to \mathcal{D}$ with $qF = p$ that preserves cartesian morphism. So for $\textbf{BiFib}$ such a fibred functor would preserve (co)cartesian morphisms. I guess in some sense the pushforward would have to give the 2-cocartesian 1-cells in some way...
I am not sure how to proceed with the proof, essentially would suffice to show that
$cod \colon \textbf{BiFib}\left((\mathcal{E \to^p A}) \to (\mathcal{D \to^q B})\right) \to \textbf{Cat}(\mathcal{A,B} )$ is a fibration. And additionally, it is a opfibration.
