[001Z] Construction 2.6·a.

Let $E$ be fibered over $B$; we may define the opposite fibered category $\OpCat{E}$ over $B$ like so:

  1. An object of $\OpCat{E}\Sub{x}$ is given by an object of $E\Sub{x}$.

  2. Given $f : x \to y\in B$, a morphism $\bar{x}\to_f \bar{y}$ in $\OpCat{E}$ is given in terms of $E$ by a cartesian map $\bar{y}\Sub{f} : \bar{y}\Sub{x} \to\Sub{f} \bar{y}$ together with a vertical map $h : \bar{y}\Sub{x}\to\Sub{\Idn{x}} \bar{y}$ as depicted below: such that $\brc{\bar{x} \leftarrow \bar{y}\Sub{x}\Sup{1}\to \bar{y}}$ is identified with $\brc{\bar{x}\leftarrow\bar{y}\Sub{x}\Sup{2}\to \bar{y}}$ when they agree up to the unique vertical isomorphism $\bar{y}\Sub{x}\Sup{1}\cong\bar{y}\Sub{x}\Sup{2}$ induced by the universal property of cartesian maps in the sense that the following diagram commutes: