#### 6.1.1. The total opposite of a displayed category [0018]

[001I] Construction 6.1.1·a.

Let $E$ be displayed over $B$; we define its total opposite $\TotOpCat{E}$ displayed over $\OpCat{B}$ as follows:

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

2. Given $f : x \to y\in \OpCat{B}$, a displayed morphism $\bar{x}\to\Sub{f} \bar{y}$ in $\TotOpCat{E}$ is given by a displayed morphism $\bar{y}\to\Sub{f} \bar{x}$ in $E$.

Warning. Do not confuse this construction with Construction 2.6·a [001Z], which produces a displayed category over $B$ and not $\OpCat{B}$.

[001J] Exercise 6.1.1·b.

Let $E$ be displayed over $B$. Prove that the total category (Section 3.1 [000A]) $\TotCat{\TotOpCat{E}}$ is $\OpCat{\prn{\TotCat{E}}}$, and its projection functor is $\OpCat{\prn{p\Sub{E}}} : \OpCat{\TotCat{E}}\to\OpCat{B}$.

[001K] Exercise 6.1.1·c.

Let $E$ be displayed over $B$, and let $f:x\to y\in B$. Prove that a morphism $\bar{f}:\bar{x}\to\Sub{f}\bar{y}$ is cartesian over $f$ in $E$ if and only if $\bar{f}:\bar{y}\to\Sub{f}\bar{x}$ is cocartesian over $f$ in $\TotOpCat{E}$.

[001L] Exercise 6.1.1·d.

Prove that a displayed category $E$ is a cartesian fibration over $B$ if and only if $\TotOpCat{E}$ is a cocartesian fibration over $\OpCat{B}$.