[002B] Lemma 2.3·c (Equivalence with Grothendieck's fibrations).

Let $E$ be displayed over $B$. Then $E$ is a cartesian fibration in the sense of Definition 2·c [0002] if and only if the following two conditions hold:

  1. Hypocartesian lifts. For each $f:x\to y\in B$ and $\bar{y}\in E\Sub{y}$ there exists a displayed object $\bar{x}\in E\Sub{x}$ and hypocartesian morphism $\bar{f}:\bar{x}\to\Sub{f}\bar{y}$.
  2. Closure under composition. If $\bar{f}:\bar{x}\to\Sub{f}\bar{y}$ and $\bar{g}:\bar{y}\to\Sub{g}\bar{z}$ are hypocartesian, then $\bar{f};\bar{g}$ is hypocartesian.

Proof. Suppose first that $E$ is a cartesian fibration in our sense. Then $E$ has hypocartesian lifts because it has cartesian lifts. For closure under composition, fix hypocartesian $\bar{f},\bar{g}$; by Lemma 2.3·b [002C] we know that $\bar{f},\bar{g}$ are also cartesian and hence by Lemma 2.2·a [001H] so is the composite $\bar{f};\bar{g}$; therefore it follows that $\bar{f};\bar{g}$ is also hypocartesian.

Conversely, suppose that $E$ is a cartesian fibration in the sense of Grothendieck, and let $\bar{f}:\bar{x}\to\Sub{f}\bar{y}$ be the hypocartesian lift of $f:x\to y$ at $\bar{y}\in E\Sub{y}$; we shall see that $\bar{f}$ is also a cartesian lift of $f$ at $\bar{y}$ by constructing a unique factorization as follows: Let $\bar{m}:\bar{u}\tick\to\Sub{m}\bar{x}$ be the hypocartesian lift of $m$ at $\bar{x}$, where $\bar{u}\tick\in E\Sub{u}$. By hypothesis, the composite $\bar{m};\bar{f} : \bar{u}\tick\to\Sub{m;f}\bar{y}$ is hypocartesian, so $\bar{h}$ factors uniquely through $\bar{m};\bar{f}$ over $\Idn{u}$: The composite $i;\bar{m} : \bar{u}\to\Sub{m}\bar{x}$ is the required (cartesian) factorization of $\bar{h}$ through $\bar{f}$ over $m$. To see that $i;\bar{m}$ is the unique such map, we observe that all morphisms $\bar{u}\to\Sub{m}\bar{x}$ factor uniquely through $\bar{m}$ over $\Idn{u}$ as a consequence of $\bar{m}$ being hypocartesian.