Proposition 5.1.4.14 (01UM)

Proposition 5.1.4.14. Let $q: X \rightarrow S$ be a morphism of simplicial sets. The following conditions are equivalent:

$(1)$: The morphism $q$ is a right fibration.
$(2)$: The morphism $q$ is a cartesian fibration and every edge of $X$ is $q$-cartesian.
$(3)$: The morphism $q$ is a cartesian fibration and, for every vertex $s \in S$, the fiber $X_{s} = \{ s\} \times _{S} X$ is a Kan complex.

Proof. The equivalence $(1) \Leftrightarrow (2)$ is immediate from the definitions. The implication $(2) \Rightarrow (3)$ follows from Propositions 5.1.4.11 and 4.4.2.1. We will complete the proof by showing that $(3)$ implies $(2)$. Assume that $q$ is a cartesian fibration and that each fiber of $q$ is a Kan complex. Let $h: x \rightarrow z$ be an edge of $X$; we wish to show that $h$ is $q$-cartesian. Since $q$ is a cartesian fibration, we can choose a $q$-cartesian edge $g: y \rightarrow z$ of $X$ satisfying $q(g) = q(h)$. The assumption that $g$ is $q$-cartesian then guarantees the existence of a $2$-simplex $\sigma $ of $X$ satisfying

\[ d^{2}_0(\sigma ) = g \quad \quad d^{2}_1(\sigma ) = h \quad \quad q(\sigma ) = s^{1}_0( q(h) ), \]

as depicted in the diagram

\[ \xymatrix@R =50pt@C=50pt{ & y \ar [dr]^{g} & \\ x \ar [ur]^{f} \ar [rr]^-{h} & & z. } \]

Set $s = q(x)$, so that $f$ is a morphism in the $\infty $-category $X_{s} = \{ s\} \times _{S} X$. Since $X_{s}$ is a Kan complex, $f$ is an isomorphism (Proposition 1.4.6.10). Applying Remark 5.1.3.8 and Corollary 5.1.3.9, we deduce that $h$ is $q$-cartesian. $\square$