$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

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


The morphism $q$ is left cofinal and a left fibration.


The morphism $q$ is right cofinal and a right fibration.


The morphism $q$ is a trivial Kan fibration.

Proof. If $q$ is a trivial Kan fibration, then it is both a left fibration and a right fibration (Example Moreover, $q$ is also a categorical equivalence of simplicial sets (Proposition, hence left and right cofinal by virtue of Corollary This proves the implications $(3) \Rightarrow (1)$ and $(3) \Rightarrow (2)$.

We will complete the proof by showing that $(1) \Rightarrow (3)$ (the proof of the implication $(2) \Rightarrow (3)$ is similar). Assume that $q$ is a left cofinal left fibration. Then composition with $q$ induces a homotopy equivalence of Kan complexes $\operatorname{Fun}_{/S}( S, X) \rightarrow \operatorname{Fun}_{/S}(X,X)$. In particular, the morphism $q$ admits a section $f: S \rightarrow X$ such that $\operatorname{id}_{X}$ and $f \circ q$ belong to the same connected component of $\operatorname{Fun}_{/S}(X,X)$. For each vertex $s \in S$, let $X_{s} = \{ s\} \times _{S} X$ be the fiber of $q$ over $s$. Then the identity map $\operatorname{id}: X_{s} \rightarrow X_{s}$ is homotopic to the constant map $X_{s} \rightarrow \{ f(s) \} \hookrightarrow X_{s}$. It follows that the Kan complex $X_{s}$ is contractible. Allowing $s$ to vary, we conclude that the left fibration $q$ is a trivial Kan fibration (Proposition $\square$