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Proposition Let $f: X \rightarrow S$ be a morphism of simplicial sets. If $f$ is a Kan fibration, then the induced map $\operatorname{Ex}^{\infty }(f): \operatorname{Ex}^{\infty }(X) \rightarrow \operatorname{Ex}^{\infty }(S)$ is also a Kan fibration.

Proof. Since the collection of Kan fibrations is stable under the formation of filtered colimits (Remark, it will suffice to show that each of the maps $\operatorname{Ex}^{n}(f): \operatorname{Ex}^{n}(X) \rightarrow \operatorname{Ex}^{n}(S)$ is a Kan fibration. Proceeding by induction on $n$, we can reduce to the case $n=1$, which follows from Corollary $\square$