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Corollary Let $f: X \rightarrow Y$ be a morphism of simplicial sets. Then $f$ is a weak homotopy equivalence if and only if $\operatorname{Ex}^{\infty }(f)$ is a weak homotopy equivalence.

Proof. We have a commutative diagram

\[ \xymatrix@R =50pt@C=50pt{ X \ar [d]^{ \rho ^{\infty }_ X} \ar [r]^-{f} & Y \ar [d]^{ \rho ^{\infty }_{Y} } \\ \operatorname{Ex}^{\infty }(X) \ar [r]^-{ \operatorname{Ex}^{\infty }(f) } & \operatorname{Ex}^{\infty }(Y), } \]

where the vertical maps are weak homotopy equivalences (Proposition The desired result now follows from the two-out-of-three property (Remark $\square$