Corollary Let $f: X \rightarrow Y$ be a morphism of simplicial sets, and let $Z$ be a Kan complex. If $f$ is a weak homotopy equivalence, then composition with $f$ induces a homotopy equivalence $\operatorname{Fun}( Y, Z) \rightarrow \operatorname{Fun}(X,Z)$.

Proof. Using Corollary, we can choose an anodyne morphism $g: Y \hookrightarrow Y'$, where $Y'$ is a Kan complex. Using Proposition, we can factor $g \circ f$ as a composition $X \xrightarrow {g'} X' \xrightarrow {f'} Y'$, where $g'$ is anodyne and $f'$ is a Kan fibration. We then have a commutative diagram

\[ \xymatrix@C =50pt@R=50pt{ X \ar [r]^-{f} \ar [d]^{g'} & Y \ar [d]^{g} \\ X' \ar [r]^-{ f' } & Y', } \]

where $f'$ is a Kan fibration between Kan complexes and the vertical maps are anodyne, and therefore weak homotopy equivalences. Using the two-out-of-three property, we deduce that $f'$ is also a weak homotopy equivalence (Remark It follows that $f'$ is a homotopy equivalence (Proposition We then obtain a commutative diagram of Kan complexes

\[ \xymatrix@C =50pt@R=50pt{ \operatorname{Fun}(X,Z) & \ar [l] \operatorname{Fun}(Y,Z) \\ \operatorname{Fun}(X', Z) \ar [u] & \operatorname{Fun}(Y', Z) \ar [l] \ar [u] } \]

where the lower horizontal map is a homotopy equivalence, and the vertical maps are trivial Kan fibrations (Corollary In particular, the vertical maps are homotopy equivalences (Proposition, so the two-out-of-three property guarantees that the upper horizontal map is also a homotopy equivalence (Remark $\square$