Kerodon

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Why not prove this result in the following easier way? It suffices to prove that, for every simplicial set $A$, the induced map $\operatorname{Fun}(A,\operatorname{Fun}(Y,Z)) \to \operatorname{Fun}(A,\operatorname{Fun}(X,Z))$ is bijective on $\pi_0$ (by Remark 00U5). But this map is isomorphic to the induced map $\operatorname{Fun}(Y,\operatorname{Fun}(A,Z)) \to \operatorname{Fun}(X,\operatorname{Fun}(A,Z))$, which is bijective on $\pi_0$ since $f \colon X \to Y$ is a weak homotopy equivalence and $\operatorname{Fun}(A,Z)$ is a Kan complex.

Yes, that sounds better. Thanks!

Why is this result a corollary? The proof does not make use of fibrant replacement.

Indeed. Thanks!