Corollary 3.5.7.14. Let $f: X \rightarrow Y$ be a morphism of Kan complexes which is $n$-connective for some integer $n$. Then the induced map $\operatorname{cosk}_{n}(f): \operatorname{cosk}_{n}(X) \rightarrow \operatorname{cosk}_{n}(Y)$ is a homotopy equivalence.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proof. We have a commutative diagram of Kan complexes
\[ \xymatrix@R =50pt@C=50pt{ X \ar [r] \ar [d]^{f} & \operatorname{cosk}_{n}(X) \ar [d]^{\operatorname{cosk}_{n}(f) } \\ Y \ar [r] & \operatorname{cosk}_{n}(Y), } \]
where the horizontal maps are $n$-connective (Remark 3.5.3.22). Applying Corollary 3.5.1.28, we deduce that $\operatorname{cosk}_{n}(f)$ is $n$-connective. Since $\operatorname{cosk}_{n}(X)$ and $\operatorname{cosk}_{n}(Y)$ are $(n-1)$-truncated (Example 3.5.7.2), Corollary 3.5.7.13 guarantees that $\operatorname{cosk}_{n}(f)$ is a homotopy equivalence. $\square$