Proof.
We first show that $(1)$ implies $(2)$. Let $Z$ be an $n$-truncated simplicial set; we wish to show that composition with $f$ induces a homotopy equivalence $\theta : \operatorname{Fun}(Y, Z ) \rightarrow \operatorname{Fun}(X, Z)$. By virtue of Proposition 3.5.7.15, we may assume without loss of generality that $Z$ is $(n+1)$-coskeletal. In this case, we can use Proposition 3.5.3.17 to identify $\theta $ with the map
\[ \operatorname{Fun}( \operatorname{cosk}_{n+1}(Y), Z ) \rightarrow \operatorname{Fun}( \operatorname{cosk}_{n+1}(X), Z ) \]
given by precomposition with $\operatorname{cosk}_{n+1}(f)$. If $f$ is $(n+1)$-connective, then Corollary 3.5.7.14 guarantees that $\operatorname{cosk}_{n+1}(f)$ is a homotopy equivalence, so that $\theta $ is also a homotopy equivalence.
The implication $(2) \Rightarrow (3)$ follows from Remark 3.1.6.5. We will complete the proof by showing that $(3)$ implies $(1)$. Let $u: X \rightarrow \operatorname{cosk}_{n+1}(X)$ be the tautological map. Then $u$ exhibits $\operatorname{cosk}_{n+1}(X)$ as an $n$-truncation of $X$ (Example 3.5.7.23). In particular, $\operatorname{cosk}_{n+1}(X)$ is an $n$-truncated Kan complex, so condition $(3)$ guarantees that there exists a morphism $g: Y \rightarrow \operatorname{cosk}_{n+1}(X)$ such that $g \circ f$ is homotopic to $u$. For every $n$-truncated Kan complex $Z$, we have a commutative diagram
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{Kan}}}( \operatorname{cosk}_{n+1}(X), Z) ) \ar [rr]^{ \circ [g] } \ar [dr]_{ \circ [u] } & & \operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{Kan}}}( Y, Z ) \ar [dl]^{ \circ [f] } \\ & \operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{Kan}}}( X, Z), & } \]
where the vertical maps are bijective. It follows that $g$ is a homotopy equivalence, so that $f$ exhibits $Y$ as an $n$-truncation of $X$ by virtue of Remarks 3.5.7.22 and 3.5.7.21.
$\square$