Corollary 3.5.2.2. Let $n$ be an integer and let $f: X \rightarrow Y$ be a morphism of simplicial sets which is bijective on $k$-simplices for $k < n$ and surjective for $k = n$. Then $f$ is $n$-connective.
Proof. For $n \leq 0$, this follows immediately from Example 3.5.1.17. We will therefore assume that $n > 0$. Our assumptions on $f$ guarantee that for $0 < m \leq n$, every lifting problem
admits a solution. Using Variant 3.1.7.12, we can factor $f$ as a composition $X \xrightarrow {i} X' \xrightarrow {f'} Y$, where $i$ is an anodyne morphism which is bijective on $k$-simplices for $k < n$, and $f'$ is a Kan fibration. Since the collection of $n$-connective morphisms is closed under composition, it will suffice to show that $f'$ is $n$-connective. By virtue of Proposition 3.5.1.22, this is equivalent to the assertion that for each vertex $y \in Y$, the fiber $X'_{y} = \{ y\} \times _{Y} X$ is an $n$-connective Kan complex. This follows from Example 3.5.1.10. $\square$