Corollary 3.5.7.18. Let $X$ be a Kan complex and let $n \geq 0$ be an integer. The following conditions are equivalent:
There exists a homotopy equivalence of Kan complexes $X \rightarrow \mathrm{K}(G,n)$. Here $G$ is a set if $n=0$, a group if $n =1$, and an abelian group if $n \geq 2$ (see Construction 2.5.6.9).
The Kan complex $X$ is $n$-truncated and $n$-connective.
Proof. We will show that $(2) \Rightarrow (1)$ (the reverse implication is clear). Assume that $X$ is $n$-truncated and $n$-connective. By virtue of Proposition 3.5.2.9 (and Remark 3.5.2.10), we can assume that $X$ has a single $m$-simplex for each $m < n$. Our assumption that $X$ is $n$-truncated guarantees that the tautological map $X \rightarrow \pi _{\leq n}(X)$ is a homotopy equivalence (Variant 3.5.7.16). It will therefore suffice to show that $\pi _{\leq n}(X)$ is an Eilenberg-MacLane space $\mathrm{K}(G,n)$, which follows from the criterion of Proposition 3.5.5.16. $\square$