Proposition 3.5.9.13. Let $f: X \rightarrow Y$ be a morphism of Kan complexes and let $n$ be an integer. Then:
If $Y$ is $n$-truncated and $f$ is $n$-truncated, then $X$ is $n$-truncated.
If $X$ is $n$-truncated and $Y$ is $(n+1)$-truncated, then $f$ is $n$-truncated.
If $X$ is $n$-truncated, $f$ is $(n-1)$-truncated, and $\pi _0(f)$ is surjective, then $Y$ is $n$-truncated.
Proof. For every integer $m \geq 0$ and every vertex $x \in X$ having image $y = f(x)$, we make the following observations:
If the morphism $\pi _{m}(f): \pi _{m}(X,x) \rightarrow \pi _{m}(Y,y)$ is injective and $\pi _{m}(Y,y)$ is a singleton, then $\pi _{m}(X,x)$ is also a singleton.
If the sets $\pi _{m}(X,x)$ and $\pi _{m+1}(Y,y)$ are singletons, then $\pi _{m}(f)$ is injective and $\pi _{m+1}(f)$ is surjective.
If $\pi _{m}(f)$ is surjective and $\pi _{m}(X,x)$ is a singleton, then $\pi _{m}(Y,y)$ is a singleton.
If $n \geq -1$, then Proposition 3.5.9.13 follows by combining these observations with Proposition 3.5.7.7 (together with Remarks 3.5.7.8 and 3.5.7.9), by allowing $m$ to range over integers $> n$ and allowing the vertex $x$ to vary. The case $n \leq -2$ then follows from the following additional observations:
If the morphism $\pi _0(f)$ is surjective and $\pi _0(Y)$ is nonempty, then $\pi _0(X)$ is also nonempty.
If $\pi _0(X)$ is nonempty and $\pi _0(Y)$ has at most one element, then $\pi _0(f)$ is surjective.
If $\pi _0(X)$ is nonempty, then $\pi _0(Y)$ is also nonempty.