Proposition 3.5.1.26. Let $f: X \rightarrow Y$ be a morphism of simplicial sets and let $n$ be an integer. Then:
If the $Y$ is $n$-connective and $f$ is $n$-connective, then $X$ is $n$-connective.
If $X$ is $n$-connective and $Y$ is $(n+1)$-connective, then $f$ is $n$-connective.
If $f$ is $n$-connective and $X$ is $(n+1)$-connective, then $Y$ is $(n+1)$-connective.
Proof. Without loss of generality, we may assume that $X$ and $Y$ are Kan complexes. We proceed by induction on $n$. If $n < 0$, then assertions $(1)$ and $(2)$ are vacuous, and $(3)$ reduces to the assertion that if $X$ is nonempty, then $Y$ is also nonempty. When $n=0$, we can restate Proposition 3.5.1.26 as follows:
If $Y$ is nonempty and $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is surjective, then $X$ is nonempty.
If $X$ is nonempty and $Y$ is connected, then the map $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is surjective.
If $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is surjective and $X$ is connected, then $Y$ is connected.
Assume that $n > 0$, and let $x \in X$ be a vertex having image $y = f(x)$. The inductive step is a consequence of the following observations:
If the morphism $\pi _{n-1}(f): \pi _{n-1}(X,x) \rightarrow \pi _{n-1}(Y,y)$ is bijective and $\pi _{n-1}(Y,y)$ is a singleton, then $\pi _{n-1}(X,x)$ is also a singleton.
If the sets $\pi _{n-1}(X,x)$ and $\pi _{n}(Y,y)$ are singletons, then $\pi _{n-1}(f)$ is injective and $\pi _{n}(f)$ is surjective.
If $\pi _{n}(f)$ is surjective and $\pi _{n}(X,x)$ is a singleton, then $\pi _{n}(Y,y)$ is a singleton.