Kerodon

Proof. Using Proposition 3.1.7.1, we can choose a commutative diagram

where the horizontal maps are anodyne, $Y'$ is a Kan complex, and $f'$ is a Kan fibration. Without loss of generality, we may assume that the map $g$ is bijective on vertices (for example, we could take $Y' = \operatorname{Ex}^{\infty }(Y)$; see Proposition 3.3.6.2). It follows from Proposition 3.3.7.1 that for each vertex $y \in Y$, the induced map of Kan complexes $X_{y} \rightarrow X'_{ g(y) }$ is a homotopy equivalence. In particular, $X_{y}$ is $n$-connective if and only if $X'_{ g(y) }$ is $n$-connective (Remark 3.5.1.5). We can therefore replace $f$ by $f'$, and thereby reduce to proving Proposition 3.5.1.22 in the special case where $X$ and $Y$ are Kan complexes.

Without loss of generality, we may assume that $n \geq 0$ (otherwise, the assertion is vacuous). Our proof proceeds by induction on $n$. In the case $n=0$, we must show that $f$ induces a surjection $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ if and only if every fiber of $f$ is nonempty, which follows from Corollary 3.2.6.3. Let us therefore assume that $n > 0$ and that $f$ has nonempty fibers. To carry out the inductive step, it will suffice to show that for every vertex $x \in X$ having image $y = f(x)$, the following conditions are equivalent:

$(a)$

The morphism $f$ induces a surjective group homomorphism $\pi _{n}(X,x) \rightarrow \pi _{n}(Y,y)$, and the kernel of the map $\pi _{n-1}(X,x) \rightarrow \pi _{n-1}(Y,y)$ is trivial (by convention, in the case $n = 1$, we define this kernel to be the inverse image of $[y] \in \pi _0(Y)$).

$(b)$

The set $\pi _{n-1}( X_{y}, x )$ consists of a single element.

This follows from Corollary 3.2.6.8 in the case $n > 1$, and from Variant 3.2.6.9 in the case $n =1$. $\square$