Proof of Theorem 11.10.6.3.
The equivalence $(1) \Leftrightarrow (2)$ follows from Remark 11.10.6.2, and the implication $(2) \Rightarrow (3)$ is immediate (Example 3.1.3.12). We will complete the proof by showing that $(3) \Rightarrow (2)$. Assume that we are given a commutative diagram of simplicial sets
\[ \xymatrix@R =50pt@C=50pt{ X \ar [rr]^{f} \ar [dr]_{q_ X} & & Y \ar [dl]^{q_ Y} \\ & S, & } \]
where $q_{X}$ and $q_{Y}$ are left fibrations and, for every vertex $s \in S$, the induced map of fibers $f_{s}: X_{s} \rightarrow Y_{s}$ is a homotopy equivalence of Kan complexes. We wish to show that, for each object $B \in (\operatorname{Set_{\Delta }})_{/S}$, the induced map $f_{B}: \operatorname{Fun}_{/S}(B,X) \rightarrow \operatorname{Fun}_{/S}(B,Y)$ is a homotopy equivalence. For each $n \geq 0$, let $\operatorname{sk}_{n}(B)$ denote the $n$-skeleton of $B$ (Construction 1.1.4.1). Then $f_{A}$ can be realized as the inverse limit of a tower of morphisms $\{ f_{ \operatorname{sk}_ n(B) }: \operatorname{Fun}_{/S}( \operatorname{sk}_ n(B), X) \rightarrow \operatorname{Fun}_{/S}( \operatorname{sk}_ n(B), Y) \} _{n \geq 0}$. Using Proposition 11.10.6.5, we see that the transition morphisms $\operatorname{Fun}_{/S}( \operatorname{sk}_{n+1}(B), X) \rightarrow \operatorname{Fun}_{/S}( \operatorname{sk}_ n(B), X)$ and $\operatorname{Fun}_{/S}( \operatorname{sk}_{n+1}(B), Y) \rightarrow \operatorname{Fun}_{/S}( \operatorname{sk}_ n(B), Y)$ are Kan fibrations. Consequently, to show that $f_{B}$ is a homotopy equivalence, it will suffice to show that each of the morphisms $f_{ \operatorname{sk}_{n}(B) }$ is a homotopy equivalence (Proposition 11.9.5.6). We may therefore replace $B$ by $\operatorname{sk}_{n}(B)$ and thereby reduce to the case where the simplicial set $B$ has dimension $\leq n$, for some $n \geq -1$.
The proof now proceeds by induction on $n$. In the case $n=-1$, the simplicial set $B$ is empty, and the morphism $f_{B}$ is an isomorphism (see Remark 11.10.6.7). Assume that $n \geq 0$, let $B'$ denote the $(n-1)$-skeleton of $B$, and let $T$ denote the collection of all nondegenerate $n$-simplices of $B$. Then Proposition 1.1.4.12 supplies a pushout diagram of simplicial sets
\[ \xymatrix@R =50pt@C=50pt{ A' \ar [r] \ar [d] & A \ar [d] \\ B' \ar [r] & B, } \]
where $A = \coprod _{ \tau \in T} \Delta ^{n}$ and $A' = \coprod _{\tau \in T} \operatorname{\partial \Delta }^ n$. This pushout square determines a commutative diagram of Kan complexes
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Fun}_{/S}(B, X) \ar [dd]^(.5){ f_{B} } \ar [rr] \ar [dr] & & \operatorname{Fun}_{/S}( B', X) \ar [dd]^(.6){ f_{B'} } \ar [dr] & \\ & \operatorname{Fun}_{/S}(A, X) \ar [rr] \ar [dd]^(.6){ f_{A} } & & \operatorname{Fun}_{/S}(A', X) \ar [dd]^(.5){ f_{A'} } \\ \operatorname{Fun}_{/S}(B, Y) \ar [rr] \ar [dr] & & \operatorname{Fun}_{/S}(B', Y) \ar [dr] & \\ & \operatorname{Fun}_{/S}(A, Y) \ar [rr] & & \operatorname{Fun}_{/S}(A', Y), } \]
where the upper and lower squares are homotopy Cartesian by virtue of Corollary 11.10.6.8. Consequently, to show that $f_{B}$ is a homotopy equivalence of Kan complexes, it will suffice to show that $f_{B'}$, $f_{A'}$, and $f_{A}$ are homotopy equivalences (Corollary 3.4.1.12). In the first two cases, this follows from our inductive hypothesis. We may therefore replace $B$ by $A$, and thereby reduce to the case where is a coproduct of simplices. Using Remarks 11.10.6.7 and 3.1.6.8, we can further reduce to the case where $B = \Delta ^ n$ is a simplex of dimension $n$. In this case, the inclusion of the initial vertex $0 \in \Delta ^ n$ is left anodyne (Example 4.3.7.10). Let $s \in S$ denote the image of the vertex $0 \in \Delta ^ n$, so that we have a commutative diagram of Kan complexes
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Fun}_{/S}( B, X) \ar [r]^-{ f_{B} } \ar [d] & \operatorname{Fun}_{/S}(B, Y) \ar [d] \\ \operatorname{Fun}_{/S}( \{ s\} , X) \ar [r]^-{f_ s} & \operatorname{Fun}_{/S}( \{ s\} , Y). } \]
Invoking Remark 11.10.6.6, we conclude that the vertical maps in this diagram are trivial Kan fibrations. Since $f_{s}$ is a homotopy equivalence by assumption, it follows that $f_{B}$ is also a homotopy equivalence.
$\square$