$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proposition 11.10.7.14. Suppose 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. Then the following conditions are equivalent:
- $(1)$
The morphism $f$ is a homotopy equivalence relative to $S$.
- $(2)$
The morphism $f$ is a covariant equivalence relative to $S$.
- $(3)$
For each vertex $s \in S$, the induced map $f_{s}: X_{s} \rightarrow Y_{s}$ is a homotopy equivalence of Kan complexes.
Proof.
The implications $(1) \Rightarrow (2)$ and $(1) \Rightarrow (3)$ follow from Example 11.10.7.3 and Remark 11.10.6.2 (and do not require the assumption that $q_{X}$ and $q_{Y}$ are left fibrations). The implication $(3) \Rightarrow (1)$ follows from Theorem 11.10.6.3. We will complete the proof by showing that $(2) \Rightarrow (1)$. Assume that $f$ is a covariant equivalence relative to $S$. Then precomposition with $f$ induces a bijection $\pi _0( \operatorname{Fun}_{/S}(Y,X) ) \rightarrow \pi _0( \operatorname{Fun}_{/S}(X,X) )$. We can therefore choose a morphism $g: Y \rightarrow X$ in the slice category $(\operatorname{Set_{\Delta }})_{/S}$ such that $g \circ f$ and $\operatorname{id}_{X}$ belong to the same connected component of $\operatorname{Fun}_{/S}(X,X)$. Then $f \circ g \circ f$ and $f$ belong to the same connected component of the simplicial set $\operatorname{Fun}_{/S}(X,Y)$. Since the map $\pi _0( \operatorname{Fun}_{/S}(Y,Y) ) \rightarrow \pi _0( \operatorname{Fun}_{/S}(X,Y) )$ is injective, it follows that $f \circ g$ and $\operatorname{id}_{Y}$ belong to the same connected component of $\operatorname{Fun}_{/S}(Y,Y)$. It follows that $f$ is a homotopy equivalence relative to $S$ (with homotopy inverse $g$).
$\square$