Proposition 10.3.4.17. Let $f: X \rightarrow Y$ be a map of Kan complexes. The following conditions are equivalent:
The map $f$ is a universal quotient morphism in the $\infty $-category $\operatorname{\mathcal{S}}$ (Definition 10.3.4.1).
The map $f$ is a quotient morphism in the $\infty $-category $\operatorname{\mathcal{S}}$ (Definition 10.3.2.1).
The map $f$ is $0$-connective: that is, it induces a surjection $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$.
Proof. The implication $(1) \Rightarrow (2)$ is a special case of Corollary 10.3.4.10. We next show that $(2)$ implies $(3)$. Assume that $f$ is a quotient morphism, so that $Y$ can be identified with the geometric realization of the Čechnerve $\operatorname{\check{C}}(X/Y)_{\bullet }$ in the $\infty $-category $\operatorname{\mathcal{S}}$ (Proposition 10.3.2.4). Note that the functor
preserves the formation of geometric realizations (since it is left adjoint to the inclusion functor). It follows that $\pi _0(Y)$ can be identified with the geometric realization of $\pi _0( \operatorname{\check{C}}(X/Y)_{\bullet } )$ in the category of sets: that is, with the coequalizer of the projection maps $\pi _0( X \times _{Y} X ) \rightrightarrows \pi _0(X)$ (see Corollary 10.2.2.12). In particular, the tautological map $\pi _0(X) \rightarrow \pi _0(Y)$ is surjective.
We now show that $(3)$ implies $(1)$. Assume that condition $(3)$ is satisfied. For every contractible Kan complex $C$, the composition map $\operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{\mathcal{S}}}}( C, X ) \xrightarrow { [f] \circ } \operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{\mathcal{S}}}}( C, Y )$ can be identified with $\pi _0(f)$, and is therefore surjective. Since the contractible Kan complexes span a dense subcategory of $\operatorname{\mathcal{S}}$ (Example 8.4.2.3), Lemma 10.3.4.16 implies that $f$ is a universal quotient morphism. $\square$