Proposition 10.2.3.14. Let $f: X_0 \rightarrow X$ be a map of Kan complexes. The following conditions are equivalent:
The map $f$ is a monomorphism in the $\infty $-category of spaces $\operatorname{\mathcal{S}}$, in the sense of Definition 10.2.3.1.
The map $f$ induces a homotopy equivalence of $X_0$ with a summand of $X$.
For every vertex $x \in X$, the homotopy fiber $\{ x\} \times ^{\mathrm{h}}_{X} X_0$ is either empty or contractible.
Proof. The implication $(1) \Rightarrow (2)$ is immediate from the definitions, and the equivalence $(2) \Leftrightarrow (3)$ follows from Remark 3.4.0.6. We will complete the proof by showing that $(2)$ implies $(1)$. Without loss of generality, we may assume that $X_0$ is a summand of $X$ and that $f$ is the inclusion map. For every Kan complex $C$, composition with $f$ induces an isomorphism of $\operatorname{Fun}(C,X_0)$ with the summand of $\operatorname{Fun}(C,X)$ spanned by those maps $g: C \rightarrow X$ which factor through $X_0$. Combining this observation with Remark 5.6.1.5, we deduce that the map $\operatorname{Hom}_{\operatorname{\mathcal{S}}}(C, X_0) \xrightarrow { [f] \circ } \operatorname{Hom}_{\operatorname{\mathcal{S}}}(C,X)$ is a homotopy equivalence of $\operatorname{Hom}_{\operatorname{\mathcal{S}}}(C,X_0)$ with a summand of $\operatorname{Hom}_{\operatorname{\mathcal{S}}}(C,X)$. $\square$