Proposition 9.2.6.7. Let $X$ be a Kan complex. Then $X$ finitely dominated if and only if it is compact when viewed as an object of the $\infty $-category $\operatorname{\mathcal{S}}$.
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Proof. If $X$ is finitely dominated, then it is a retract of an essentially finite Kan complex $Y$. Since $Y$ is a compact object of $\operatorname{\mathcal{S}}$ (Corollary 9.2.6.5), it follows that $X$ is also a compact object of $\operatorname{\mathcal{S}}$ (Remark 9.2.5.18).
Conversely, suppose that $X$ is a compact object of $\operatorname{\mathcal{S}}$. Let $\{ X_{\alpha } \} _{\alpha \in A}$ be the collection of all finite simplicial subsets of $X$, and let $\operatorname{Ex}^{\infty }: \operatorname{Set_{\Delta }}\rightarrow \operatorname{Set_{\Delta }}$ be the functor defined in Construction 3.3.6.1. Then $\operatorname{Ex}^{\infty }(X)$ can be identified with the filtered colimit $\varinjlim _{\alpha \in A} \operatorname{Ex}^{\infty }( X_{\alpha } )$ in the ordinary category of simplicial sets, and therefore also in the $\infty $-category $\operatorname{\mathcal{S}}$ (Variant 9.1.6.4). Since $X$ is compact, the functor
\[ \operatorname{\mathcal{S}}\rightarrow \operatorname{N}_{\bullet }(\operatorname{Set}) \quad \quad Y \mapsto \pi _0( \operatorname{Hom}_{\operatorname{\mathcal{S}}}(X,Y) ) = \operatorname{Hom}_{\operatorname {h}\! \mathit{\operatorname{Kan}}}( X,Y) \]
preserves filtered colimits, and therefore induces a bijection
\[ \varinjlim _{\alpha \in A} \operatorname{Hom}_{\operatorname {h}\! \mathit{\operatorname{Kan}} }( X, \operatorname{Ex}^{\infty }( X_{\alpha } ) ) \rightarrow \operatorname{Hom}_{\operatorname {h}\! \mathit{\operatorname{Kan}}}( X, \operatorname{Ex}^{\infty }(X) ). \]
In particular, the homotopy equivalence $X \rightarrow \operatorname{Ex}^{\infty }(X)$ of Proposition 3.3.6.7 factors through $\operatorname{Ex}^{\infty }(X_{\alpha } )$ for some $\alpha \in A$. It follows that $X$ is a retract of the essentially finite Kan complex $\operatorname{Ex}^{\infty }( X_{\alpha } )$, and is therefore finitely dominated. $\square$