Kerodon

Proof. By virtue of Remark 3.6.1.8, we can write $X$ as a filtered colimit of finite simplicial subsets $Y \subseteq X$. If $X$ is a compact object of $\operatorname{Set_{\Delta }}$, then the identity map $\operatorname{id}_ X: X \rightarrow X$ factors through some finite simplicial subset $Y \subseteq X$. It follows that $Y = X$, so that $X$ is a finite simplicial set. To prove the converse, assume that $X$ is finite. Using Proposition 3.6.1.7, we can choose an epimorphism of simplicial sets $U \twoheadrightarrow X$, where $U$ is a finite coproduct of standard simplices. In particular, $U$ is also a finite simplicial set (Example 3.6.1.2 and Remark 3.6.1.4). The fiber product $U \times _{X} U$ can be regarded as a simplicial subset of $U \times U$, and is therefore also finite (Remarks 3.6.1.6 and 3.6.1.3). Applying Proposition 3.6.1.7 again, we can choose an epimorphism of simplicial sets $V \twoheadrightarrow U \times _{X} U$, where $V$ is a finite coproduct of standard simplices. It follows that $X$ can be realized as the coequalizer of a pair of maps $f_0, f_1: V \rightarrow U$. Consequently, to show that $X$ is compact, it will suffice to show that $U$ and $V$ are compact. Since the collection of compact objects of $\operatorname{Set_{\Delta }}$ is closed under the formation of finite coproducts and coequalizers, we are reduced to showing that each standard simplex $\Delta ^{n}$ is a compact object of $\operatorname{Set_{\Delta }}$. This is an immediate consequence of Proposition 1.1.0.12 and Remark 1.1.0.8. $\square$