Kerodon

Proof. If $X$ is finite, then it has dimension $\leq n$ for some integer $n \gg 0$. It follows that every nondegenerate simplex of $X$ has dimension $\leq n$. Since $X$ has only finitely many (nondegenerate) simplices of each dimension, it follows that $X$ has only finitely many nondegenerate simplices. This proves that $(c) \Rightarrow (a)$. The implication $(b) \Rightarrow (c)$ follows from Example 3.6.1.2 together with Remarks 3.6.1.4 and 3.6.1.5. We will complete the proof by showing that $(a)$ implies $(b)$. Let $\{ \sigma _{i}: \Delta ^{n_ i} \rightarrow X \} _{i \in I}$ be the collection of all nondegenerate simplices of $X$, and amalgamate the morphisms $\sigma _{i}$ to a single map $f: Y = \coprod _{i \in I} \Delta ^{n_ i} \rightarrow X$. By construction, every nondegenerate simplex of $X$ belongs to the image of $f$ and therefore every simplex of $f$ belongs to the image of $f$ (see Proposition 1.1.3.8). It follows that $f$ is an epimorphism of simplicial sets. If condition $(a)$ is satisfied, then the set $I$ is finite, so that $f: Y \twoheadrightarrow X$ satisfies the requirements of $(b)$. $\square$

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. $\square$

Proof. Since colimits in the category of simplicial sets are computed levelwise (Remark 1.1.0.8), it will suffice to that the functor

commutes with filtered colimits for each $n \geq 0$. This is a special case of Proposition 3.6.1.9, since the product $\Delta ^ n \times X$ is also a finite simplicial set (Remark 3.6.1.6 and Example 3.6.1.2. $\square$

Proof. The geometric realization $|X|$ can be described as the colimit $\varinjlim _{\sigma : \Delta ^{n} \rightarrow X} | \Delta ^{n} |$, indexed by the category of simplices of $X$ (see Construction 1.1.3.9). Consequently, to show that the subset $S \subseteq | X |$ is closed, it will suffice to show that the inverse image $| \sigma |^{-1}(S) \subseteq | \Delta ^{n} |$ is closed, for every $n$-simplex $\sigma : \Delta ^{n} \rightarrow X$. We proceed by induction on $n$. Using Proposition 1.1.3.8, we can reduce to the case where $\sigma $ is nondegenerate. In this case, our inductive hypothesis guarantees that $| \sigma |^{-1}(S)$ has closed intersection with the boundary $| \operatorname{\partial \Delta }^{n} | \subseteq | \Delta ^{n} |$. Since $| \sigma |^{-1}(S)$ contains only finitely many points in the interior of $| \Delta ^ n |$, it is closed. $\square$

Proof of Proposition 3.6.1.11. Let $X$ be a simplicial set. If $X' \subseteq X$ is a finite simplicial subset, then the geometric realization $| X' |$ is a continuous image of a finite disjoint union $\coprod _{i \in I} | \Delta ^{n_ i} |$ (Proposition 3.6.1.7), and is therefore compact. It follows that any closed subset $K \subseteq |X'|$ is also compact. Conversely suppose that $K \subseteq |X|$ is compact. Since $|X|$ is Hausdorff, the set $K$ is closed. We wish to show that $K$ is contained in $|X'|$ for some finite simplicial subset $X' \subseteq X$. Suppose otherwise. Then we can choose an infinite collection of nondegenerate simplices $\{ \sigma _{j}: \Delta ^{n_ j} \rightarrow X \} _{j \in J}$ for which each of the corresponding cells $\mathring {| \Delta ^{n_ j} |} \hookrightarrow |X|$ contains some point $x_ j \in K$. Applying Lemma 3.6.1.13, we deduce that for every subset $J' \subseteq J$, the set $\{ x_ j \} _{j \in J'}$ is closed in $|X|$. In particular, $\{ x_ j \} _{j \in J}$ is an infinite closed subset of $K$ endowed with the discrete topology, contradicting our assumption that $K \subseteq |X|$ is compact. $\square$