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3.5.1 Digression: Finite Simplicial Sets

We now introduce a finiteness condition on simplicial sets.

Definition 3.5.1.1. We say that a simplicial set $X$ is finite if it satisfies the following pair of conditions:

  • For every integer $n \geq 0$, the set of $n$-simplices $X_{n} \simeq \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^ n, X)$ is finite.

  • The simplicial set $X$ is finite-dimensional (Definition 1.1.3.9): that is, there exists an integer $m$ such that every nondegenerate simplex has dimension $\leq m$.

Example 3.5.1.2. For each integer $n \geq 0$, the standard $n$-simplex $\Delta ^{n}$ is finite.

Remark 3.5.1.3. Let $X$ be a finite simplicial set. Then any simplicial subset $Y \subseteq X$ is also finite. In particular, any retract of $X$ is finite.

Remark 3.5.1.4. If $X$ and $Y$ are finite simplicial sets, then the coproduct $X \coprod Y$ is also finite.

Remark 3.5.1.5. Let $f: X \twoheadrightarrow Y$ be an epimorphism of simplicial sets. If $X$ is finite, then $Y$ is also finite.

Remark 3.5.1.6. Let $X$ and $Y$ be finite simplicial sets. Then the product $X \times Y$ is finite (see Proposition 1.1.3.11).

Proposition 3.5.1.7. Let $X$ be a simplicial set. The following conditions are equivalent:

$(a)$

The simplicial set $X$ has only finitely many nondegenerate simplices.

$(b)$

There exists an epimorphism of simplicial sets $f: Y \twoheadrightarrow X$, where $Y \simeq \coprod _{i \in I} \Delta ^{n_ i}$ is a finite coproduct of standard simplices.

$(c)$

The simplicial set $X$ is finite (Definition 3.5.1.1).

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 3.5.1.2 together with Remarks 3.5.1.4 and 3.5.1.5. 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 1.1.3.4). 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 $(b)$. $\square$

Remark 3.5.1.8. Every simplicial set $X$ can be realized as a union $\bigcup _{X' \subseteq X} X'$, where $X'$ ranges over the collection of finite simplicial subsets of $X$ (to prove this, we observe that every $n$-simplex $\sigma $ is contained in a finite simplicial subset $X' \subseteq X$: in fact, we can take $X'$ to be the image of $\sigma : \Delta ^{n} \rightarrow X$). Moreover, the collection of finite simplicial subsets of $X$ is closed under finite unions. It follows that realization $X \simeq \bigcup _{X' \subseteq X} X'$ exhibits $X$ as a filtered direct limit of its finite simplicial subsets.

Let $X$ be a simplicial set having geometric realization $|X|$. For every simplicial subset $X' \subseteq X$, the inclusion of $X'$ into $X$ induces a homeomorphism from $|X'|$ onto a closed subset of $|X|$. In what follows, we will abuse notation by identifying $|X'|$ with its image in $|X|$.

Proposition 3.5.1.9. Let $X$ be a simplicial set. Then a subset $K \subseteq |X|$ is compact if and only if it is closed and contained in $|X'| \subseteq |X|$, for some finite simplicial subset $X' \subseteq X$.

Corollary 3.5.1.10. A simplicial set $X$ is finite if and only if the topological space $|X|$ is compact.

The proof of Proposition 3.5.1.9 is based on the following observation:

Lemma 3.5.1.11. Let $X$ be a simplicial set and let $S$ be a subset of the geometric realization $|X|$. Suppose that, for every nondegenerate $n$-simplex $\sigma $ of $X$, the inverse image of $S$ under the composite map $| \Delta ^{n} | \rightarrow |X|$ contains only finitely many points of the interior $\mathring {| \Delta ^{n} |} \subseteq | \Delta ^{n} |$. Then $S$ is closed.

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 1.1.8.19). 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 1.1.3.4, 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 3.5.1.9. 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 3.5.1.7), 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 $s_ j \in S$. Applying Lemma 3.5.1.11, we deduce that for every subset $J' \subseteq J$, the set $\{ s_ j \} _{j \in J'}$ is closed in $|X|$. In particular, $\{ s_ j \} _{j \in J}$ is an infinite closed subset of $S$ endowed with the discrete topology, contradicting our assumption that $S \subseteq |X|$ is compact. $\square$