Proposition 3.6.1.7. Let $X$ be a simplicial set. The following conditions are equivalent:
The simplicial set $X$ has only finitely many nondegenerate simplices.
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.
The simplicial set $X$ is finite (Definition 3.6.1.1).
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 3.6.1.2 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 $(b)$. $\square$