Lemma 14.18.2. Let $U$ be a simplicial set. Then $U$ has a unique splitting with $N(U_ m)$ equal to the set of nondegenerate $m$-simplices.

Proof. From the definition it follows immediately, that if there is a splitting then $N(U_ m)$ has to be the set of nondegenerate simplices. Let $x \in U_ n$. Suppose that there are surjections $\varphi : [n] \to [k]$ and $\psi : [n] \to [l]$ and nondegenerate simplices $y \in U_ k$, $z \in U_ l$ such that $x = U(\varphi )(y)$ and $x = U(\psi )(z)$. Choose a right inverse $\xi : [l] \to [n]$ of $\psi$, i.e., $\psi \circ \xi = \text{id}_{[l]}$. Then $z = U(\xi )(x)$. Hence $z = U(\xi )(x) = U(\varphi \circ \xi )(y)$. Since $z$ is nondegenerate we conclude that $\varphi \circ \xi : [l] \to [k]$ is surjective, and hence $l \geq k$. Similarly $k \geq l$. Hence we see that $\varphi \circ \xi : [l] \to [k]$ has to be the identity map for any choice of right inverse $\xi$ of $\psi$. This easily implies that $\psi = \varphi$. $\square$

There are also:

• 6 comment(s) on Section 14.18: Splitting simplicial objects

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).