## Tag `017R`

Chapter 14: Simplicial Methods > Section 14.18: Splitting simplicial objects

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 the 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 $\psi = \varphi$. $\square$

The code snippet corresponding to this tag is a part of the file `simplicial.tex` and is located in lines 1784–1788 (see updates for more information).

```
\begin{lemma}
\label{lemma-splitting-simplicial-sets}
Let $U$ be a simplicial set. Then $U$ has a unique splitting
with $N(U_m)$ equal to the set of nondegenerate $m$-simplices.
\end{lemma}
\begin{proof}
From the definition it follows immediately, that if there is a
splitting then $N(U_m)$ has the 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 $\psi = \varphi$.
\end{proof}
```

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