# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Proposition 1.1.3.4. Let $\sigma : \Delta ^{n} \rightarrow S_{\bullet }$ be a map of simplicial sets. Then $\sigma$ can be factored as a composition

$\Delta ^{n} \xrightarrow {\alpha } \Delta ^{m} \xrightarrow { \tau } S_{\bullet },$

where $\alpha$ corresponds to a surjective map of linearly ordered sets $[n] \rightarrow [m]$ and $\tau$ is a nondegenerate $m$-simplex of $S_{\bullet }$. Moreover, this factorization is unique.

Proof. Let $m$ be the smallest nonnegative integer for which $\sigma$ can be factored as a composition $\Delta ^{n} \xrightarrow {\alpha } \Delta ^{m} \xrightarrow {\tau } S_{\bullet }$. It follows from the minimality of $m$ that $\alpha$ must induce a surjection of linearly ordered sets $[n] \rightarrow [m]$ (otherwise, we could replace $[m]$ by the image of $\alpha$) and that the $m$-simplex $\tau$ is nondegenerate. This proves the existence of the desired factorization.

To establish uniqueness, let us suppose we are given another factorization of $\sigma$ as a composition $\Delta ^{n} \xrightarrow {\alpha '} \Delta ^{m'} \xrightarrow {\tau '} S_{\bullet }$. By assumption, $\alpha$ and $\alpha '$ determine surjections of linearly ordered sets $[n] \rightarrow [m]$ and $[n] \rightarrow [m']$, and therefore admit sections which we will denote by $\beta$ and $\beta '$, respectively. The equality $\sigma = \tau \circ \alpha$ then gives

$\tau = \sigma \circ \beta = \tau ' \circ \alpha ' \circ \beta .$

Our assumption that $\tau$ is nondegenerate then guarantees that the map $\alpha ' \circ \beta : [m] \rightarrow [m']$ is injective, so that $m \leq m'$. The same argument shows that $m' \leq m$, so we must have $m = m'$. Since the only nondecreasing injection from $[m]$ to itself is the identity map, we conclude that $\alpha ' \circ \beta = \operatorname{id}_{[m]}$. The desired uniqueness now follows from the calculations

$\tau = \tau ' \circ \alpha ' \circ \beta = \tau ' \quad \quad \alpha = \alpha ' \circ \beta \circ \alpha = \alpha '.$
$\square$