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 $[n] \twoheadrightarrow [m]$ (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.

We now establish uniqueness. Suppose we are given another factorization of $\sigma $ as a composition $\Delta ^{n} \xrightarrow {\alpha '} \Delta ^{m'} \xrightarrow {\tau '} S_{\bullet }$, and assume that $\alpha '$ induces a surjection $[n] \twoheadrightarrow [m']$. We first claim that, for any pair of integers $0 \leq i < j \leq n$ satisfying $\alpha '(i) = \alpha '(j)$, we also have $\alpha (i) = \alpha (j)$. Assume otherwise. Then $\alpha $ admits a section $\beta : \Delta ^{m} \hookrightarrow \Delta ^ n$ whose images include $i$ and $j$. We then have

\[ \tau = \tau \circ \alpha \circ \beta = \sigma \circ \beta = \tau ' \circ \alpha ' \circ \beta . \]

Our assumption that $\alpha '(i) = \alpha '(j)$ guarantees that the map $(\alpha ' \circ \beta ): \Delta ^{m} \rightarrow \Delta ^{m'}$ is not injective on vertices, contradicting our assumption that $\tau $ is nondegenerate.

It follows from the preceding argument that $\alpha $ factors uniquely as a composition $\Delta ^{n} \xrightarrow {\alpha '} \Delta ^{m'} \xrightarrow {\alpha ''} \Delta ^{m}$, for some morphism $\alpha '': \Delta ^{m'} \rightarrow \Delta ^ m$ (which is also surjective on vertices). Let $\beta '$ be a section of $\alpha '$, and note that we have

\[ \tau ' = \tau ' \circ \alpha ' \circ \beta ' = \sigma \circ \beta ' = \tau \circ \alpha \circ \beta ' = \tau \circ \alpha '' \circ \alpha ' \circ \beta ' = \tau \circ \alpha ''. \]

Consequently, if the simplex $\tau '$ is nondegenerate, then $\alpha ''$ must also be injective on vertices. It follows that $m' = m$ and $\alpha ''$ is the identity map, so that $\alpha = \alpha '$ and $\tau = \tau '$.
$\square$