Kerodon

Proof. Let $m$ be the smallest nonnegative integer for which $\sigma $ can be factored as a composition $\Delta ^{n} \xrightarrow {\alpha } \Delta ^{m} \xrightarrow {\tau } S$. 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 $\Delta ^{n} \xrightarrow {\alpha '} \Delta ^{m'} \xrightarrow {\tau '} S$, 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

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

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 $\tau = \tau '$. $\square$