Kerodon

Proof. Assertion $(1')$ is immediate (the morphism $u$ is bijective on vertices by construction). For each $m \geq 0$, the category $\operatorname{Path}[Q]_ m$ can be described concretely as follows:

• The objects of $\operatorname{Path}[Q]_ m$ are the elements of $Q$.

• If $x$ and $y$ are elements of $Q$, then a morphism from $x$ to $y$ in $\operatorname{Path}[Q]_ m$ is a chain

  \[ \overrightarrow {J} = (J_0 \supseteq J_1 \supseteq \cdots \supseteq J_ m) \]

  of finite linearly ordered subsets of $Q$, where each $J_ i$ has least element $x$ and greatest element $y$.

Note that a morphism $\overrightarrow {J}$ from $x$ to $y$ is indecomposable (in the sense of Definition 1.3.7.8) if and only if $x < y$ and $J_ m = \{ x,y\} $. Moreover, an arbitrary morphism $\overrightarrow {J}$ from $x$ to $y$ with $J_ m = \{ x = x_0 < x_1 < \cdots < x_ k = y \} $ decomposes uniquely as a composition of indecomposable morphisms

where $J(a)_{b} = \{ z \in J_ b: x_{a-1} \leq z \leq x_ a \} $. Applying Proposition 1.3.7.11, we deduce that the category $\operatorname{Path}[Q]_ m$ is free, which proves $(2')$. To prove $(3')$, we observe that every indecomposable morphism $\overrightarrow {J}$ can be written uniquely in the form $u( \sigma , \overrightarrow {I} )$, where $(\sigma , \overrightarrow {I} )$ is an element of the set $E(S,m)$ of Notation 2.4.4.9. Writing $J_0 = \{ x = x_0 < \cdots < x_ n = y \} $, we see that $\sigma $ must be the nondegenerate $n$-simplex of $\operatorname{N}_{\bullet }(Q)$ given by the map

and $\overrightarrow {I}$ must be the chain $( \sigma ^{-1}(J_0) \supseteq \sigma ^{-1}(J_{1}) \supseteq \cdots \supseteq \sigma ^{-1}(J_ m) )$ of subsets of $[n]$. $\square$