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Proposition Let $K_{\bullet }$ be a simplicial set of dimension $\leq 1$, corresponding to a directed graph $G$ which satisfies conditions $(a)$ and $(b)$ of Definition Let $\operatorname{\mathcal{C}}$ be an ordinary category, and let $\sigma : K_{\bullet } \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ be a diagram. Then:


There is a partial ordering $\leq $ on the vertex set $\operatorname{Vert}(G)$, where we have $v \leq w$ if and only if there exists a sequence of vertices $(v = v_0, v_1, \ldots , v_ n = w)$ with the property that the edges $( v_{i-1}, v_ i) \in \operatorname{Edge}(G)$ exist for $1 \leq i \leq n$.


There is a unique monomorphism of simplicial sets $K_{\bullet } \hookrightarrow \operatorname{N}_{\bullet }( \operatorname{Vert}(G) )$ which carries each vertex to itself.


The diagram $\sigma $ extends to a map $\overline{\sigma }: \operatorname{N}_{\bullet }( \operatorname{Vert}(G) ) \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ (that is, to a functor $\operatorname{Vert}(G) \rightarrow \operatorname{\mathcal{C}}$) if and only if it is commutative, in the sense of Definition Moreover, if the extension $\overline{\sigma }$ exists, then it is unique.

Proof. It follows immediately from the definitions that the relation $\leq $ defined in $(1)$ is reflexive and transitive. Antisymmetry follows from our assumption that the graph $G$ has no directed loops (condition $(b)$ of Definition By construction, we have $v \leq w$ whenever $v$ and $w$ are connected by an edge $(v,w) \in \operatorname{Edge}(G)$. From the description of the simplicial set $K_{\bullet }$ given in Remark, we immediately see that there is a unique map of simplicial sets $i: K_{\bullet } \rightarrow \operatorname{N}_{\bullet }( \operatorname{Vert}(G) )$ which is the identity on vertices. It follows from assumption $(a)$ of Definition that the map $i$ is a monomorphism. Let us henceforth identify $K_{\bullet }$ with a simplicial subset of $\operatorname{N}_{\bullet }( \operatorname{Vert}(G) )$ given by the image of $i$. Let us identify $\sigma $ with a pair $( \{ C_ v \} _{v \in \operatorname{Vert}(G)}, \{ f_{w,v}: C_ v \rightarrow C_ w \} _{(v,w) \in \operatorname{Edge}(G) } )$. Suppose that the diagram $\sigma $ extends to a functor $\overline{\sigma }: \operatorname{N}_{\bullet }(\operatorname{Vert}(G)) \rightarrow \operatorname{\mathcal{C}}$. If $v$ and $w$ are a pair of vertices of $G$ with $v \leq w$, then we can choose a directed path $(v = v_0, v_1, \ldots , v_ n = w)$ from $v$ to $w$. The compatibility of $\overline{\sigma }$ with composition then guarantees that $\overline{\sigma }$ must carry the edge $(v,w)$ of $\operatorname{N}_{\bullet }( \operatorname{Vert}(G) )$ to the iterated composition $f_{ v_{n}, v_{n-1} } \circ f_{ v_{n-1}, v_{n-2} } \circ \cdots \circ f_{ v_1, v_0} \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}( C_{v}, C_ w )$. Since the morphism $\overline{\sigma }(v, w)$ is independent of the choice of directed path, it follows that the diagram $\sigma $ is commutative. Conversely, if $\sigma $ is commutative, then we can define $\overline{\sigma }$ on morphisms by the formula $\overline{\sigma }(v, w) = f_{ v_{n}, v_{n-1} } \circ f_{ v_{n-1}, v_{n-2} } \circ \cdots \circ f_{ v_1, v_0}$ to obtain the desired extension of $\sigma $. $\square$