Kerodon

Proof. This is a special case of Proposition 1.2.3.15, since the category $\operatorname{Cat_{\Delta }}$ admits small colimits (Proposition 2.4.1.13). Explicitly, the simplicial path category of a simplicial set $S_{\bullet }$ is given by the generalized geometric realization

where $\operatorname{Path}[-]_{\bullet }$ denotes the cosimplicial object of $\operatorname{Cat_{\Delta }}$ defined in Notation 2.4.3.1. $\square$

Proof. Unwinding the definitions, we must show that for every simplicial category $\operatorname{\mathcal{D}}_{\bullet }$, the composite map

is a bijection. Here the first map is bijective because the simplicial category $\underline{\operatorname{Path}[G]}_{\bullet }$ is constant (Remark 2.4.2.6), the second by virtue of Proposition 1.3.7.5, and the third because $G_{\bullet }$ has dimension $\leq 1$ and the comparison map $\operatorname{N}_{\bullet }(\operatorname{\mathcal{D}}) \rightarrow \operatorname{N}_{\bullet }^{\operatorname{hc}}(\operatorname{\mathcal{D}})$ is an isomorphism on simplices of dimension $\leq 1$ (Example 2.4.3.9). $\square$

Proof. The first assertion follows immediately from Example 2.4.3.9, since $G_{\bullet }$ is a simplicial set of dimension $\leq 1$. To prove the second assertion, we note that Theorem 2.4.4.10 guarantees that $\operatorname{Path}[S]_0$ is a free category, whose objects can be identified with the vertices of $S_{\bullet }$ and whose indecomposable morphisms can be identified with elements of the set $E(S,0)$ of Notation 2.4.4.9. By definition, $E(S,m)$ consists of pairs $(\sigma , \overrightarrow {I} )$, where $\sigma $ is a nondegenerate $n$-simplex of $S_{\bullet }$ for $n > 0$ and $\overrightarrow {I} = (I_0 \supseteq \cdots \supseteq I_ m)$ is a chain of subsets of $[n]$ satisfying $I_0 = [n]$ and $I_ m = \{ 0, n \} $. In the case $m =0$, the equality $I_0 = I_ m$ forces $n = 1$, so that $E(S,0)$ can be identified (via the morphism $u_0$) with the collection of nondegenerate $1$-simplices of $S_{\bullet }$: that is, with the collection of edges of the graph $G$. The freeness of $\operatorname{Path}[ S ]_0$ now guarantees that the induced map $\operatorname{Path}[G] \simeq \operatorname{Path}[S]_0$ is an isomorphism of categories (see Proposition 1.3.7.11). $\square$

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 $[n]$. $\square$

Proof of Theorem 2.4.4.10. Let $m$ be a nonnegative integer, which we regard as fixed throughout the proof. For each simplicial set $S$, let $G( S )$ denote the directed graph given by

where we regard each element

as an edge of $G(S)$ having source $\sigma (0) \in \operatorname{Vert}( G(S) )$ and target $\sigma (n) \in \operatorname{Vert}( G(S))$. Let $u_{S}: S \rightarrow \operatorname{N}_{\bullet }^{\operatorname{hc}}( \operatorname{Path}[ S ] )$ exhibit the simplicial category $\operatorname{Path}_{\bullet }[S]$ as a path category of $S$. Then $u_{S}$ induces a map of simplicial sets $G(S)_{\bullet } \rightarrow \operatorname{N}_{\bullet }( \operatorname{Path}[S]_{m} )$, which we can identify with a functor of ordinary categories $F_{S}: \operatorname{Path}[ G(S) ] \rightarrow \operatorname{Path}[ S ]_{m}$. Let us say that the simplicial set $S$ is good if $F_{S}$ is an isomorphism of categories. Theorem 2.4.4.10 is equivalent to the assertion that every simplicial set is good (for every choice of nonnegative integer $m$). We will prove this by verifying that the collection of good simplicial sets satisfies the hypotheses of Lemma 1.2.3.13:

• Suppose we are given a pushout diagram of simplicial sets $\sigma :$

  \[ \xymatrix@R =50pt@C=50pt{ S \ar [r] \ar [d] & T \ar [d] \\ S' \ar [r] & T', } \]

  where the horizontal maps are monomorphisms. Suppose that $S$, $T$, and $S'$ are good; we wish to show that $T'_{\bullet }$ is good. Note that the horizonal maps induce monomorphisms of directed graphs

  \[ G(S) \hookrightarrow G(T ) \quad \quad G(S' ) \hookrightarrow G(T'). \]

  Define subgraphs $G_0(S) \subseteq G(S )$ and $G_0( T) \subseteq G(T )$ by the formulae

  \[ \operatorname{Vert}( G_0(S ) ) = \operatorname{Vert}( G(S) ) = S_0 \quad \quad \operatorname{Vert}( G_0(T ) ) = \operatorname{Vert}( G(T) ) = T_0 \]
  \[ \operatorname{Edge}( G_0( S) ) = \emptyset \quad \quad \operatorname{Edge}( G_0( T ) ) = \operatorname{Edge}( G(T) ) \setminus \operatorname{Edge}( G( S) ). \]

  We then have a commutative diagram of categories

  \[ \xymatrix@R =50pt@C=50pt{ \operatorname{Path}[ G_0(S)] \ar [r] \ar [d] & \operatorname{Path}[ G_0(T) ] \ar [d] \\ \operatorname{Path}[ G( S' ) ] \ar [r] \ar [d]^{ F_{S'} } & \operatorname{Path}[ G( T' )] \ar [d]^{ F_{T'} } \\ \operatorname{Path}[ S' ]_{m} \ar [r] & \operatorname{Path}[ T' ]_{m}. } \]

  We wish to show that the functor $F_{T'}$ is an isomorphism of categories, and the map $F_{S' }$ is an isomorphism by assumption. It will therefore suffice to show that the lower square in this diagram is a pushout. Note that the upper square is a pushout (since it is obtained from a pushout diagram in the category of directed graphs by passing to path categories). We are therefore reduced to showing that the outer rectangle is a pushout. We can rewrite this as the outer rectangle in another commutative diagram of categories

  \[ \xymatrix@R =50pt@C=50pt{ \operatorname{Path}[ G_0(S)] \ar [r] \ar [d] & \operatorname{Path}[ G_0(T) ] \ar [d] \\ \operatorname{Path}[ G( S ) ] \ar [r] \ar [d]^{ F_{S} } & \operatorname{Path}[ G(T )] \ar [d]^{ F_{T} } \\ \operatorname{Path}[ S ]_{m} \ar [r] \ar [d] & \operatorname{Path}[ T ]_{m} \ar [d] \\ \operatorname{Path}[ S' ]_{m} \ar [r] & \operatorname{Path}[ T' ]_{m}. } \]

  We now conclude by observing that the upper square in this diagram is a pushout (because it is obtained from a pushout diagram of directed graphs by passing to path categories), the middle square is a pushout (since $F_{S}$ and $F_{T}$ are isomorphisms), and the lower square is a pushout (since the construction $X_{\bullet } \mapsto \operatorname{Path}[ X ]_{m}$ preserves colimits).

• Suppose we are given a sequence of monomorphisms of simplicial sets

  \[ S(0) \hookrightarrow S(1) \hookrightarrow S(2) \hookrightarrow \cdots \]

  with colimit $S$. Then the functor $F_{S}: \operatorname{Path}[ G(S) ] \rightarrow \operatorname{Path}[ S ]_{m}$ can be written as a filtered colimit of functors $F_{S(i)}: \operatorname{Path}[ G(S(i))] \rightarrow \operatorname{Path}[ S(i) ]_{m}$. Consequently, if each $S(i)$ is good, then $S$ is good.

• Let $S$ be a simplicial set which can be written as a coproduct $\coprod _{i \in I} \Delta ^ n$; we must show that $S$ is good. Without loss of generality, we may assume that $I$ is a singleton, so that $S = \Delta ^ n$. In this case, Example 2.4.4.6 supplies an equivalence of simplicial categories $\operatorname{Path}[S]_{\bullet } \simeq \operatorname{Path}[n]_{\bullet }$. The desired result now follows from Lemma 2.4.4.16.

Proof. Using Proposition 2.4.4.15, we can identify the pushout $\operatorname{Path}[ Q_{-} ]_{\bullet } \coprod _{ \{ q\} } \operatorname{Path}[ Q_{+} ]_{\bullet }$ with the simplicial path category of the simplicial set $S = \operatorname{N}_{\bullet }(Q_{-} ) \coprod _{ \{ q\} } \operatorname{N}_{\bullet }( Q_{+} )$. The tautological map $S \rightarrow \operatorname{N}_{\bullet }(Q)$ is a monomorphism of simplicial sets, and therefore induces an equivalence from $\operatorname{Path}[S]_{\bullet }$ to a simplicial subcategory $\operatorname{\mathcal{C}}\subseteq \operatorname{Path}[Q]_{\bullet }$ (Remark 2.4.4.12). It is clear that this subcategory contains both $\operatorname{Path}[ Q_{-} ]_{\bullet }$ and $\operatorname{Path}[ Q_{+} ]_{\bullet }$. To complete the proof, it will suffice to show that if $\operatorname{\mathcal{D}}$ is any other simplicial subcategory of $\operatorname{Path}[Q]_{\bullet }$ which contains $\operatorname{Path}[Q_{-} ]_{\bullet }$ and $\operatorname{Path}[ Q_{+} ]_{\bullet }$, then $\operatorname{\mathcal{D}}$ contains $\operatorname{\mathcal{C}}$. This is clear: the universal property of $\operatorname{\mathcal{C}}$ guarantees that there is a unique simplicial functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ which is the identity on both $\operatorname{Path}[ Q_{-} ]_{\bullet }$ and $\operatorname{Path}[ Q_{+} ]_{\bullet }$. Invoking the universal property of $\operatorname{\mathcal{C}}$ again, we deduce that the composite functor $\operatorname{\mathcal{C}}\xrightarrow {F} \operatorname{\mathcal{D}}\hookrightarrow \operatorname{Path}[Q]_{\bullet }$ coincides with the inclusion map, so that $\operatorname{\mathcal{C}}\subseteq \operatorname{\mathcal{D}}$. $\square$

Proof. By virtue of Corollary 2.4.4.18, we can identify the pushout $\operatorname{Path}[ Q_{-} ]_{\bullet } \coprod _{ \{ q\} } \operatorname{Path}[ Q_{+} ]_{\bullet }$ with a simplicial subcategory $\operatorname{\mathcal{C}}\subseteq \operatorname{Path}[Q]_{\bullet }$; we wish to show that there is a unique simplicial functor $R: \operatorname{Path}[Q]_{\bullet } \rightarrow \operatorname{\mathcal{C}}$ satisfying $R|_{\operatorname{\mathcal{C}}} = \operatorname{id}_{\operatorname{\mathcal{C}}}$. Our assumption that $Q = Q_{-} \cup Q_{+}$ guarantees that $\operatorname{\mathcal{C}}$ contains every object of $\operatorname{Path}[Q]_{\bullet }$. To prove existence, we take the simplicial functor $R$ to be the identity on objects and given on morphisms by the maps

where $P_{a,b}$ and the subset $P'_{a,b} \subseteq P_{a,b}$ are defined as in Remark 2.4.4.19.

To prove uniqueness, let $R': \operatorname{Path}[Q]_{\bullet } \rightarrow \operatorname{\mathcal{C}}$ be another simplicial functor satisfying $R'|_{\operatorname{\mathcal{C}}} = \operatorname{id}_{\operatorname{\mathcal{C}}}$; we wish to show that $R' = R$. It is clear that $R$ and $R'$ agree at the level of objects. For every pair of elements $a,b \in Q$, the simplicial functors $R$ and $R'$ induce maps $\theta , \theta ': \operatorname{Hom}_{ \operatorname{Path}[Q] }(a,b)_{\bullet } \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}}(a,b)_{\bullet }$; we wish to show that $\theta = \theta '$. Since $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(a,b)_{\bullet }$ can be identified with the nerve of the partially ordered set $P'_{a,b}$, it will suffice to show that $\theta $ and $\theta '$ agree on vertices. For every finite linearly ordered subset $J \subseteq Q$ having least element $a$ and greatest element $b$, let $f_{J}: a \rightarrow b$ denote the corresponding morphism in the path category $\operatorname{Path}[Q]$; we wish to show that $\theta ( f_{J} ) = \theta '( f_{J} )$. Without loss of generality, we may assume that the morphism $f_{J}$ is indecomposable: that is, that we have $a \neq b$ and that $J = \{ a < b \} $. We may further assume that $a < q < b$ (otherwise, $f_{J}$ is a morphism in the category $\operatorname{\mathcal{C}}$ and we have $\theta (f_ J) = f_{J} = \theta '(f_ J)$). Set $J^{+} = \{ a < q < b \} $, so that $\theta ( f_ J ) = f_{J^{+} }$. Write $\theta '( f_ J ) = f_{K}$ where $K \subseteq Q$ is a finite linearly ordered subset having least element $a$ and greatest element $b$. Since $f_{J^{+}}$ is a morphism of $\operatorname{\mathcal{C}}$, we have $\theta '( f_{J^{+}} ) = f_{J^{+} }$. The inclusion $J \subseteq J^{+}$ then implies that $K \subseteq J^{+}$. On the other hand, $f_{K}$ is also a morphism of $\operatorname{\mathcal{C}}$, so we must have $q \in K$. It follows that $K = J^{+}$, so that $\theta ( f_ J) = f_{J^{+}} = f_{K} = \theta '( f_ J )$ as desired. $\square$