Kerodon

Proof. Using Proposition 8.1.3.7, we can rewrite (8.9) as a lifting problem

Let $P$ denote the set of all ordered pairs $(i,j)$, where $i$ and $j$ are integers satisfying $0 \leq i \leq j \leq n$. We regard $P$ as a partially ordered set by identifying it with its image in the product $[n]^{\operatorname{op}} \times [n]$ (so that $(i,j) \leq (i',j')$ if and only if $i' \leq i$ and $j \leq j'$). In what follows, we will identify $\operatorname{Tw}( \Delta ^ n )$ with the nerve $\operatorname{N}_{\bullet }(P)$; under this identification, $\operatorname{Tw}( \Lambda ^{n}_{0} )$ corresponds to a simplicial subset $K_0 \subseteq \operatorname{N}_{\bullet }(P)$.

Let $S = \{ (i_0, j_0) < (i_1, j_1) < \cdots < (i_ d, j_ d) \} $ be a nonempty linearly ordered subset of $P$, so that we have inequalities $0 \leq i_ d \leq i_{d-1} \leq \cdots \leq i_0 \leq j_0 \leq j_{1} \leq \cdots \leq j_ d \leq n$. In this case, we write $\tau _{S}$ for the corresponding nondegenerate $d$-simplex of $\operatorname{N}_{\bullet }(P)$. We will say that $S$ is basic if $\tau _ S$ is contained in $K_0$. Equivalently, $S$ is basic if the set $\{ i_0, i_1, \cdots , i_ d, j_0, j_1, \cdots , j_ d \} $ does not contain $\{ 1 < 2 < \cdots < n \} $. If $S$ is not basic, we let $\mathrm{pr}(S)$ denote the largest integer $j$ such that $S$ contains the pair $(i,j)$ for some $i \neq 0$. If no such integer exists, we define $\mathrm{pr}(S) = 0$. We will refer to $\mathrm{pr}(S)$ as the priority of $S$. We say that $S$ is prioritized if it is not basic and contains the pair $(0, \mathrm{pr}(S) )$.

Let $\{ S_1, S_2, \cdots , S_ m \} $ be an enumeration of the collection of all prioritized linearly ordered subsets of $P$ which satisfies the following conditions:

• The sequence of priorities $\mathrm{pr}(S_1), \mathrm{pr}(S_2), \cdots , \mathrm{pr}( S_ m)$ is nondecreasing. That is, if $1 \leq k \leq \ell \leq m$, then we have $\mathrm{pr}( S_{k} ) \leq \mathrm{pr}(S_{\ell })$.

• If $\mathrm{pr}( S_{k} ) = \mathrm{pr}( S_{\ell } )$ for $k \leq \ell $, then $| S_{k} | \leq | S_{\ell } |$.

For $1 \leq \ell \leq m$, let $\tau _{\ell } \subseteq \operatorname{N}_{\bullet }(P)$ denote the simplex $\tau _{S_{\ell }}$ and let $K_{\ell } \subseteq \operatorname{N}_{\bullet }(P)$ denote the union of $K_0$ with the simplices $\{ \tau _1, \tau _2, \cdots , \tau _{\ell } \} $, so that we have inclusion maps

We claim that $K_ m = \operatorname{N}_{\bullet }(P)$: that is, $K_ m$ contains $\tau _{S}$ for every nonempty linearly ordered subset $S \subseteq P$. If $S$ is basic, there is nothing to prove. We may therefore assume that $S$ has priority $p$ for some integer $p \geq 0$. The union $S \cup \{ (0,p) \} $ is then a prioritized linearly ordered subset of $P$, and therefore coincides with $S_{\ell }$ for some $1 \leq \ell \leq m$. In this case, we have $\tau _{S} \subseteq \tau _{\ell } \subseteq K_{\ell } \subseteq K_ m$.

We will complete the proof by constructing a compatible sequence of maps $F_{\ell }: K_{\ell } \rightarrow \operatorname{\mathcal{C}}$ extending $F_0$ and satisfying $U \circ F_{\ell } = \overline{F}|_{ K_{\ell } }$. Fix an integer $1 \leq \ell \leq m$, and suppose that $F_{\ell -1}$ has already been constructed. Write $S_{\ell } = \{ (i_0, j_0) < (i_1, j_1) < \cdots < (i_ d, j_ d) \} $, so that the simplex $\tau _{\ell }$ has dimension $d$. Let $p$ be the priority of $S_{\ell }$. Since $S_{\ell }$ is prioritized, it contains $(0,p)$; we can therefore write $(0,p) = (i_{ d' }, j_{d'} )$ for some integer $0 \leq d' \leq d$. Let $L \subseteq \Delta ^ d$ denote the inverse image of $K_{\ell -1}$ under the map $\tau _{\ell }: \Delta ^{d} \rightarrow \operatorname{N}_{\bullet }(P)$. We will show that $L$ coincides with the horn $\Lambda ^{d}_{d'}$, so that the pullback diagram of simplicial sets

is also a pushout square (Lemma 3.1.2.11). This can be stated more concretely as follows:

$(\ast )$

Let $(i,j)$ be an element of $S_{\ell }$, and set $S' = S_{\ell } \setminus \{ (i,j) \} $. Then the simplex $\tau _{S'}$ is contained in $K_{\ell -1}$ if and only if $(i,j) \neq (0,p)$.

We first prove $(\ast )$ in the case where $(i,j) \neq (0,p)$; in this case, we wish to show that $\tau _{S'}$ is contained in $K_{\ell -1}$. If $S'$ is basic, then $\tau _{S'}$ is contained in $K_0$ and there is nothing to prove. Let us therefore assume that $S'$ is not basic. Let $p' = \mathrm{pr}(S')$ denote the priority of $S'$. Then the union $S' \cup \{ (0,p') \} $ is a prioritized subset of $P$, and therefore has the form $S_{k}$ for some $1 \leq k \leq m$. By construction, we have $\mathrm{pr}(S_ k) = p' \leq p = \mathrm{pr}( S_{\ell } )$. Moreover, if $p' = p$, then our assumption $(i,j) \neq (0,p)$ guarantees that $S_ k = S'$, so that $| S_{k} | < | S_{\ell } |$. It follows that $k < \ell $, so that we have $\tau _{S'} \subseteq \tau _ k \subseteq K_{k} \subseteq K_{\ell -1}$.

We now prove $(\ast )$ in the case $(i,j) = (0,p)$; in this case, we wish to show that $\tau _{S'}$ is not contained in $K_{\ell -1}$. Assume otherwise. Then, since $S'$ is not basic, it is contained in $S_{k}$ for some $k < \ell $. The inequalities

ensure that $S_{k}$ has priority $p$. Since $S_{k}$ is prioritized, it contains $(0,p)$, and therefore contains the union $S_{\ell } = S' \cup \{ (0,p) \} $. The inequality $|S_ k | \leq | S_{\ell } |$ then forces $k = \ell $, contradicting our assumption that $k < \ell $. This completes the proof of $(\ast )$.

Let $\rho _0$ denote the composite map $\Lambda ^{d}_{d'} = L \xrightarrow { \tau _{\ell } } K_{\ell -1} \xrightarrow { F_{\ell -1} } \operatorname{\mathcal{C}}$. To complete the proof, it will suffice to show that the lifting problem

admits a solution. We consider three cases:

• If $0 < d' < d$, then the lifting problem (8.11) admits a solution by virtue of our assumption that $U$ is an inner fibration of simplicial sets.

• Suppose that $d' = 0$: that is, the pair $(0,p)$ is the smallest element of $S_{\ell }$. Then $S_{\ell }$ does not contain any pairs $(i,j)$ with $i \neq 0$, so we have $p = 0$. Since the set $S_{\ell }$ is not basic, we must have $S_{\ell } = \{ (0,0) < (0,1) < \cdots < (0,n-1) < (0,n) \} $. In this case, the lifting problem (8.11) admits a solution by virtue of assumption $(a)$.

• Suppose that $d' = d$: that is, the pair $(0,p)$ is the largest element of $S_{\ell }$. Our assumption that $S_{\ell }$ is not basic then guarantees that $p =n$ and $(1,n) \in S_{\ell }$: that is, we have $S_{\ell } = \{ (i_0, j_0) < (i_1, j_1) < \cdots < (1,n) < (0,n) \} $. In this case, the lifting problem (8.11) admits a solution by virtue of assumption $(b)$.