# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Proposition 8.1.1.15. Let $U: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an inner fibration of simplicial sets. Then the projection maps of Notation 8.1.1.6 determine a left fibration of simplicial sets

$\operatorname{Tw}(\operatorname{\mathcal{C}}) \rightarrow (\operatorname{\mathcal{C}}^{\operatorname{op}} \times \operatorname{\mathcal{C}}) \times _{ ( \operatorname{\mathcal{D}}^{\operatorname{op}} \times \operatorname{\mathcal{D}}) } \operatorname{Tw}(\operatorname{\mathcal{D}}).$

Proof. Fix a pair of integers $0 < i \leq n$; we wish to show that every lifting problem

8.2
$$\begin{gathered}\label{equation:twisted-arrow-left-fibration} \xymatrix@R =50pt@C=50pt{ \Lambda ^{n}_{n-i} \ar [r] \ar [d] & \operatorname{Tw}(\operatorname{\mathcal{C}}) \ar [d] \\ \Delta ^ n \ar [r] \ar@ {-->}[ur] & (\operatorname{\mathcal{C}}^{\operatorname{op}} \times \operatorname{\mathcal{C}}) \times _{ ( \operatorname{\mathcal{D}}^{\operatorname{op}} \times \operatorname{\mathcal{D}}) } \operatorname{Tw}(\operatorname{\mathcal{D}}) } \end{gathered}$$

For each nonempty subset $S \subseteq [2n+1] = \{ 0 < 1 < \cdots < 2n+1 \}$, let $\sigma _{S}$ denote the corresponding nondegenerate simplex of $\Delta ^{2n+1}$. Let us say that $S$ is basic if it satisfies one of the following conditions:

$(a)$

The set $S$ is contained in $\{ 0 < 1 < \cdots < n \}$.

$(b)$

The set $S$ is contained in $\{ n+1 < n +2 < \cdots < 2n+1 \}$.

$(c)$

There exists an integer $j \neq i$ such that $0 \leq j \leq n$ and $S \cap \{ j, 2n+1- j \} = \emptyset$.

Let $K_0 \subseteq \Delta ^{2n+1}$ be the simplicial subset whose nondegenerate simplices have the form $\sigma _{S}$, where $S$ is basic. Unwinding the definitions, we can rewrite (8.2) as a lifting problem

$\xymatrix@R =50pt@C=50pt{ K_0 \ar [d] \ar [r] & \operatorname{\mathcal{C}}\ar [d]^{U} \\ \Delta ^{2n+1} \ar [r] \ar@ {-->}[ur] & \operatorname{\mathcal{D}}. }$

Since $U$ is an inner fibration, it will suffice to show that the inclusion $K_0 \hookrightarrow \Delta ^{2n+1}$ is an inner anodyne map of simplicial sets.

We now introduce two more collections of subsets of $[2n+1]$.

• We say that a subset $S \subseteq [2n+1]$ is primary if it is not basic, the intersection $S \cap \{ 0, 1, \cdots , i-1\}$ is empty, and $2n+1-i \in S$.

• We say that a subset $S \subseteq [2n+1]$ is secondary if it is not basic, the intersection $S \cap \{ 0, 1, \cdots , i-1 \}$ is nonempty, and $i \in S$.

Let $\{ S_1, S_2, \cdots , S_ m \}$ be an ordering of the collection of all subsets of $[2n+1]$ which are either primary or secondary, satisfying the following conditions:

• The sequence of cardinalities $| S_1 |, |S_2 |, \cdots , |S_ m |$ is nondecreasing. That is, for $1 \leq p \leq q \leq m$, we have $|S_ p | \leq | S_{q} |$.

• If $|S_ p| = |S_ q|$ for $p \leq q$ and $S_{q}$ is primary, then $S_ p$ is also primary.

For $1 \leq q \leq m$, let $\sigma _{q} \subseteq \Delta ^{2n+1}$ denote the simplex spanned by the vertices of $S_ q$, and let $K_{q}$ denote the union of $K_0$ with the simplices $\{ \sigma _1, \sigma _2, \cdots , \sigma _ q \}$. We have inclusion maps

$K_0 \hookrightarrow K_1 \hookrightarrow K_{2} \hookrightarrow \cdots \hookrightarrow K_{m}.$

Note that we have $\sigma _{m} = K_{m} = \Delta ^{2n+1}$ (since the set $[2n+1]$ is secondary). It will therefore suffice to show that for $1 \leq q \leq m$, the inclusion map $K_{q-1} \hookrightarrow K_{q}$ is inner anodyne.

In what follows, we regard $q$ as fixed. Let $d$ be the dimension of the simplex $\sigma _{q}$. Let us abuse notation by identifying $\sigma _{q}$ with a morphism of simplicial sets $\Delta ^{d} \rightarrow K_{q} \subseteq \Delta ^{2n+1}$, and set $L = \sigma _{q}^{-1} K_{q-1} \subseteq \Delta ^{d}$. To complete the proof, it will suffice to show that $L$ is an inner horn of $\Delta ^ d$, so that the diagram of simplicial sets

$\xymatrix@R =50pt@C=50pt{ L \ar [r] \ar [d] & K_{q-1} \ar [d] \\ \Delta ^{d} \ar [r]^-{ \sigma _ q} & K_{q}. }$

is a pushout square by virtue of Lemma 3.1.2.11.

We first consider the case where the set $S_{q} = \{ j_0 < j_1 < \cdots < j_ d \}$ is primary, so that we have $j_0 \geq i$ and $j_ k = 2n+1-i$ for some $0 \leq k \leq d$. Note that we must have $k > 0$ (otherwise $S_ q$ satisfies condition $(b)$) and $k < d$ (otherwise, $S_ q$ satisfies condition $(c)$, since it is disjoint from $\{ 0, 2n+1\}$). In this case, we will show that $L$ coincides with the inner horn $\Lambda ^{d}_{k} \subset \Delta ^{d}$. This can be restated as follows:

$(\ast )$

Let $j$ be an element of $S_{q}$, and set $S' = S_{q} \setminus \{ j\}$. Then $\sigma _{S'}$ is contained in $K_{q-1}$ if and only if $j \neq 2n+1-i$.

Assume first that $j \neq 2n+1-i$. Then the set $S'$ contains $2n+1-i$ and satisfies $S' \cap \{ 0, 1, \cdots , i-1 \} = \emptyset$. Consequently, the set $S'$ is either primary (and therefore coincides with $S_{q'}$ for some $q' < q$) or basic. In either case, the simplex $\sigma _{S'}$ belongs to the simplicial subset $K_{q-1} \subseteq \Delta ^{2n+1}$.

We now prove $(\ast )$ in the case $j = 2n+1-i$. Since $S_{q}$ does not satisfy conditions $(b)$ or $(c)$, the set $S'$ also does not satisfy conditions $(b)$ or $(c)$. It also cannot satisfy condition $(a)$: if $S'$ were contained in the set $\{ 0, 1, \cdots , n \}$, then $S_{q}$ would be contained in the set $\{ i, i+1, \cdots , n, 2n+1-i \}$, and would therefore satisfy condition $(c)$. It follows that $S'$ is not basic. Assume, for a contradiction, that $\sigma _{S'}$ is contained in $K_{q-1}$. We then have $\sigma _{S'} \subseteq \sigma _{q'}$ for some $q' < q$. Since $S'$ is neither primary nor secondary, this must be a proper inclusion: that is, we must have

$\dim ( \sigma _{q} ) - 1 = \dim ( \sigma _{S'} ) < \dim ( \sigma _{q'} ) \leq \dim ( \sigma _{q} ).$

It follows that the second inequality must be an equality: that is, we have $|S_{q'}| = |S_{q} |$ and therefore $S_{q'}$ is also primary. In particular, the set $S_{q'}$ contains $2n+1-i$, and therefore contains the union $S' \cup \{ 2n+1-i\} = S_{q}$. Since $S_{q}$ and $S_{q'}$ have the same cardinality, it follows that $S_{q} = S_{q'}$ and therefore $q = q'$, contradicting our assumption that $q' < q$.

We now consider the case where $S_{q} = \{ j_0 < j_1 < \cdots < j_ d \}$ is secondary, so that we have $j_0 < i$ and $j_ k = i$ for some $0 < k \leq d$. Note that we must have $k < d$ (otherwise, $S_ q$ satisfies condition $(a)$). In this case, we will show that $L$ coincides with the inner horn $\Lambda ^{d}_{k} \subset \Delta ^{d}$. This can be restated as follows:

$(\ast ')$

Let $j \in S_ q$ and set $S' = S_ q \setminus \{ j \}$. Then the simplex $\sigma _{S'}$ is contained in $K_{q-1}$ if and only if $j = i$.

We first treat the case where $j \neq i$, so that $i \in S'$. If $S'$ is basic, then $\sigma _{S'} \subseteq K \subseteq K_{q-1}$. We may therefore assume that $S'$ is not basic. If the intersection $S' \cap \{ 0, 1, \cdots , i-1 \}$ is nonempty, then $S'$ is secondary and has smaller cardinality than $S_{q}$. It follows that $S' = S_{q'}$ for some $q' < q$, so that $\sigma _{S'} \subseteq K_{q'} \subseteq K_{q-1}$. We may therefore assume that the intersection $S' \cap \{ 0, 1, \cdots , i-1\}$ is empty. In this case, the union $S' \cup \{ 2n+1 - i \}$ is a primary set of cardinality $\leq |S_ q |$, and therefore has the form $S_{q'}$ for some $q' < q$. From this, we again conclude that $\sigma _{S'} \subseteq K_{q'} \subseteq K_{q-1}$.

We now prove $(\ast ')$ in the case $j=i$. Since $S_{q}$ does not satisfy conditions $(a)$ or $(c)$, it follows that $S'$ also does not satisfy conditions $(a)$ or $(c)$. The set $S'$ also does not satisfy condition $(b)$, since the intersection $S' \cap \{ 0, \cdots , i-1 \}$ is nonempty. It follows that $S'$ is not basic. Assume, for a contradiction, that $\sigma _{S'}$ is contained in $K_{q-1}$. We then have $\sigma _{S'} \subseteq \sigma _{q'}$ for some $q' < q$. Since the intersection $S_{q'} \cap \{ 1, \cdots , i-1 \}$ is nonempty, the set $S_{q'}$ cannot be primary and is therefore secondary. In particular, the set $S_{q'}$ contains the element $i$ and therefore contains the union $S' \cup \{ i\} = S_{q}$. Combining this observation with the inequality $| S_{q'} | \leq | S_{q} |$, we deduce that $S_{q'} = S_{q}$ and therefore $q' = q$, contradicting our assumption that $q' < q$. $\square$