Kerodon

Proof. Let $\sigma $ be an $n$-simplex of the simplicial set $\operatorname{N}_{\bullet }( \operatorname{Tw}(\operatorname{\mathcal{C}}) )$, which we identify with a diagram

in the category $\operatorname{Tw}(\operatorname{\mathcal{C}})$. Here each $f_ i: X_ i \rightarrow Y_ i$ denotes a morphism in $\operatorname{\mathcal{C}}$, and each $(u_ i, v_ i)$ is a pair of morphisms in $\operatorname{\mathcal{C}}$ which determine a commutative diagram

In this case, we can regard the chain of morphisms

as a $(2n+1)$-simplex of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$, which we identify with an $n$-simplex $T(\sigma )$ of $\operatorname{Tw}( \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) )$. The construction $\sigma \mapsto T(\sigma )$ then determines a morphism of simplicial sets $T: \operatorname{N}_{\bullet }( \operatorname{Tw}(\operatorname{\mathcal{C}}) ) \rightarrow \operatorname{Tw}( \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) )$, which satisfies conditions $(1)$ and $(2)$ by construction.

We now claim that $T$ is an isomorphism of simplicial sets. Let $\tau $ be an $n$-simplex of $\operatorname{Tw}( \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) )$; we wish to show that there is a unique $n$-simplex $\sigma $ of $\operatorname{N}_{\bullet }( \operatorname{Tw}(\operatorname{\mathcal{C}}) )$ satisfying $T(\sigma ) = \tau $. Let us identify $\tau $ with a diagram of the form (8.1) in the category $\operatorname{\mathcal{C}}$. We wish to show that there is a unique collection of morphisms $\{ f_{i}: X_{i} \rightarrow Y_{i} \} _{1 \leq i \leq n}$ satisfying the identities $f_{i} = v_{i} \circ f_{i-1} \circ u_{i}$, which follows immediately by induction on $i$.

We now complete the proof by establishing the uniqueness of $T$. Suppose that $T': \operatorname{N}_{\bullet }( \operatorname{Tw}(\operatorname{\mathcal{C}}) ) \xrightarrow {\sim } \operatorname{Tw}( \operatorname{N}_{\bullet }( \operatorname{\mathcal{C}}) )$ is another morphism of simplicial sets satisfying conditions $(1)$ and $(2)$. Then $T^{-1} \circ T'$ determines a functor $F$ from the twisted arrow category $\operatorname{Tw}(\operatorname{\mathcal{C}})$ to itself. Because $T$ and $T'$ both satisfy condition $(1)$, the functor $F$ carries each object of $\operatorname{Tw}(\operatorname{\mathcal{C}})$ to itself. Since the forgetful functor $\operatorname{Tw}(\operatorname{\mathcal{C}}) \rightarrow \operatorname{\mathcal{C}}^{\operatorname{op}} \times \operatorname{\mathcal{C}}$ is faithful, condition $(2)$ guarantees that $F$ also carries each morphism of $\operatorname{Tw}(\operatorname{\mathcal{C}})$ to itself. It follows that $F$ is the identity functor, so that $T' = T$. $\square$

Proof. Combine Proposition 8.1.1.11 with Remark 4.1.1.9. $\square$

Proof. Combine Proposition 8.1.1.11 with Corollary 4.4.3.8. $\square$

Proof. Let $\pi _{-}: \operatorname{\mathcal{C}}^{\operatorname{op}} \times \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}^{\operatorname{op}}$ and $\pi _{+}: \operatorname{\mathcal{C}}^{\operatorname{op}} \times \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}$ denote the projection maps. Then $\pi _{-}$ and $\pi _{+}$ are cocartesian fibrations of simplicial sets. Moreover, a morphism $(e_{-}, e_{+})$ of $\operatorname{\mathcal{C}}^{\operatorname{op}} \times \operatorname{\mathcal{C}}$ is $\pi _{-}$-cocartesian if and only if $e_{+}$ is an isomorphism in $\operatorname{\mathcal{C}}$, and $\pi _{+}$-cocartesian if and only if $e_{-}$ is an isomorphism in $\operatorname{\mathcal{C}}^{\operatorname{op}}$ (this follows immediately from Remark 5.1.4.6 and Example 5.1.1.4). Corollary 8.1.1.14 now follows by applying Proposition 8.1.1.11 to left and right sides of the diagram

since the vertical map in the center is a left fibration (Proposition 8.1.1.11). $\square$

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

admits a solution.

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

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

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

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

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 $q' < q$. $\square$