Kerodon

Proof of Proposition 8.1.2.1. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $f: X \rightarrow Y$ be an isomorphism in $\operatorname{\mathcal{C}}$; we wish to show that $f$ is initial when viewed as an object of the $\infty $-category $\{ X\} \times _{ \operatorname{\mathcal{C}}^{\operatorname{op}} } \operatorname{Tw}(\operatorname{\mathcal{C}})$. Fix an integer $n > 0$ and a morphism $\rho _0: \operatorname{\partial \Delta }^{n} \rightarrow \{ X\} \times _{ \operatorname{\mathcal{C}}^{\operatorname{op}} } \operatorname{Tw}(\operatorname{\mathcal{C}})$ satisfying $\rho _0( 0 ) = f$; we wish to show that $\rho _0$ can be extended to an $n$-simplex of $\{ X\} \times _{ \operatorname{\mathcal{C}}^{\operatorname{op}} } \operatorname{Tw}(\operatorname{\mathcal{C}})$.

We now use a variation on the proof of Proposition 8.1.1.15. For every nonempty subset $S \subseteq [2n+1]$, 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 \} $.

$(b)$

There exists an integer $0 \leq i \leq n$ such that $S \cap \{ i, 2n+1-i \} = \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 identify $\rho _0$ with a morphism of simplicial sets $\theta _0: K \rightarrow \operatorname{\mathcal{C}}$, where the composition $\Delta ^{n} \hookrightarrow K \xrightarrow {\theta _0} \operatorname{\mathcal{C}}$ is the constant map taking the value $X$ and the composition

is the morphism $f$. To complete the proof, we must show that $\theta _0$ admits an extension $\theta : \Delta ^{2n+1} \rightarrow \operatorname{\mathcal{C}}$.

Let $S$ be a nonempty subset of $[2n+1]$ which is not basic. Then there exists an integer $0 \leq i \leq n$ such that $2n+1-i$ belongs to $S$. We denote the largest such integer by $\mathrm{pr}(S)$ and refer to it as the priority of $S$. We say that $S$ is prioritized if it also contained the integer $\mathrm{pr}(S)$. Let $\{ S_1, S_2, \cdots , S_ m \} $ be an ordering of the collection of all prioritized (non-basic) subsets of $[2n+1]$ 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 $1 \leq p \leq q \leq m$, then we have $\mathrm{pr}( S_ p ) \leq \mathrm{pr}(S_ q)$.

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

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 $K_{q} \subseteq \Delta ^{2n+1}$ denote the union of $K_0$ with the simplices $\{ \sigma _1, \sigma _2, \cdots , \sigma _ q \} $, so that we have inclusion maps

Note that the set $S = [2n+1]$ is prioritized (with priority $n$), and is therefore equal to $S_ m$. It follows that $K_{m} = \Delta ^{2n+1}$. We will complete the proof by showing that $\theta _0$ admits a compatible sequence of extensions $\{ \theta _ q: K_ q \rightarrow \operatorname{\mathcal{C}}\} _{0 \leq q \leq m}$, so that $\theta = \theta _ m$ is an extension of $\theta _0$ to $\Delta ^{2n+1}$.

For the remainder of the proof, we fix an integer $1 \leq q \leq m$, and suppose that the morphism $\theta _{q-1}: K_{q-1} \rightarrow \operatorname{\mathcal{C}}$ has already been constructed. Let $d$ denote 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 $L = \sigma _{q}^{-1} K_{q-1} \subseteq \Delta ^{d}$. Let $i = \mathrm{pr}(S_ q)$ denote the priority of $S_{q}$, so that $S_{q}$ contains both $i$ and $2n+1-i$. Write $S_{q} = \{ j_0 < j_1 < \cdots < j_ d \} $, so that $i = j_{k}$ for some integer $0 \leq k \leq d$. We will prove below that $L$ is equal to the horn $\Lambda ^{d}_{k} \subseteq \Delta ^{d}$, so that the diagram of simplicial sets

is a pushout square (Lemma 3.1.2.11). Let $\tau _0$ denote the composite map $L \xrightarrow { \sigma _ q } K_{q-1} \xrightarrow { \theta _{q-1} } \operatorname{\mathcal{C}}$. We will complete the proof by showing that $\tau _0$ admits an extension $\tau : \Delta ^{d} \rightarrow \operatorname{\mathcal{C}}$ (which then determines a morphism $\theta _{q}: K_{q} \rightarrow \operatorname{\mathcal{C}}$ extending $\theta _{q-1}$). The proof splits into four cases:

• Suppose that $0 < k < d$. Then $\Lambda ^{d}_{k} \subseteq \Delta ^{d}$ is an inner horn, so that $\tau _0$ admits an extension $\tau : \Delta ^{d} \rightarrow \operatorname{\mathcal{C}}$ by virtue of our assumption that $\operatorname{\mathcal{C}}$ is an $\infty $-category.

• Suppose that $k = d$. Then $S_{q}$ is contained in $\{ 0, 1, \cdots , n \} $, contradicting our assumption that $S_{q}$ is not basic.

• Suppose $k = 0$ and $i < n$, so that $i$ is the least element of $S_ q$. Our assumption $\mathrm{pr}(S_ q) = i$ guarantees that $2n-i \notin S$. Since $S_{q}$ does not satisfy $(b)$, we must also have $i+1 \in S_ q$. It follows that $d \geq 2$ (otherwise, $S_ q$ would satisfy $(a)$), and that $\tau _0: \Lambda ^{d}_{0} \rightarrow \operatorname{\mathcal{C}}$ carries the initial edge $\operatorname{N}_{\bullet }( \{ 0 < 1 \} )$ to the identity morphism $\operatorname{id}_{X}$. In this case, the existence of the extension $\tau $ follows from Theorem 4.4.2.6.

• Suppose $k = 0$ and $i = n$, so that $i = n$ is the least element of $S_ q$. Since $S_{q}$ has priority $n$, the element $n+1$ also belongs to $S_{q}$. We must then have $d \geq 2$ (otherwise, $S_ q$ would satisfy condition $(b)$). It follows that $\tau _0: \Lambda ^{d}_{0} \rightarrow \operatorname{\mathcal{C}}$ carries the initial edge $\operatorname{N}_{\bullet }( \{ 0 < 1 \} )$ to the morphism $f$, which is an isomorphism in $\operatorname{\mathcal{C}}$. In this case, the existence of the extension $\tau $ again follows from Theorem 4.4.2.6.

It remains to prove that $L = \Lambda ^{d}_{k}$, which we can formulate more concretely 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 $j \neq i$.

We first treat the case $j = i$; in this case, we wish to show that $\sigma _{S'}$ is not contained in $K_{q-1}$. Note that $S'$ cannot be basic: it cannot be contained in $\{ 0, 1, \cdots , n \} $ (otherwise $S_{q} =S' \cup \{ i\} $ would have the same property) and cannot have empty intersection with a set of the form $\{ i', 2n+1- i'\} $ (otherwise $S_ q$ would have the same property; here we use the fact that $2n+1-i$ is contained in $S_ q$). Moreover, we have $\mathrm{pr}(S') = i \notin S'$, so that $S'$ is not prioritized. Assume, for a contradiction, that $\sigma _{S'}$ is contained in $K_{q-1}$. Then we must have $S' \subseteq S_{q'}$, for some $1 \leq q' < q$. Note that $2n+1-i \in S' \subseteq S_{q'}$, so that $S_{q'}$ has priority $\geq i$. Since $q' < q$, it follows that $S_{q'}$ has priority $i$ and that $|S_{q'}| \leq | S_{q} |$. Since $S_{q'}$ is prioritized, it contains the element $i$, and therefore contains $S_{q} = S' \cup \{ i\} $. It follows that $S_{q'} = S_{q}$, contradicting our assumption that $q' < q$.

We now treat the case $j \neq i$; in this case, we wish to show that $\sigma _{S'}$ is contained in $K_{q-1}$. We may assume without loss of generality that $S'$ is not basic (otherwise, the simplex $\sigma _{S'}$ is already contained in $K_0$). Let $i'= \mathrm{pr}( S' )$ denote the priority of $S'$; note that the inclusion $S' \subseteq S_{q}$ guarantees that $i' \leq i$. If $i' < i$, then $S' \cup \{ i' \} $ is a prioritized set of priority $< i$, and therefore of the form $S_{q'}$ for some $q' < q$. If $i' = i$, then $S'$ is a prioritized set of priority $i$ and cardinality $| S_{q} | - 1$, and therefore of the form $S_{q'}$ for some $q' < q$. In either case, we obtain $\sigma _{S'} \subseteq \sigma _{q'} \subseteq K_{q-1}$. $\square$