Kerodon

Proof. We prove $(a)$ and $(b)$ by simultaneous induction on the length $n = \ell (f)$. If $n = 1$, then assertion $(a)$ follows from condition $(2)$ of Definition 6.2.5.4 and assertion $(b)$ is vacuous. We therefore assume that $n \geq 2$. We first prove $(b)$. Choose a factorization

where $s' \in S$ and $\ell (g') = n-1$. If the factorization (6.3) is $S$-optimal, then we can choose a $3$-simplex

Condition $(2)$ of Definition 6.2.5.4 guarantees that $h$ belongs to $S$. Our inductive hypothesis guarantees that the left half of the diagram satisfies assertion $(a)$; so that $\ell (g) \leq \ell (g') = \ell (f) - 1$. The reverse inequality follows immediately from the definition.

We now prove $(a)$. Choose an $S$-optimal factorization

It follows from the preceding argument that $\ell (g'') = n-1$. We then have a commutative diagram

and condition $(2)$ of Definition 6.2.5.4 guarantees that $j$ belongs to $S$. We therefore obtain $\ell (g) \leq \ell (g'') + \ell (j) \leq (n-1) + 1 = n$, as desired. $\square$

Proof of Theorem 6.2.5.10. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $S$ be a class of short morphisms for $\operatorname{\mathcal{C}}$. For every nonnegative integer $p$, let $\operatorname{\mathcal{C}}^{\leq p}$ denote the smallest simplicial subset of $\operatorname{\mathcal{C}}$ which contains all simplices of priority $\leq p$ (so that a nondegenerate simplex of $\operatorname{\mathcal{C}}$ belongs to $\operatorname{\mathcal{C}}^{\leq p}$ if and only if it has priority $\leq p$). Then $\operatorname{\mathcal{C}}$ is the colimit of the sequence of inclusion maps

We will complete the proof by showing that each of these inclusion maps is inner anodyne. For the remainder of the proof, we fix a positive integer $p$; our goal is to show that the inclusion map $\operatorname{\mathcal{C}}^{\leq p-1} \hookrightarrow \operatorname{\mathcal{C}}^{\leq p}$ is inner anodyne.

Choose a function $\sigma \mapsto \sigma ^{+}$ satisfying the requirements of Lemma 6.2.5.15. For each integer $n \geq 0$, let $\operatorname{\mathcal{C}}^{\leq p}(n)$ denote the simplicial subset of $\operatorname{\mathcal{C}}^{\leq p}$ generated by $\operatorname{\mathcal{C}}^{\leq p-1}$ together with all simplices of the form $\sigma ^{+}$, where $\sigma $ is a simplex of $\operatorname{\mathcal{C}}$ having priority $p$ and dimension $\leq n$. By virtue of requirement $(2)$ of Lemma 6.2.5.15, it suffices to allow $\sigma $ to range over nondegenerate simplices which satisfy these conditions. Note that each $\operatorname{\mathcal{C}}^{\leq p}(n)$ contains the $n$-skeleton of $\operatorname{\mathcal{C}}^{\leq p}$, so that $\operatorname{\mathcal{C}}^{\leq p}$ can be realized as the colimit of the sequence

It will therefore suffice to show that each of these inclusions is inner anodyne. For the remainder of the proof, we fix an integer $n > 0$; our goal is to show that the inclusion map $\operatorname{\mathcal{C}}^{\leq p}(n-1) \hookrightarrow \operatorname{\mathcal{C}}^{\leq p}(n)$ is inner anodyne.

Let $\{ \sigma _{t} \} _{t \in T}$ denote the collection of $n$-simplices of $\operatorname{\mathcal{C}}$ which have priority $p$ but are not contained in $\operatorname{\mathcal{C}}^{\leq p}(n-1)$. We first claim that $\operatorname{\mathcal{C}}^{\leq p}(n)$ is generated by $\operatorname{\mathcal{C}}^{\leq p}(n-1)$ together with the collection of $(n+1)$-simplices $\{ \sigma _{t}^{+} \} _{t \in T}$. To prove this, it suffices to show that if $\sigma $ is an $n$-simplex of $\operatorname{\mathcal{C}}$ which belongs to $\operatorname{\mathcal{C}}^{\leq p}(n-1)$, then $\sigma ^{+}$ also belongs to $\operatorname{\mathcal{C}}^{\leq p}(n-1)$. If $\sigma $ is degenerate, then we can write $\sigma ^{+} = s^{n}_ i( \sigma _0^{+} )$ when $\sigma _0$ is an $(n-1)$-simplex of $\operatorname{\mathcal{C}}$ having priority $\leq p$ (Lemma 6.2.5.15), and the desired conclusion follows from the observation that $\sigma _0^{+}$ is contained in $\operatorname{\mathcal{C}}^{\leq p}(n-1)$. We may therefore assume that $\sigma $ is a nondegenerate $n$-simplex of $\operatorname{\mathcal{C}}$. If $\sigma $ has priority $< p$, then $\sigma ^{+}$ also has priority $< p$ and is therefore contained in $\operatorname{\mathcal{C}}^{\leq p-1}$. We may therefore assume that $\sigma $ has priority $p$, and must therefore be of the form $\tau ^{+}$ where $\tau $ is an $(n-1)$-simplex of $\operatorname{\mathcal{C}}$ having priority $p$. Condition $(3)$ of Lemma 6.2.5.15 then guarantees that $\sigma ^{+}$ can be obtained from $\sigma $ by applying a degeneracy operator, and is therefore contained in $\operatorname{\mathcal{C}}^{\leq p}(n-1)$ as desired.

For each $t \in T$, we define the complexity of $t$ to be the integer $c(t) = \sum _{ q = p}^{n} \ell ( \sigma _{t}(p-1) \rightarrow \sigma _{t}(q) )$. Using Proposition 4.7.1.35, we can choose a well-ordering on $T$ for which the complexity function

is nondecreasing. For each $t \in T$, let $\operatorname{\mathcal{C}}^{\leq p}_{\leq t}(n)$ denote the simplicial subset of $\operatorname{\mathcal{C}}^{\leq p}(n)$ generated by $\operatorname{\mathcal{C}}^{\leq p}(n-1)$ together with the simplices $\sigma _{s}^{+}$ for $s \leq t$, and define $\operatorname{\mathcal{C}}^{\leq p}_{< t}(n)$ similarly. Then the inclusion map $\operatorname{\mathcal{C}}^{\leq p}(n-1) \hookrightarrow \operatorname{\mathcal{C}}^{\leq p}(n)$ can be realized as a transfinite composition of inclusion maps $\{ \operatorname{\mathcal{C}}^{\leq p}_{< t}(n) \hookrightarrow \operatorname{\mathcal{C}}^{\leq p}_{\leq t}(n) \} _{t \in T}$. It will therefore suffice to show that each of these inclusion maps is inner anodyne.

Fix an element $t \in T$ and let $L_{t} \subseteq \Delta ^{n+1}$ be the inverse image of $\operatorname{\mathcal{C}}^{\leq p}_{< t}(n)$ under the map $\sigma _{t}^{+}: \Delta ^{n+1} \rightarrow \operatorname{\mathcal{C}}$, so that we have a pullback diagram of simplicial sets

We will complete the proof by showing that $L_{t}$ coincides with the inner horn $\Lambda ^{n+1}_{p} \subseteq \Delta ^{n+1}$, so that the diagram (6.10) is also a pushout square (Lemma 3.1.2.11). This is equivalent to the following more concrete assertion:

$(\ast _{t})$

For $0 \leq i \leq n+1$, the $n$-simplex $d^{n+1}_{i}( \sigma _{t}^{+} )$ belongs to $\operatorname{\mathcal{C}}^{\leq p}_{< t}(n)$ if and only if $i \neq p$.

Our proof proceeds by induction on $t$. We consider several cases:

• For $0 \leq i < p$, the $n$-simplex $d^{n+1}_ i( \sigma _{t}^{+} )$ has priority $< p$, and is therefore contained in $\operatorname{\mathcal{C}}^{< p} \subseteq \operatorname{\mathcal{C}}^{\leq p}(n-1) \subseteq \operatorname{\mathcal{C}}^{\leq p}_{< t}(n)$.

• For $i = p$, the $n$-simplex $d^{n+1}_ i( \sigma _{t}^{+} )$ coincides with $\sigma _{t}$ (Lemma 6.2.5.15), which is not contained in $\operatorname{\mathcal{C}}^{\leq p}(n-1)$. Consequently, if $\sigma _{t}$ is contained in $\operatorname{\mathcal{C}}^{\leq p}_{< t}(n-1)$, then there exists some $t' < t$ such that $\sigma _{t}$ is contained in $\operatorname{\mathcal{C}}^{\leq p}_{\leq t'}(n-1)$ but not in $\operatorname{\mathcal{C}}^{\leq p}_{< t'}(n-1)$. Applying our inductive hypothesis, we deduce that $\sigma _{t} = \sigma _{t'}$, which contradicts the inequality $t' < t$.

• For $p < i \leq n$, condition $(1)$ of Lemma 6.2.5.15 implies that $d^{n+1}_ i( \sigma _{t}^{+} )$ coincides with $d^{n}_{i-1}( \sigma _{t} )^{+}$, and therefore belongs to $\operatorname{\mathcal{C}}^{\leq p}(n-1) \subseteq \operatorname{\mathcal{C}}^{\leq p}_{< t}(n)$.

• Suppose that $i = n+1$ and set $\tau = d^{n+1}_{i}( \sigma _{t}^{+} )$; we wish to show that $\tau $ is contained in $\operatorname{\mathcal{C}}^{\leq p}_{< t}(n)$. Note that $\tau $ has priority $\leq p$. If $\tau $ is contained in $\operatorname{\mathcal{C}}^{\leq p}(n-1)$, there is nothing to prove. We may therefore assume without loss of generality that $\tau $ is contained in $T$: that is, we have $\tau = \sigma _{t'}$ for some $t' \in T$. Set $X = \sigma _{t}(p-1) = \sigma _{t'}(p-1)$. For $p \leq q \leq n$, let $f_{q}: X \rightarrow \sigma _{t}(q)$ be the morphism determined by $\sigma _{t}$, and define $f'_{q}: X \rightarrow \sigma _{t'}(q)$ similarly. By construction, the morphism $f_{q}$ coincides with $f'_{q+1}$ for $p \leq q < n$. Moreover, the restriction of $\sigma _{t}^{+}$ to the $2$-simplex $\operatorname{N}_{\bullet }( \{ p-1 < p < n+1 \} $ determines a diagram

  \[ \xymatrix { & \sigma _{t'}(p) \ar [dr] & \\ X \ar [ur]^{ f'_{p} } \ar [rr]^{ f_{n} } & & \sigma _{t}(n) } \]

  which is an $S$-optimal factorization of $f_ n$, so that $\ell ( f'_{p} ) = \ell ( f_ n ) - 1$ by virtue of Lemma 6.2.5.13. It follows that the complexity $c(\sigma _{t'} )$ is given by

  \begin{eqnarray*} c(\sigma _{t'} ) & = & \sum _{q = p}^{n} \ell ( f'_{q} ) \\ & = & \ell (f'_ p) + \sum _{ q = p+1}^{n} \ell (f'_ q) \\ & = & \ell ( f_ n ) - 1 + \sum _{ q = p}^{n-1} \ell ( f_ q ) \\ & = & ( \sum _{q = p}^{n} \ell (f_ q) ) - 1 \\ & = & c(\sigma _{t} ) - 1. \end{eqnarray*}

  We therefore have $t' < t$, so that $\tau = \sigma _{t'} = d^{n+1}_ p( \sigma _{t'}^{+} )$ is contained in $\operatorname{\mathcal{C}}^{\leq p}_{\leq t'}(n) \subseteq \operatorname{\mathcal{C}}^{\leq p}_{< t}(n)$.

Proof of Lemma 6.2.5.15. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $S$ be a class of short morphisms for $\operatorname{\mathcal{C}}$. Our construction proceeds by recursion. Fix an integer $n \geq 0$. Assume that we have constructed a function $\tau \mapsto \tau ^{+}$ on simplices of $\operatorname{\mathcal{C}}$ having dimension $< n$ and priority $> 0$, satisfying conditions $(1)$ through $(4)$ of Lemma 6.2.5.15. Let $\sigma $ be an $n$-simplex of $\operatorname{\mathcal{C}}$ having priority $> 0$; we wish to show that there is an $(n+1)$-simplex $\sigma ^{+}$ which also satisfies conditions $(1)$ through $(4)$. Let us say that $\sigma $ of $\operatorname{\mathcal{C}}$ is free if it is not of the form $\tau ^{+}$, where $\tau $ is an $(n-1)$-simplex of priority $> 0$. We divide the construction into three cases:

$(a)$

The $n$-simplex $\sigma $ is not free.

$(b)$

The $n$-simplex $\sigma $ is free and degenerate.

$(c)$

The $n$-simplex $\sigma $ is free and nondegenerate.

We begin with case $(a)$. Assume that $\sigma = \tau ^{+}$, where $\tau $ is an $(n-1)$-simplex of $\operatorname{\mathcal{C}}$ having priority $p > 0$. It follows from our inductive hypothesis that $\sigma $ has the same priority $p$, and that $\tau = d^{n}_ p(\sigma )$. In particular, $\tau $ is uniquely determined by $\sigma $. In this case, we define $\sigma ^{+} = s^{n}_ p(\sigma )$, so that condition $(3)$ is satisfied by construction. Since $p \leq n-1 < n$, condition $(4)$ is vacuous. Note that the faces $d^{n+1}_ p( \sigma ^{+} )$ and $d^{n+1}_{p+1}( \sigma ^{+} )$ coincide with $\sigma = \tau ^{+} = d^{n}_ p( \sigma )^{+}$, so that condition $(1)$ is satisfied for $i = p$ and $i = p+1$. For $p+1 < i \leq n$, we compute

It remains to verify condition $(2)$. Suppose that $\sigma = s^{n-1}_ j(\sigma ')$ for some $(n-1)$-simplex $\sigma '$ of $\operatorname{\mathcal{C}}$. Note that, since $\sigma $ has priority $p$, we must have $j \neq p-1$ (see Notation 6.2.5.14). We first consider the case $j < p-1$, so that $\sigma '$ has priority $p-1$. In this case, we wish to show that $\sigma ^{+} = s^{n}_ j( \sigma '^{+})$. Set $\tau ' = d^{n-1}_{p-1}( \sigma ' )$. We then have

so that $\sigma = \tau ^{+} = s^{n-1}_ j( \tau '^{+} )$. Applying the face operator $d^{n}_ j$, we obtain $\sigma ' = \tau '^{+}$, so that $\sigma '^{+} = s^{n-1}_{p-1}( \sigma ' )$. The desired result now follows from the calculation

We now treat the case $j \geq p$, so that $\sigma '$ has priority $p$. In this case, we wish to show that $\sigma ^{+} = s^{n}_{j+1}( \sigma '^{+} )$. If $j = p$, this follows from the calculation

Let us therefore assume that $j > p$, and set $\tau ' = d_{p}^{n-1}( \sigma ' )$. We then have

so that $\sigma = \tau ^{+} = s^{n-1}_ j( \tau '^{+} )$. Applying the face operator $d^{n}_ j$, we deduce that $\sigma ' = \tau '^{+}$, so that $\sigma '^{+} = s^{n-1}_{p}( \sigma ' )$. The desired result now follows from the calculation

This completes our treatment of case $(a)$.

We now consider case $(b)$. Assume that $\sigma $ is a free simplex of $\operatorname{\mathcal{C}}$ of the form $s^{n-1}_ j( \tau )$. Choose $j$ as small as possible and let $p$ be the priority of $\tau $. We first treat the case where $j < p$, so that $\sigma $ has priority $p+1$ (see Notation 6.2.5.14). In this case, we define $\sigma ^{+} = s^{n}_ j( \tau ^{+} )$, so that

For $p+1 < i \leq n$, a similar calculation gives

which proves $(1)$.

To verify $(2)$, suppose that $\sigma = s^{n-1}_{j'}( \tau ' )$, for some $(n-1)$-simplex $\tau '$ of $\operatorname{\mathcal{C}}$. Note that we must have $j' \geq j$. Since $\sigma $ has priority $p+1$, we also have $j' \neq p$. Assume first that $j' < p$, so that $\tau '$ has priority $p$. In this case, we wish to show that $\sigma ^{+} = s^{n}_{j'} ( \tau '^{+} )$. If $j' = j$, this is immediate. We may therefore assume that $j' > j$, so that we can write $\tau ' = s^{n-2}_ j( \tau '' )$ and $\tau = s^{n-2}_{j'-1}( \tau '' )$ for some unique $(n-2)$-simplex $\tau ''$ of $\operatorname{\mathcal{C}}$. In this case, the desired result follows from the calculation

If $j' > p$, then $\tau '$ instead has priority $p+1$, and the desired result follows instead from the calculation

Condition $(3)$ is vacuous (since we have assumed that $\sigma $ is free). To prove $(4)$, we note that if $\sigma $ has priority $n$, then $\tau $ has priority $(n-1)$; the desired result now follows from the observation that the restriction of $\sigma ^{+}$ to $\operatorname{N}_{\bullet }( \{ n-1 < n < n+1 \} )$ coincides with the restriction of $\tau ^{+}$ to $\operatorname{N}_{\bullet }( \{ n-2 < n-1 < n \} )$, and is therefore an $S$-optimal factorization. This completes the construction in the case $j < p$.

We now treat the case $j \geq p$, so that the simplex $\sigma = s^{n-1}_ j(\tau )$ has priority $p$. In this case, we set $\sigma ^{+} = s^{n}_{j+1}(\tau ^{+} )$. Condition $(3)$ again vacuous (since $\sigma $ is assumed to be free), and condition $(4)$ is vacuous since $p < n$. We next prove $(1)$. Note that we have

To complete the proof of $(1)$, we must show that $d^{n+1}_ i( \sigma ^{+} ) = d^{n}_{i-1}(\sigma )^{+}$ for $p < i \leq n$. For $i \leq j$, this follows from the calculation

For $j+2 < i \leq n$, it follows instead from the calculation

It will therefore suffice to treat the case $i \in \{ j+1, j+2 \} $, in which case we have

To verify condition $(2)$, suppose that $\sigma = s^{n-1}_{j'}( \tau ' )$. By construction, we then have $j' \geq j \geq p$, so that the simplex $\tau '$ has priority $p$. We wish to show that $\sigma ^{+} = s^{n}_{j'+1}( \tau '^{+} )$. If $j' = j$, this is immediate. We may therefore assume that $j' > j$, so that we can write $\tau ' = s^{n-2}_ j( \tau '' )$ and $\tau = s^{n-2}_{j'-1}( \tau '' )$ as above. In this case, the desired result follows from the calculation

This completes the treatment of case $(b)$.

We now consider case $(c)$. For the remainder of the proof, we assume that the simplex $\sigma $ is free and nondegenerate, of priority $p > 0$. Let us decompose $\Delta ^{n+1}$ as a join $\Delta ^{p-1} \star \Delta ^{n-p} \star \{ z\} $. In what follows, we write $x$ for the final vertex of $\Delta ^{p-1}$ (corresponding to the element $p-1 \in [n+1]$) and $y$ for the initial vertex of $\Delta ^{n-p}$ (corresponding to the element $p \in [n+1]$). Note that the $n$-simplices $\sigma $ and $\{ d^{n+1}_ i(\sigma )^{+} \} _{ p \leq i < n}$ determine a morphism of simplicial sets $\sigma ^{\dagger }: \Delta ^{p-1} \star \operatorname{\partial \Delta }^{n-p} \star \{ z\} \rightarrow \operatorname{\mathcal{C}}$. Unwinding the definitions, we see that an $(n+1)$-simplex $\sigma ^{+}$ of $\operatorname{\mathcal{C}}$ satisfying condition $(1)$ can be identified with an extension of $\sigma ^{\dagger }$ to the join $\Delta ^{p-1} \star \Delta ^{n-p} \star \{ z\} \simeq \Delta ^{n+1}$. We wish to show that such an extension can always be found, which additionally satisfies condition $(4)$ in the case $p=n$ (note that conditions $(2)$ and $(3)$ are vacuous, by virtue of our assumption that $\sigma $ is free and nondegenerate).

Let $\overline{\sigma }^{\dagger }$ denote the restriction of $\sigma ^{\dagger }$ to $\{ x\} \star \operatorname{\partial \Delta }^{n-p} \star \{ z\} $. Since the inclusion $\{ x\} \hookrightarrow \Delta ^{p-1}$ is right anodyne (see Example 4.3.7.11), it will suffice to show that $\overline{\sigma }^{\dagger }$ can be extended to an $(n+2-p)$-simplex $\overline{\sigma }^{+}$ of $\operatorname{\mathcal{C}}$, having the additional property that $\overline{\sigma }^{+}$ is an $S$-optimal factorization in the case $p=n$. If $p = n$, the existence of $\overline{\sigma }^{+}$ follows from our assumption that $S$ is a class of short morphisms for $\operatorname{\mathcal{C}}$. We therefore assume that $p < n$. Set $Z = \sigma ^{\dagger }(z)$, so that we can identify $\overline{\sigma }^{\dagger }$ with a morphism of simplicial sets $\rho _0: \Lambda ^{n+1-p}_{0} \rightarrow \operatorname{\mathcal{C}}_{/Z}$; we wish to extend $\rho _0$ to an $(n+1-p)$-simplex of $\operatorname{\mathcal{C}}_{/Z}$. For $0 < i \leq n+1-p$, the image $\rho _0(i)$ belongs to the full subcategory $\operatorname{\mathcal{C}}_{/Z}^{\mathrm{short}} \subseteq \operatorname{\mathcal{C}}_{/Z}$. By virtue of Proposition 6.2.2.10, it will suffice to show that the restriction of $\rho _0$ to $\Delta ^1$ exhibits $\rho _0(1)$ as a $\operatorname{\mathcal{C}}_{/Z}^{\mathrm{short}}$-reflection of $\rho _0(0)$. This is equivalent to the assertion that the $2$-simplex

is an $S$-optimal factorization of the lower horizontal morphism (Remark 6.2.5.9), which follows from our inductive hypothesis. $\square$