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
\begin{eqnarray*} d^{n+1}_ i( \sigma ^{+} ) & = & d^{n+1}_ i( s^{n}_ p( \tau ^{+} ) ) \\ & = & s^{n-1}_ p( d^{n}_{i-1}( \tau ^{+} ) ) \\ & = & s^{n-1}_ p( d^{n-1}_{i-2}(\tau )^{+} ) \\ & = & (d^{n-1}_{i-2}(\tau )^{+})^{+} \\ & = & d^{n}_{i-1}( \tau ^{+} )^{+} \\ & = & d^{n}_{i-1}(\sigma )^{+}. \end{eqnarray*}
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
\[ \tau = d^{n}_ p( \sigma ) = d^{n}_ p( s^{n-1}_ j( \sigma ' ) ) = s^{n-2}_{j}( d^{n-1}_{p-1}( \sigma ' ) ) = s^{n-2}_ j(\tau ' ), \]
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
\[ \sigma ^{+} = s^{n}_ p( \sigma ) = s^{n}_ p( s^{n-1}_ j( \sigma ' ) ) = s^{n}_ j( s^{n-1}_{p-1}( \sigma ' ) ) = s^{n}_ j( \sigma '^{+} ). \]
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
\begin{eqnarray*} \sigma ^{+} & = & s^{n}_ p( \sigma ) \\ & = & s^{n}_ p( s^{n-1}_ p( \sigma ' )) \\ & = & s^{n}_{p+1}( s^{n-1}_ p( \sigma ' ) ) \\ & = & s^{n}_{p+1}(\sigma ) \\ & = & s^{n}_{p+1}( \tau ^{+} ). \end{eqnarray*}
Let us therefore assume that $j > p$, and set $\tau ' = d_{p}^{n-1}( \sigma ' )$. We then have
\[ \tau = d^{n}_ p( \sigma ) = d^{n}_ p( s^{n-1}_ j( \sigma ' ) ) = s^{n-2}_{j-1}( d^{n-1}_{p}( \sigma ' ) ) = s^{n-2}_{j-1}(\tau ' ), \]
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
\[ \sigma ^{+} = s^{n}_ p( \sigma ) = s^{n}_ p( s^{n-1}_ j( \sigma ' ) ) = s^{n}_{j+1}( s^{n-1}_{p}( \sigma ' ) ) = s^{n}_{j+1}( \sigma '^{+} ). \]
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
\[ d^{n+1}_{p+1}( \sigma ^{+} ) = d^{n+1}_{p+1}( s^{n}_ j( \tau ^{+} ) ) = s^{n-1}_ j( d^{n}_ p( \tau ^{+} ) ) = s^{n-1}_ j( \tau ) = \sigma . \]
For $p+1 < i \leq n$, a similar calculation gives
\begin{eqnarray*} d^{n+1}_ i( \sigma ^{+} ) & = & d^{n+1}_ i( s^{n}_ j( \tau ^{+} ) ) \\ & = & s^{n-1}_ j( d^{n}_{i-1}( \tau ^{+} ) ) \\ & = & s^{n-1}_ j( d^{n-1}_{i-2}(\tau )^{+} ) \\ & = & s^{n-1}_ j(d^{n-1}_{i-2}(\tau ))^{+} \\ & = & d^{n}_{i-1}( s^{n-1}_ j(\tau ) )^{+} \\ & = & d_{i-1}^{n}( \sigma )^{+}, \end{eqnarray*}
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
\begin{eqnarray*} \sigma ^{+} & = & s^{n}_ j( \tau ^{+}) \\ & = & s^{n}_ j( s^{n-2}_{j'-1}(\tau '')^{+} ) \\ & = & s^{n}_ j( s^{n-1}_{j'-1}( \tau ''^+ )) \\ & = & s^{n}_{j'} s^{n-1}_ j( \tau ''^{+} ) \\ & = & s^{n}_{j'} (s^{n-2}_ j(\tau '')^{+}) \\ & = & s^{n}_{j'} (\tau '^{+} ). \end{eqnarray*}
If $j' > p$, then $\tau '$ instead has priority $p+1$, and the desired result follows instead from the calculation
\begin{eqnarray*} \sigma ^{+} & = & s^{n}_ j( \tau ^{+}) \\ & = & s^{n}_ j( s^{n-2}_{j'-1}(\tau '')^{+} ) \\ & = & s^{n}_ j( s^{n-1}_{j'}( \tau ''^+ )) \\ & = & s^{n}_{j'+1} s^{n-1}_ j( \tau ''^{+} ) \\ & = & s^{n}_{j'+1} (s^{n-2}_ j(\tau '')^{+}) \\ & = & s^{n}_{j'+1} (\tau '^{+} ). \end{eqnarray*}
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
\[ d^{n+1}_ p( \sigma ^{+} ) = d^{n+1}_ p( s^{n}_{j+1}( \tau ^{+} ) ) = s^{n-1}_{j}( d^{n}_ p( \tau ^{+} ) ) = s^{n-1}_ j( \tau ) = \sigma . \]
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
\begin{eqnarray*} d^{n+1}_ i( \sigma ^{+} ) & = & d^{n+1}_ i( s^{n}_{j+1}( \tau ^{+} ) ) \\ & = & s^{n-1}_ j( d^{n}_ i( \tau ^{+} ) ) \\ & = & s^{n-1}_ j( d^{n-1}_{i-1}(\tau )^{+} ) \\ & = & s^{n-2}_{j-1}( d^{n-1}_{i-1}(\tau ) )^{+} \\ & = & d^{n}_{i-1}( s^{n-1}_ j( \tau ) )^{+} \\ & = & d^{n}_{i-1}(\sigma )^{+}. \end{eqnarray*}
For $j+2 < i \leq n$, it follows instead from the calculation
\begin{eqnarray*} d^{n+1}_ i( \sigma ^{+} ) & = & d^{n+1}_ i( s^{n}_{j+1}( \tau ^{+} ) ) \\ & = & s^{n-1}_{j+1}( d^{n}_{i-1}( \tau ^{+} ) ) \\ & = & s^{n-1}_{j+1}( d^{n-1}_{i-2}(\tau )^{+} ) \\ & = & s^{n-2}_{j}( d^{n-1}_{i-2}(\tau ) )^{+} \\ & = & d^{n}_{i-1}( s^{n-1}_ j( \tau ) )^{+} \\ & = & d^{n}_{i-1}(\sigma )^{+}. \end{eqnarray*}
It will therefore suffice to treat the case $i \in \{ j+1, j+2 \} $, in which case we have
\begin{eqnarray*} d^{n+1}_ i( \sigma ^+ ) & = & d^{n+1}_ i( s^{n}_{j+1}( \tau ^{+} ) ) \\ & = & \tau ^{+} \\ & = & d^{n}_{i-1}( s^{n-1}_ j( \tau ) )^{+} \\ & = & d^{n}_{i-1}( \sigma )^{+}. \end{eqnarray*}
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
\begin{eqnarray*} \sigma ^{+} & = & s^{n}_{j+1}( \tau ^{+}) \\ & = & s^{n}_{j+1}( s^{n-2}_{j'-1}(\tau '')^{+} ) \\ & = & s^{n}_{j+1}( s^{n-1}_{j'}( \tau ''^+ )) \\ & = & s^{n}_{j'+1} s^{n-1}_{j+1}( \tau ''^{+} ) \\ & = & s^{n}_{j'+1} (s^{n-2}_{j}(\tau '')^{+}) \\ & = & s^{n}_{j'+1} (\tau '^{+} ). \end{eqnarray*}
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
\[ \xymatrix { & \sigma ^{\dagger }(y) \ar [dr] & \\ \sigma ^{\dagger }(x) \ar [rr] \ar [ur] & & \sigma ^{\dagger }(z) } \]
is an $S$-optimal factorization of the lower horizontal morphism (Remark 6.2.5.9), which follows from our inductive hypothesis.
$\square$