Lemma 8.1.7.1. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $\sigma $ be a $2$-simplex of $\operatorname{Corr}(\operatorname{\mathcal{C}})$, which we identify with a diagram $\varphi : \operatorname{Tw}( \Delta ^2) \rightarrow \operatorname{\mathcal{C}}$. If $\sigma $ is thin, then $\varphi $ is a colimit diagram.

### 8.1.7 Thin $2$-Simplices of $\operatorname{Cospan}(\operatorname{\mathcal{C}})$

Let $\operatorname{\mathcal{C}}$ be an $\infty $-category. Our goal in this section is to prove Proposition 8.1.6.2, which provides necessary and sufficient conditions for a $2$-simplex of the simplicial set $\operatorname{Cospan}(\operatorname{\mathcal{C}})$ to be thin. By virtue of Remark 8.1.6.5, it will suffice to prove the following pair of assertions:

Lemma 8.1.7.2. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $\sigma $ be a $2$-simplex of $\operatorname{Corr}(\operatorname{\mathcal{C}})$, which we identify with a diagram $\varphi : \operatorname{Tw}( \Delta ^2) \rightarrow \operatorname{\mathcal{C}}$. If $\varphi $ is a colimit diagram, then $\sigma $ is thin.

**Proof of Lemma 8.1.7.1.**
Let $Q$ denote the partially ordered set appearing in the proof of Corollary 8.1.6.4, so that we can identify $\operatorname{Tw}( \Lambda ^{2}_{1} )$ with the nerve of $Q$. Set $\varphi _0 = \varphi |_{ \operatorname{N}_{\bullet }(Q) }$. Assume that $\sigma $ is thin. We wish to show that the restriction map $\operatorname{\mathcal{C}}_{ \varphi /} \rightarrow \operatorname{\mathcal{C}}_{ \varphi _0/}$ is a trivial Kan fibration: that is, every lifting problem

admits a solution.

Let $K$ denote the coproduct $( \operatorname{Tw}( \Delta ^2) \star \operatorname{\partial \Delta }^ n) \coprod _{ ( \operatorname{N}_{\bullet }(Q) \star \operatorname{\partial \Delta }^ n) } (\operatorname{N}_{\bullet }(Q) \star \Delta ^ n)$, which we regard as a simplicial subset of $\operatorname{Tw}( \Delta ^2) \star \Delta ^ n$. Unwinding the definitions, we can identify the lifting problem (8.12) with a morphism of simplicial sets $\tau _0: K \rightarrow \operatorname{\mathcal{C}}$ satisfying $\tau _0|_{ \operatorname{Tw}( \Delta ^2) } = \varphi $. We wish to show that $\tau _0$ can be extended to a morphism $\tau : \operatorname{Tw}( \Delta ^{2} ) \star \Delta ^ n\rightarrow \operatorname{\mathcal{C}}$.

Let $\iota : \operatorname{Tw}( \Delta ^2) \star \Delta ^ n \rightarrow \operatorname{Tw}( \Delta ^{n+3} )$ be the morphism of simplicial sets given on vertices by the formula

The morphism $\iota $ has a left inverse $\rho : \operatorname{Tw}( \Delta ^{n+3}) \rightarrow \operatorname{Tw}(\Delta ^2) \star \Delta ^ n$, given on vertices by the formula

We observe that $\iota $ and $\rho $ restrict to morphisms of simplicial subsets

so that we have a commutative diagram of simplicial sets

where the horizontal compositions are equal to the identity.

The composition $\tau _0 \circ \rho _0$ can be identified with a morphism of simplicial sets $\psi _0: \Lambda ^{m+3}_{1} \rightarrow \operatorname{Cospan}(\operatorname{\mathcal{C}})$ having the property that the composition

coincides with $\sigma $. Since $\sigma $ is thin, we can extend $\psi _0$ to an $(n+3)$-simplex $\psi $ of $\operatorname{Cospan}(\operatorname{\mathcal{C}})$, which we identify with a map $\tau ': \operatorname{Tw}( \Delta ^{n+3} ) \rightarrow \operatorname{\mathcal{C}}$. It follows that the composition $\tau = \tau ' \circ \iota $ is a morphism $\operatorname{Tw}( \Delta ^{2} ) \star \Delta ^ n \rightarrow \operatorname{\mathcal{C}}$ satisfying $\tau |_{K} = \tau _0$. $\square$

**Proof of Lemma 8.1.7.2.**
Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $\sigma $ be a $2$-simplex of $\operatorname{Cospan}(\operatorname{\mathcal{C}})$, which we identify with a diagram $\varphi : \operatorname{Tw}( \Delta ^2 ) \rightarrow \operatorname{\mathcal{C}}$. Assume that $\varphi $ is a colimit diagram. We wish to show that $\sigma $ is thin. We proceed by a (somewhat more complicated) variation on the proof of Lemma 8.1.6.6.

Fix integers $0 < q < n$ with $n \geq 3$, and suppose that we are given a morphism $f_0: \Lambda ^{n}_{q} \rightarrow \operatorname{Cospan}(\operatorname{\mathcal{C}})$ for which the composition

is equal to $\sigma $; we wish to show that $f_0$ can be extended to an $n$-simplex of $\operatorname{Cospan}(\operatorname{\mathcal{C}})$. Using Proposition 8.1.4.7, we can identify $f_0$ with a morphism of simplicial sets $F_0: \operatorname{Tw}( \Lambda ^{n}_{q}) \rightarrow \operatorname{\mathcal{C}}$; we wish to show that $F_0$ admits an extension $\operatorname{Tw}( \Delta ^ n ) \rightarrow \operatorname{\mathcal{C}}$.

Let $P$ denote the set of all ordered pairs $(i,j)$, where $i$ and $j$ are integers satisfying $0 \leq i \leq j \leq n$. We regard $P$ as a partially ordered set by identifying it with its image in the product $[n]^{\operatorname{op}} \times [n]$ (so that $(i,j) \leq (i',j')$ if and only if $i' \leq i$ and $j \leq j'$). In what follows, we will identify $\operatorname{Tw}( \Delta ^ n )$ with the nerve $\operatorname{N}_{\bullet }(P)$; under this identification, $\operatorname{Tw}( \Lambda ^{n}_{q} )$ corresponds to a simplicial subset $K_0 \subseteq \operatorname{N}_{\bullet }(P)$.

Let $S = \{ (i_0, j_0) < (i_1, j_1) < \cdots < (i_ d, j_ d) \} $ be a nonempty linearly ordered subset of $P$, so that we have inequalities $0 \leq i_ d \leq i_{d-1} \leq \cdots \leq i_0 \leq j_0 \leq j_{1} \leq \cdots \leq j_ d \leq n$. In this case, we write $\tau _{S}$ for the corresponding nondegenerate $d$-simplex of $\operatorname{N}_{\bullet }(P)$. Let $C(S)$ denote the set of integers $\{ i_0, i_1, \cdots , i_ d, j_0, j_1, \cdots , j_ d \} $, which we regard as a subset of $[n] = \{ 0 < 1 < \cdots < n \} $; we will refer to $C(S)$ as the *content* of $S$. We will say that:

The set $S$ is

*basic*if $C(S) \cup \{ q \} \neq [n]$. Equivalently, $S$ is basic if the simplex $\tau _{S}$ is contained in $K_0$.The set $S$ is

*low*if it is not basic and it contains an ordered pair $(p, q+1)$ for some $p \leq q-2$. In this case, we denote the largest such integer $p$ by $\mathrm{pr}( S)$ and refer to it as the*priority*of $S$.The set $S$ is

*high*if it is not basic and it contains an ordered pair $(q-1,r)$ for some $r \geq q+2$.

Note that the set $S$ cannot be both low and high, since the elements $(p,q+1)$ and $(q-1,r)$ are incomparable in $Q$ when $p < q < r$. Moreover, if $S$ is low, then any nonempty subset $S' \subseteq S$ is either low (and satisfies $\mathrm{pr}(S') \leq \mathrm{pr}(S)$) or satisfies $q+1 \notin C(S')$, so that $S'$ is basic (the set $S'$ cannot contain any elements of the form $(q+1, r)$, since these are incomparable with $(p,q+1)$). Consequently, the collection of simplices of the form $\tau _{S}$ where $S$ is either basic or low determine a simplicial subset $K_{\mathrm{low}} \subseteq \operatorname{N}_{\bullet }(P)$. Similarly, the collection of simplices of the form $\tau _{S}$ where $S$ is either basic or high determine a simplicial subset $K_{\mathrm{hiugh}} \subseteq \operatorname{N}_{\bullet }(P)$, and the intersection $K_{\mathrm{low}} \cap K_{\mathrm{high}}$ is equal to $K_0$.

We will prove the following:

- $(\ast )$
The inclusion maps $K_0 \hookrightarrow K_{\mathrm{low}}$ and $K_0 \hookrightarrow K_{\mathrm{high}}$ are inner anodyne.

We will prove that the inclusion map $K_0 \hookrightarrow K_{\mathrm{low}}$ is inner anodyne; the analogous assertion for the inclusion $K_0 \hookrightarrow K_{\mathrm{high}}$ follows by a similar argument. Let us say that a linearly ordered subset $S \subseteq Q$ is *prioritized* if it low and contains the ordered pair $(p,q)$, where $p = \mathrm{pr}(S)$ denotes the priority of $S$.

Let $\{ S_1, S_2, \cdots , S_ m \} $ be an enumeration of the collection of all prioritized low subsets of $P$ 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 $1 \leq k \leq \ell \leq m$, then we have $\mathrm{pr}( S_{k} ) \leq \mathrm{pr}(S_{\ell })$.

If $\mathrm{pr}( S_{k} ) = \mathrm{pr}( S_{\ell } )$ for $k \leq \ell $, then $| S_{k} | \leq | S_{\ell } |$.

For $1 \leq \ell \leq m$, let $\tau _{\ell } \subseteq \operatorname{N}_{\bullet }(P)$ denote the simplex $\tau _{S_{\ell }}$ and let $K_{\ell }$ denote the union of $K_0$ with the simplices $\{ \tau _1, \tau _2, \cdots , \tau _{\ell } \} $, so that we have inclusion maps

We claim that $K_{m} = K_{\mathrm{low}}$: that is, every low subset $S \subseteq P$ is contained in $S_{\ell }$ for some $1 \leq \ell \leq m$. This is clear: if $p = \mathrm{pr}(S)$ is the priority of $S$, then the union $S \cup \{ (p, q) \} $ is a prioritized low subset of $Q$ (having the same priority $p$).

We will prove $(\ast )$ by showing that, for $1 \leq \ell \leq m$, the inclusion map $K_{\ell -1} \hookrightarrow K_{\ell }$ is inner anodyne. Set $S_ q = \{ (i_0, j_0) < (i_1, j_1) < \cdots < (i_ d, j_ d) \} $. Let $p$ denote the priority of $S_{\ell }$. Since $S_{\ell }$ is prioritized, it contains the ordered pair $(p, q)$. We therefore have $(p,q) = (i_{c}, j_{c} )$ for some $0 \leq c \leq d$. Note that since $p \leq r-2$, we must have $c > 0$: otherwise, we have $q-1 \notin C(S_{\ell })$, contradicting our assumption that $S_{\ell }$ is not basic. Since $S_{\ell }$ also contains $(p, q+1)$, we must also have $c < d$. Let $L \subseteq \Delta ^ d$ denote the inverse image of $K_{\ell -1}$ under the map $\tau _{\ell }: \Delta ^{d} \rightarrow \operatorname{N}_{\bullet }(P)$, so that we have a pushout diagram of simplicial sets

We will complete the proof of $(\ast )$ by showing that $L$ coincides with the inner horn $\Lambda ^{d}_{c}$. This can be stated more concretely as follows:

- $(\ast ')$
Let $(i,j)$ be an element of $S_{\ell }$, and set $S' = S_{\ell } \setminus \{ (i,j) \} $. Then the simplex $\tau _{S'}$ is contained in $K_{\ell -1}$ if and only if $(i,j) \neq (p,q)$.

We first prove $(\ast ')$ in the case where $(i,j) \neq (p,q)$; in this case, we wish to show that $\tau _{S'}$ is contained in $K_{\ell -1}$. If $S'$ is basic, then $\tau _{S'}$ is contained in $K_0$ and there is nothing to prove. We may therefore assume that $S'$ is not basic, and is therefore low. If $(i,j) \neq (p,q+1)$, then $S'$ is a prioritized low subset of $Q$ satisfying $\mathrm{pr}(S') = p = \mathrm{pr}(S_{\ell })$ and $| S' | < | S_{\ell } |$. It follows that $S' = S_{k}$ for some $k < \ell $, so that $\tau _{S'}$ is contained in $K_{k} \subseteq K_{\ell -1}$. In the case $(i,j) = (p,q+1)$, the set $S'$ has priority $p' = \mathrm{pr}(S') < p$. It follows that $S' \cup \{ (p', q) \} $ is a prioritized low subset of $Q$ having priority $p' < \mathrm{pr}(S_ q)$, and is therefore of the form $S_{k}$ for some $k < \ell $. In this case, we again conclude that $\tau _{S'}$ is contained in $K_{k} \subseteq K_{\ell -1}$.

We now prove $(\ast ')$ in the case where $(i,j) = (p,q)$; in this case, we wish to show that $\tau _{S'}$ is not contained in $K_{\ell -1}$. Note that, since $S'$ contains $(p,q+1)$, we have $C(S') \cup \{ q\} = C(S_{\ell }) \cup \{ q\} $. Since $S_{\ell }$ is not basic, it follows that $S'$ is not basic. Assume, for a contradiction, that $\tau _{S'}$ is contained in $K_{\ell -1}$; it follows that we have $S' \subseteq S_ k$ for some $1 \leq k \leq < \ell $. We then have $\mathrm{pr}( S_ k ) \leq \mathrm{pr}( S_{\ell } ) = p$. Since $S_ k$ contains $(p,q+1)$, we must have $\mathrm{pr}(S_ k) = p$. Since $S_ k$ is prioritized, it contains $(p,q)$, and therefore contains $S' \cup \{ (p,q) \} = S_{\ell }$. The inequality $k < \ell $ guarantees that $|S_{k} | \leq | S_{\ell } |$. It follows that $S_{k} = S_{\ell }$, contradicting our assumption that $k < \ell $. This completes the proof of $(\ast )$.

Since $\operatorname{\mathcal{C}}$ is an $\infty $-category, assertion $(\ast )$ guarantees that the morphism $F_0: K_0 \rightarrow \operatorname{\mathcal{C}}$ admits an extension $F_{\mathrm{low}}: K_{\mathrm{low}} \rightarrow \operatorname{\mathcal{C}}$ (Proposition 1.4.6.7). Similarly, the morphism $F_0$ admits an extension $F_{\mathrm{high}}: K_{\mathrm{high}} \rightarrow \operatorname{\mathcal{C}}$. Let $K$ denote the union of $K_{\mathrm{low}}$ with $K_{\mathrm{high}}$ (as simplicial subsets of $\operatorname{N}_{\bullet }(P)$). Since the intersection $K_{\mathrm{low}} \cap K_{\mathrm{high}}$ coincides with $K_{0}$, we can amalgamate $F_{\mathrm{low}}$ with $F_{\mathrm{high}}$ to obtain a morphism of simplicial sets $F: K \rightarrow \operatorname{\mathcal{C}}$. We will complete the proof of Proposition 8.1.6.2 by showing that $F$ can be extended to a morphism $\operatorname{N}_{\bullet }(P) \rightarrow \operatorname{\mathcal{C}}$.

Set $P_{-} = \{ (i,j) \in P: (i,j) < (q-1, q+1) \} $ and $P_+ = \{ (i,j) \in P: (i,j) > (q-1, q+1) \} $. Let us say that a nonempty linearly ordered subset $S \subseteq P$ is *decomposable* if the union $S \cup \{ ( q-1, q+1) \} $ is also linearly ordered. In this case, we can write $S$ (uniquely) as a union $S_{-} \cup S_0 \cup S_{+}$, where $S_{-} \subseteq P_{-}$, $S_0 \subseteq \{ ( q-1, q+1) \} $, and $S_{+} \subseteq P_{+}$. The collection of simplices $\tau _{S}$, where $S$ is decomposable, span a simplicial subset of $\operatorname{N}_{\bullet }(P)$ which will identify with the join $\operatorname{N}_{\bullet }( P_{-} ) \star \{ (q-1, q+1) \} \star \operatorname{N}_{\bullet }(P_{+} )$.

We next claim that $\operatorname{N}_{\bullet }(P)$ is the union of $K$ with the join $\operatorname{N}_{\bullet }( P_{-} ) \star \{ (q-1, q+1) \} \star \operatorname{N}_{\bullet }(P_{+} )$. In other words, if a nonempty linearly ordered subset $S \subseteq P$ is not decomposable, then $\tau _{S}$ is contained in $K$. Choose an element $(i,j) \in S$ which is incomparable with $(q-1, q+1)$ in the partially ordered set $P$. Without loss of generality, we may assume that $i < q-1$ and $j < q+1$. If $S$ is basic, there is nothing to prove. We may therefore assume that $q+1$ belongs to the content $C(S)$. Note that ordered pairs of the form $(q+1, r)$ are incomparable with $(i,j)$, and therefore cannot be contained in $S$. It follows that $S$ contains an element of the form $(p, q+1)$. Since $(p,q+1)$ is comparable with $(i,j)$ in $P$, we must have $p \leq i \leq q-2$. It follows that $S$ is low, so that $\tau _{S}$ is contained in $K_{ \mathrm{low} } \subseteq K$.

Let $K' \subseteq K$ denote the intersection of $K$ with the join $\operatorname{N}_{\bullet }( P_{-} ) \star \{ (q-1, q+1) \} \star \operatorname{N}_{\bullet }(P_{+} )$, so that we have a pushout diagram of simplicial sets

Set $F' = F|_{K'}$. To complete the proof, it will suffice to show that $F'$ can be extended to a morphism of simplicial sets $\operatorname{N}_{\bullet }( P_{-} ) \star \{ (q-1, q+1) \} \star \operatorname{N}_{\bullet }(P_{+} ) \rightarrow \operatorname{\mathcal{C}}$.

We now give a more explicit description of $K'$. Let us say that a linearly ordered subset $S_{+} \subseteq P_{+}$ is *old* if the simplex $\tau _{ S_{+} \cup \{ (q-1, q+1) \} }$ is contained in $K$. Let $S \subseteq P$ be an arbitrary decomposable subset, and write $S = S_{-} \cup S_{0} \cup S_{+}$ as above. We then make the following observations:

- $(a)$
Suppose that $S_0 = \{ (q-1, q+1) \} $. Then replacing $S$ by the subset $S_0 \cup S_{+}$ does not change the set $C(S) \cup \{ q\} $. In particular, $S$ is basic if and only if $S_0 \cup S_{+}$ is basic. Moreover, since $S_{-}$ does not contain any pairs of the form $(p,q+1)$ for $p \leq q-1$, it follows that $S$ is low if and only if $S_0 \cup S_{+}$ is low. Similarly, $S$ is high if and only if $S_0 \cup S_{+}$ is high. It follows that $\tau _{S}$ is contained in $K$ if and only if $\tau _{S_0 \cup S_{+}}$ is contained in $K$: that is, if and only if $S_{+}$ is old.

- $(b)$
Suppose that $S_{0} = \emptyset $. In this case, we claim that $\tau _{S}$ is automatically contained in $K$. Assume otherwise. Then $S$ is not basic, so the set $C(S)$ contains the element $q+1$. Since ordered pairs of the form $(q+1, q')$ are incomparable with $(q-1, q+1)$ for $q' > q+1$, it follows that $S$ contains an ordered pair of the form $(p, q+1)$ for some $p \leq q+1$. Since $S$ is not low, we must have $p \geq q-1$. By assumption, $S$ does not contain $(q-1, q+1)$, so we must have $p \in \{ q, q+1\} $. By the same reasoning (using the fact that $C(S)$ contains $q-1$ and $S$ is not high), we conclude that $S$ contains an element of the form $(q-1, r)$ for $r \in \{ q-1, q \} $. This is a contradiction, since $S$ is linearly ordered and the ordered pairs $(p,q+1)$ and $(q-1, r)$ are incomparable in $P$.

Let $A \subseteq \operatorname{N}_{\bullet }(P_{+})$ denote the simplicial subset spanned by the simplices $\tau _{S_{+}}$, where $S_{+}$ is old. Combining $(a)$ and $(b)$, we deduce that $K'$ can be identified with the pushout

Note that, since the restriction of $f_0$ to the simplex $\operatorname{N}_{\bullet }( \{ q-1 < q < q+1 \} )$ coincides with $\sigma $, the restriction of $F'$ to the simplicial subset $\operatorname{N}_{\bullet }( P_{-}) \star \{ (q-1, q+1) \} \subseteq K$ can be identified with the diagram $\varphi : \operatorname{Tw}( \Delta ^2) \rightarrow \operatorname{\mathcal{C}}$. Let $\varphi _0$ denote the restriction of $\varphi $ to $\operatorname{N}_{\bullet }(P_-)$. Unwinding the definitions, we see that the problem of extension of $F'$ to a morphism $\operatorname{N}_{\bullet }( P_{-} ) \star \{ (q-1, q+1) \} \star \operatorname{N}_{\bullet }(P_{+} ) \rightarrow \operatorname{\mathcal{C}}$ can be rewritten as a lifting problem

This lifting problem admits a solution by virtue of our assumption that $\varphi $ is a colimit diagram in $\operatorname{\mathcal{C}}$. $\square$