Lemma 8.1.7.12. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category, let $L$ and $R$ be collections of morphisms of $\operatorname{\mathcal{C}}$ which have the left two-out-of-three property, and let $\operatorname{Fun}'( \operatorname{Tw}(\operatorname{\mathcal{D}}), \operatorname{\mathcal{C}})$ denote the full subcategory of $\operatorname{Fun}( \operatorname{Tw}(\operatorname{\mathcal{D}}), \operatorname{\mathcal{C}})$ spanned by those objects which correspond to diagrams $\operatorname{\mathcal{D}}\rightarrow \operatorname{Cospan}^{L,R}(\operatorname{\mathcal{C}}) \subseteq \operatorname{Cospan}(\operatorname{\mathcal{C}})$. Let $R'$ be the collection of morphisms in $\operatorname{Fun}'( \operatorname{Tw}(\operatorname{\mathcal{D}}), \operatorname{\mathcal{C}})$ which satisfy the equivalent conditions of Remark 8.1.7.11, and define $L'$ similarly. Then the isomorphism $\operatorname{Fun}( \operatorname{\mathcal{D}}, \operatorname{Cospan}( \operatorname{\mathcal{C}}) ) \simeq \operatorname{Cospan}( \operatorname{Fun}(\operatorname{Tw}(\operatorname{\mathcal{D}}), \operatorname{\mathcal{C}}) )$ of Remark 8.1.3.9 restricts to an isomorphism of simplicial subsets $\operatorname{Fun}(\operatorname{\mathcal{D}}, \operatorname{Cospan}^{L,R}( \operatorname{\mathcal{C}}) ) \simeq \operatorname{Cospan}^{L',R'}( \operatorname{Fun}'( \operatorname{Tw}(\operatorname{\mathcal{D}}), \operatorname{\mathcal{C}}) )$.
Proof. Writing $\operatorname{Cospan}^{L,R}(\operatorname{\mathcal{C}})$ as the intersection $\operatorname{Cospan}^{L, \mathrm{all}}(\operatorname{\mathcal{C}}) \cap \operatorname{Cospan}^{\mathrm{all}, R}( \operatorname{\mathcal{C}})$ (see Variant 8.1.7.4), we can reduce to the case where either $L$ or $R$ is the collection of all morphisms of $\operatorname{\mathcal{C}}$. Let us assume that $L$ is the collection of all morphisms of $\operatorname{\mathcal{C}}$, so that $L'$ is the collection of all morphisms of $\operatorname{Fun}'( \operatorname{Tw}(\operatorname{\mathcal{D}}), \operatorname{\mathcal{C}})$. Suppose we are given another simplicial set $\operatorname{\mathcal{E}}$ and a diagram $F: \operatorname{Tw}(\operatorname{\mathcal{D}}) \times \operatorname{Tw}(\operatorname{\mathcal{E}}) \rightarrow \operatorname{\mathcal{C}}$. We can identify $F$ with a morphism of simplicial sets $\operatorname{\mathcal{E}}\rightarrow \operatorname{Fun}( \operatorname{\mathcal{D}}, \operatorname{Cospan}( \operatorname{\mathcal{C}}) )$. By virtue of Remark 8.1.6.3, this morphism factors through the simplicial subset $\operatorname{Fun}( \operatorname{\mathcal{D}}, \operatorname{Cospan}^{L,R}(\operatorname{\mathcal{C}}) )$ if and only if, for every edge $u: D' \rightarrow D$ of $\operatorname{\mathcal{D}}$ and every edge $v: E' \rightarrow E$ of $\operatorname{\mathcal{E}}$, the morphism $F$ satisfies the following condition:
- $(1_{u,v})$
Let $u_{R}: \operatorname{id}_{D} \rightarrow u$ and $v_{R}: \operatorname{id}_{E} \rightarrow v$ be the edges of $\operatorname{Tw}(\operatorname{\mathcal{D}})$ and $\operatorname{Tw}(\operatorname{\mathcal{E}})$ described in Example 8.1.3.6. Then the morphism $F(u_{R}, v_{R} )$ belongs to $R$.
Identifying $F$ with a morphism $f: \operatorname{\mathcal{E}}\rightarrow \operatorname{Cospan}( \operatorname{Fun}( \operatorname{Tw}(\operatorname{\mathcal{D}}), \operatorname{\mathcal{C}}) )$, we see that $f$ factors through $\operatorname{Cospan}( \operatorname{Fun}'( \operatorname{Tw}(\operatorname{\mathcal{D}}), \operatorname{\mathcal{C}}) )$ if and only if it satisfies condition $(1_{u,v} )$ whenever $v$ is a degenerate edge of $\operatorname{\mathcal{E}}$. Under this assumption, $f$ factors through $\operatorname{Cospan}^{L',R'}( \operatorname{Fun}'( \operatorname{Tw}(\operatorname{\mathcal{D}}), \operatorname{\mathcal{C}}) )$ if and only if, for every edge $u: D' \rightarrow D$ of $\operatorname{\mathcal{D}}$ and every edge $v: E' \rightarrow E$ of $\operatorname{\mathcal{E}}$, the diagram $F$ satisfies the following condition:
- $(2_{u,v})$
The morphism $F( \operatorname{id}_{u}, v )$ belongs to $R$.
To complete the proof, it suffices to observe that if condition $(1_{ u, \operatorname{id}_{E} } )$ is satisfied, then condition $(1_{u,v} )$ is equivalent to condition $(2_{u,v} )$. This follows by applying the left two-out-of-three property to the upper triangle appearing in the diagram