Proposition 2.4.5.9. Let $\operatorname{\mathcal{E}}_{\bullet }$ be a simplicial category containing a pair of objects $x,y \in \operatorname{Ob}( \operatorname{\mathcal{E}}_{\bullet } )$. Assume that, for each object $z \in \operatorname{Ob}( \operatorname{\mathcal{E}}_{\bullet } )$, we have
\[ \operatorname{Hom}_{\operatorname{\mathcal{E}}}( z, x)_{\bullet } = \begin{cases} \{ \operatorname{id}_ x \} & \text{ if } z = x \\ \emptyset & \text{ otherwise. } \end{cases} \quad \quad \operatorname{Hom}_{\operatorname{\mathcal{E}}}( y, z)_{\bullet } = \begin{cases} \{ \operatorname{id}_ y \} & \text{ if } z = y \\ \emptyset & \text{ otherwise. } \end{cases} \]
Let $\operatorname{\mathcal{D}}_{\bullet } \subseteq \operatorname{\mathcal{E}}_{\bullet }$ denote a simplicial subcategory having the same objects, which satisfies
\[ \operatorname{Hom}_{\operatorname{\mathcal{D}}}( a, b )_{\bullet } = \operatorname{Hom}_{\operatorname{\mathcal{E}}}( a, b )_{\bullet } \]
unless $(a,b) = (x,y)$. Let $F: \operatorname{\mathcal{D}}_{\bullet } \rightarrow \operatorname{\mathcal{C}}_{\bullet }$ be a functor of simplicial categories carrying $x$ to an object $X = F(x)$ and $y$ to an object $Y \in F(y)$, so that $F$ induces a map of simplicial sets $F_{x,y}: \operatorname{Hom}_{\operatorname{\mathcal{D}}}(x,y)_{\bullet } \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{\bullet }$. Then the construction $\overline{F} \mapsto \overline{F}_{x,y}$ induces a bijection
\[ \xymatrix@R =50pt@C=50pt{ \{ \textnormal{Simplicial functors $\overline{F}: \operatorname{\mathcal{E}}_{\bullet } \rightarrow \operatorname{\mathcal{C}}_{\bullet }$ extending $F$} \} \ar [d]^{\sim } \\ \{ \textnormal{Maps $\lambda : \operatorname{Hom}_{\operatorname{\mathcal{E}}}(x,y)_{\bullet } \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{\bullet }$ extending $F_{x,y}$} \} . } \]
Proof.
Fix a map of simplicial sets $\lambda : \operatorname{Hom}_{\operatorname{\mathcal{E}}}(x,y)_{\bullet } \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{\bullet }$ which extends $F_{x,y}$. We wish to show that there is a unique simplicial functor $\overline{F}: \operatorname{\mathcal{E}}_{\bullet } \rightarrow \operatorname{\mathcal{C}}_{\bullet }$ such that $F = \overline{F}|_{\operatorname{\mathcal{D}}_{\bullet }}$ and $\overline{F}_{x,y} = \lambda $. The uniqueness is clear: the simplicial functor $\overline{F}$ must coincide with $F$ on objects and satisfy $\overline{F}_{x',y'} = F_{x',y'}$ for $(x',y') \neq (x,y)$. To prove existence, one must show that this prescription defines a simplicial functor: that is, that for every triple of objects $a,b,c \in \operatorname{Ob}(\operatorname{\mathcal{E}}_{\bullet })$, the resulting diagram of simplicial sets
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\operatorname{\mathcal{E}}}(b,c)_{\bullet } \times \operatorname{Hom}_{\operatorname{\mathcal{E}}}(a,b)_{\bullet } \ar [r] \ar [d]^{ \overline{F}_{a,b} \otimes \overline{F}_{b,c} } & \operatorname{Hom}_{\operatorname{\mathcal{E}}}(a,c)_{\bullet } \ar [d]^{ \overline{F}_{a,c} } \\ \operatorname{Hom}_{\operatorname{\mathcal{C}}}( F(b), F(c) )_{\bullet } \times \operatorname{Hom}_{\operatorname{\mathcal{C}}}( F(a), F(b) )_{\bullet } \ar [r] & \operatorname{Hom}_{\operatorname{\mathcal{C}}}( F(a), F(c) )_{\bullet } } \]
is commutative. We consider several cases:
Suppose that $(a,b) = (x,y)$. If $c \neq y$, then the simplicial set $\operatorname{Hom}_{\operatorname{\mathcal{E}}}(b,c)_{\bullet }$ is empty and the commutativity of the diagram is automatic. If $c = y$, then both compositions can be identified with the map
\[ \{ \operatorname{id}_ y \} \times \operatorname{Hom}_{\operatorname{\mathcal{E}}}(x,y)_{\bullet } \simeq \operatorname{Hom}_{\operatorname{\mathcal{E}}}(x,y)_{\bullet } \xrightarrow {\lambda } \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{\bullet }. \]
Suppose that $(b,c) = (x,y)$. If $a \neq x$, then the simplicial set $\operatorname{Hom}_{\operatorname{\mathcal{E}}}(a,b)_{\bullet }$ is empty and the commutativity of the diagram is automatic. If $a = x$, then both compositions can be identified with the map
\[ \operatorname{Hom}_{\operatorname{\mathcal{E}}}(x,y)_{\bullet } \times \{ \operatorname{id}_ x \} \simeq \operatorname{Hom}_{\operatorname{\mathcal{E}}}(x,y)_{\bullet } \xrightarrow {\lambda } \operatorname{Hom}_{\operatorname{\mathcal{C}}}( X, Y)_{\bullet }. \]
If neither $(a,b) = (x,y)$ or $(b,c) = (x,y)$, then the desired result follows from the commutativity of the diagram
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\operatorname{\mathcal{D}}}(b,c)_{\bullet } \times \operatorname{Hom}_{\operatorname{\mathcal{D}}}(a,b)_{\bullet } \ar [r] \ar [d]^{ F_{a,b} \otimes F_{b,c} } & \operatorname{Hom}_{\operatorname{\mathcal{D}}}(a,c)_{\bullet } \ar [d]^{ F_{a,c} } \\ \operatorname{Hom}_{\operatorname{\mathcal{C}}}( F(b), F(c) )_{\bullet } \times \operatorname{Hom}_{\operatorname{\mathcal{C}}}( F(a), F(b) )_{\bullet } \ar [r] & \operatorname{Hom}_{\operatorname{\mathcal{C}}}( F(a), F(c) )_{\bullet } } \]
(since $F$ is assumed to be a simplicial functor).
$\square$