Lemma 9.1.2.11. Let $\kappa $ be an infinite cardinal, let $\operatorname{\mathcal{C}}$ be a $\kappa $-filtered $\infty $-category containing objects $X$ and $Y$, and let $K$ be a $\kappa $-small simplicial set equipped with a morphism $f: K \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$. Then there exists a morphism $v: Y \rightarrow Z$ of $\operatorname{\mathcal{C}}$ for which the composition $K \xrightarrow {f} \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) \xrightarrow { v \circ } \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Z)$ is nullhomotopic.
Proof. Let $\Sigma (K)$ denote the iterated coproduct
so that we can identify $f$ with a morphism of simplicial sets $F: \Sigma (K) \rightarrow \operatorname{\mathcal{C}}$ satisfying $F(x) = X$ and $F(y) = Y$. Note that $\Sigma (K)$ is $\kappa $-small, so our assumption that $\operatorname{\mathcal{C}}$ is $\kappa $-filtered guarantees that we can extend $F$ to a morphism of simplicial set $\overline{F}: \Sigma (K) \star \{ z\} \rightarrow \operatorname{\mathcal{C}}$. Set $Z = \overline{F}(z)$. Then $\overline{F}$ carries $\{ x\} \star \{ z\} $ and $\{ y\} \star \{ z\} $ to morphisms $u: X \rightarrow Z$ and $v: Y \rightarrow Z$ in $\operatorname{\mathcal{C}}$. Moreover, the natural map $\Delta ^1 \times K \rightarrow \Sigma (K)$ admits a unique extension $q: \Delta ^2 \times K \rightarrow \Sigma (K) \star \{ z\} $ carrying $\{ 2\} \times K$ to the vertex $z$, and the composition
determines a morphism from $K$ to the simplicial set $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y,Z)$ of Construction 4.6.9.9. Unwinding the definitions, we see that the diagram of simplicial sets
is strictly commutative, from which we immediately deduce (from the definition of the composition law on $\operatorname{\mathcal{C}}$) that the composition $K \xrightarrow {f} \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y) \xrightarrow {v \circ } \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Z)$ is homotopic to the constant map taking the value $u$. $\square$