Proof.
Assume that $(2)$ is satisfied; we will prove $(1)$ (the reverse implication is immediate from the definitions). Let $X$ be a small Kan complex; we wish to show composition with $\beta $ induces a homotopy equivalence $\theta _{X}: \operatorname{Hom}_{\operatorname{\mathcal{S}}}( X, Y ) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{S}}) }( \underline{X}_{\operatorname{\mathcal{C}}}, \mathscr {F} )$. Choose a simplicial set $K$ equipped with a weak homotopy equivalence $u: K \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{S}}}( \Delta ^0, X )$ (for example, we can take $K = X$ and $u$ to be the homotopy equivalence of Remark 5.5.1.5). It follows from Example 7.1.2.10 that $u$ exhibits $X$ as a copower of $\Delta ^0$ by $K$ in the $\infty $-category $\operatorname{\mathcal{S}}$, so that the composite map
\[ \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X, Y ) \times K \xrightarrow {\operatorname{id}\times u} \operatorname{Hom}_{\operatorname{\mathcal{S}}}(X,Y) \times \operatorname{Hom}_{\operatorname{\mathcal{S}}}( \Delta ^0,X) \xrightarrow {\circ } \operatorname{Hom}_{\operatorname{\mathcal{C}}}( \Delta ^0, Y) \]
determines a homotopy equivalence $T: \operatorname{Hom}_{\operatorname{\mathcal{S}}}(X,Y) \rightarrow \operatorname{Fun}(K, \operatorname{Hom}_{\operatorname{\mathcal{S}}}( \Delta ^0, Y) )$. Similarly, the composite map
\[ K \xrightarrow {u} \operatorname{Hom}_{\operatorname{\mathcal{S}}}( \Delta ^0, X) \rightarrow \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{S}})}( \underline{ \Delta ^0}_{\operatorname{\mathcal{C}}}, \underline{X}_{\operatorname{\mathcal{C}}} ) \]
exhibits $\underline{X}_{\operatorname{\mathcal{C}}}$ as a copower of $\underline{ \Delta ^0 }_{\operatorname{\mathcal{C}}}$ by $K$ in the $\infty $-category $\operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{S}})$ (see Proposition 7.1.7.3), so that $u$ also determines a homotopy equivalence $T': \operatorname{Hom}_{\operatorname{Fun}(\operatorname{\mathcal{C}},\operatorname{\mathcal{S}})}( \underline{X}_{\operatorname{\mathcal{C}}}, \mathscr {F} ) \rightarrow \operatorname{Fun}(K, \operatorname{Hom}_{ \operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{S}}) }( \underline{ \Delta ^0 }_{\operatorname{\mathcal{C}}}, \mathscr {F} )$. Note that we have a homotopy commutative diagram of Kan complexes
\[ \xymatrix { \operatorname{Hom}_{\operatorname{\mathcal{S}}}(X, Y) \ar [d]^{T}_{\sim } \ar [r]^{\theta _{X}} & \operatorname{Hom}_{\operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{S}})}( \underline{X}_{\operatorname{\mathcal{C}}}, \mathscr {F} ) \ar [d]^{T'}_{\sim } \\ \operatorname{Fun}(K, \operatorname{Hom}_{\operatorname{\mathcal{S}}}( \Delta ^0, Y) ) \ar [r] & \operatorname{Fun}(K, \operatorname{Hom}_{\operatorname{Fun}(\operatorname{\mathcal{C}}, \operatorname{\mathcal{S}})}( \underline{\Delta ^0}_{\operatorname{\mathcal{C}}}, \mathscr {F} ) } \]
where the bottom horizontal map is obtained by applying the functor $\operatorname{Fun}(K, \bullet )$ to $\theta $ and is therefore a homotopy equivalence by virtue of assumption $(2)$. It follows that $\theta _{X}$ is also a homotopy equivalence.
$\square$