We construct this simplicial object as follows. For $n \geq 0$ let $\Hom(U, V)_n$ denote the object of $\mathcal{C}$ representing the functor $$ X \longmapsto \Mor_{\text{Simp}(\mathcal{C})}(X \times U \times \Delta[n], V) $$ This exists by Lemma \ref{lemma-exists-hom-0-from-simplicial-set-finite} because $U \times \Delta[n]$ is a simplicial set with finite sets of simplices and no nondegenerate simplices in high enough degree, see Lemma \ref{lemma-product-degenerate}. For $\varphi : [m] \to [n]$ we obtain an induced map of simplicial sets $\varphi : \Delta[m] \to \Delta[n]$. Hence we obtain a morphism $X \times U \times \Delta[m] \to X \times U \times \Delta[n]$ functorial in $X$, and hence a transformation of functors, which in turn gives $$ \Hom(U, V)(\varphi) : \Hom(U, V)_n \longrightarrow \Hom(U, V)_m. $$ Clearly this defines a contravariant functor $\Hom(U, V)$ from $\Delta$ into the category $\mathcal{C}$. In other words, we have a simplicial object of $\mathcal{C}$. \medskip\noindent We have to show that $\Hom(U, V)$ satisfies the desired universal property $$ \Mor_{\text{Simp}(\mathcal{C})}(W, \Hom(U, V)) = \Mor_{\text{Simp}(\mathcal{C})}(W \times U, V) $$ To see this, let $f : W \to \Hom(U, V)$ be given. We want to construct the element $f' : W \times U \to V$ of the right hand side. By construction, each $f_n : W_n \to \Hom(U, V)_n$ corresponds to a morphism $f_n : W_n \times U \times \Delta[n] \to V$. Further, for every morphism $\varphi : [m] \to [n]$ the diagram $$ \xymatrix{ W_n \times U \times \Delta[m] \ar[rr]_{W(\varphi)\times \text{id} \times \text{id}} \ar[d]_{\text{id} \times \text{id} \times \varphi} & & W_m \times U \times \Delta[m] \ar[d]^{f_m} \\ W_n \times U \times \Delta[n] \ar[rr]^{f_n} & & V } $$ is commutative. For $\psi : [n] \to [k]$ in $(\Delta[n])_k$ we denote $(f_n)_{k, \psi} : W_n \times U_k \to V_k$ the component of $(f_n)_k$ corresponding to the element $\psi$. We define $f'_n : W_n \times U_n \to V_n$ as $f'_n = (f_n)_{n, \text{id}}$, in other words, as the restriction of $(f_n)_n : W_n \times U_n \times (\Delta[n])_n \to V_n$ to $W_n \times U_n \times \text{id}_{[n]}$. To see that the collection $(f'_n)$ defines a morphism of simplicial objects, we have to show for any $\varphi : [m] \to [n]$ that $V(\varphi) \circ f'_n = f'_m \circ W(\varphi) \times U(\varphi)$. The commutative diagram above says that $(f_n)_{m, \varphi} : W_n \times U_m \to V_m$ is equal to $(f_m)_{m, \text{id}} \circ W(\varphi) : W_n \times U_m \to V_m$. But then the fact that $f_n$ is a morphism of simplicial objects implies that the diagram $$ \xymatrix{ W_n \times U_n \times (\Delta[n])_n \ar[r]_-{(f_n)_n} \ar[d]_{\text{id} \times U(\varphi) \times \varphi} & V_n \ar[d]^{V(\varphi)} \\ W_n \times U_m \times (\Delta[n])_m \ar[r]^-{(f_n)_m} & V_m } $$ is commutative. And this implies that $(f_n)_{m, \varphi} \circ U(\varphi)$ is equal to $V(\varphi) \circ (f_n)_{n, \text{id}}$. Altogether we obtain $ V(\varphi) \circ (f_n)_{n, \text{id}} = (f_n)_{m, \varphi} \circ U(\varphi) = (f_m)_{m, \text{id}} \circ W(\varphi)\circ U(\varphi) = (f_m)_{m, \text{id}} \circ W(\varphi)\times U(\varphi) $ as desired. \medskip\noindent On the other hand, given a morphism $f' : W \times U \to V$ we define a morphism $f : W \to \Hom(U, V)$ as follows. By Lemma \ref{lemma-morphism-from-coproduct} the morphisms $\text{id} : W_n \to W_n$ corresponds to a unique morphism $c_n : W_n \times \Delta[n] \to W$. Hence we can consider the composition $$ W_n \times \Delta[n] \times U \xrightarrow{c_n} W \times U \xrightarrow{f'} V. $$ By construction this corresponds to a unique morphism $f_n : W_n \to \Hom(U, V)_n$. We leave it to the reader to see that these define a morphism of simplicial sets as desired. \medskip\noindent We also leave it to the reader to see that $f \mapsto f'$ and $f' \mapsto f$ are mutually inverse operations.