Proof.
We construct this simplicial object as follows. For $n \geq 0$ let $\mathop{\mathrm{Hom}}\nolimits (U, V)_ n$ denote the object of $\mathcal{C}$ representing the functor
\[ X \longmapsto \mathop{\mathrm{Mor}}\nolimits _{\text{Simp}(\mathcal{C})}(X \times U \times \Delta [n], V) \]
This exists by Lemma 14.17.3 because $U \times \Delta [n]$ is a simplicial set with finite sets of simplices and no nondegenerate simplices in high enough degree, see Lemma 14.11.5. 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
\[ \mathop{\mathrm{Hom}}\nolimits (U, V)(\varphi ) : \mathop{\mathrm{Hom}}\nolimits (U, V)_ n \longrightarrow \mathop{\mathrm{Hom}}\nolimits (U, V)_ m. \]
Clearly this defines a contravariant functor $\mathop{\mathrm{Hom}}\nolimits (U, V)$ from $\Delta $ into the category $\mathcal{C}$. In other words, we have a simplicial object of $\mathcal{C}$.
We have to show that $\mathop{\mathrm{Hom}}\nolimits (U, V)$ satisfies the desired universal property
\[ \mathop{\mathrm{Mor}}\nolimits _{\text{Simp}(\mathcal{C})}(W, \mathop{\mathrm{Hom}}\nolimits (U, V)) = \mathop{\mathrm{Mor}}\nolimits _{\text{Simp}(\mathcal{C})}(W \times U, V) \]
To see this, let $f : W \to \mathop{\mathrm{Hom}}\nolimits (U, V)$ be given. We want to construct the element $f' : W \times U \to V$ of the right hand side. By construction, each $f_ n : W_ n \to \mathop{\mathrm{Hom}}\nolimits (U, V)_ n$ 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.
On the other hand, given a morphism $f' : W \times U \to V$ we define a morphism $f : W \to \mathop{\mathrm{Hom}}\nolimits (U, V)$ as follows. By Lemma 14.13.4 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 $f_ n : W_ n \to \mathop{\mathrm{Hom}}\nolimits (U, V)_ n$. We leave it to the reader to see that these define a morphism of simplicial sets as desired.
We also leave it to the reader to see that $f \mapsto f'$ and $f' \mapsto f$ are mutually inverse operations.
$\square$
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