>**Puzzle 197.** Suppose \\(\mathcal{V}\\) is a closed commutative monoidal poset and \\(\mathcal{X}\\) is any \\(\mathcal{V}\\)-enriched category. Show that there is a \\(\mathcal{V}\\)-enriched functor, the **hom functor**

>\[ \mathrm{hom} \colon \mathcal{X}^{\text{op}} \times \mathcal{X} \to \mathcal{V} \]

>defined on any object \\( (x,x') \\) of \\(\mathcal{X}^{\text{op}} \times \mathcal{X} \\) by

>\[ \mathrm{hom}(x,x') = \mathcal{X}(x,x') .\]

Use the definition of a \\(\mathcal{V}\\)-enriched profunctor,

>**Definition.** Suppose \\(\mathcal{V}\\) is a closed commutative monoidal poset and \\(\mathcal{X},\mathcal{Y}\\) are \\(\mathcal{V}\\)-enriched categories. Then a \\(\mathcal{V}\\)-enriched profunctor

>\[ \Phi : \mathcal{X} \nrightarrow \mathcal{Y} \]

>is a \\(\mathcal{V}\\)-enriched functor

>\[ \Phi: \mathcal{X}^{\text{op}} \times \mathcal{Y} \to \mathcal{V} .\]

we show \\(\mathrm{hom}\\) is a \\(\mathcal{V}\\)-enriched profunctor by setting,

\\[
\Phi \mapsto \mathrm{hom},\\\\ \mathcal{X}^{\text{op}}\mapsto\mathcal{X}^{\text{op}}, \\\\
\mathcal{Y} \mapsto \mathcal{X},
\\]

then since \\(\mathrm{hom} \colon \mathcal{X}^{\text{op}} \nrightarrow \mathcal{X} \\) is a\\(\mathcal{V}\\)-enriched profunctor, then by definition of a \\(\mathcal{V}\\)-enriched profunctor \\(\mathrm{hom}\\) is also a \\(\mathcal{V}\\)-enriched functor,

\\[ \mathrm{hom} \colon \mathcal{X}^{\text{op}} \times \mathcal{X} \to \mathcal{V}. \\]

\\(\blacksquare\\)