>**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\$$