(**Edit**: I had originally tried to prove this by assuming \\(x \leq y \iff x = y\\). I have updated this proof to fully answer the puzzle based on Anindya's comments below.)

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


Going back to [Lecture 32](https://forum.azimuthproject.org/discussion/2169/lecture-32-chapter-2-enriched-functors), we have the following definition of a \\(\mathcal{V}\\)-functor.

> **Definition.** Let \\(\mathcal{X}\\) and \\(\mathcal{Y}\\) be \\(\mathcal{V}\\)-categories. A **\\(\mathcal{V}\\)-functor from \\(\mathcal{X}\\) to \\(\mathcal{Y}\\)**, denoted \\(F\colon\mathcal{X}\to\mathcal{Y}\\), is a function
> \[ F\colon\mathrm{Ob}(\mathcal{X})\to \mathrm{Ob}(\mathcal{Y})\]
> such that
> \[ \mathcal{X}(x,x') \leq \mathcal{Y}(F(x),F(x')) \tag{X} \]
> for all \\(x,x' \in\mathrm{Ob}(\mathcal{X})\\).

We want \\(F = \mathrm{hom}\\) and \\(\mathcal{Y} = \mathcal{V}\\).

Based on [Lecture 60](https://forum.azimuthproject.org/discussion/2287/lecture-60-chapter-4-closed-monoidal-posets), to make \\(\mathrm{hom}\\) a \\(\mathcal{V}\\)-enriched functor we want to make the action of \\(\mathcal{Y}\\) on objects be \\((a,b) \mapsto a \multimap b \\).

All that is left is to prove \\(\mathrm{hom}\\) satisfies the rules of a \\(\mathcal{V}\\)-enriched functor.

To see \\(\mathrm{hom} = \mathcal{X}\\) works, we know:

\[ (\mathcal{X}^{\text{op}} \times \mathcal{X})((a,b),(c,d)) = \mathcal{X}(c,a) \otimes \mathcal{X}(b,d) \]

Hence, using commutativity and the fact that \\(A \otimes - \dashv A \multimap - \\), then \\(\text{(X)}\\) says that \\(\mathrm{hom}\\) is a \\(\mathcal{V}\\)-Functor if and only if:

& (\mathcal{X}^{\text{op}} \times \mathcal{X})((a,b),(c,d)) \leq \mathrm{hom}(a,b) \multimap \mathrm{hom}(c,d) \\\\
& \iff \\\\
& \mathcal{X}(c,a) \otimes \mathcal{X}(b,d) \leq \mathrm{hom}(a,b) \multimap \mathrm{hom}(c,d) \\\\
& \iff \\\\
& \mathrm{hom}(a,b) \otimes \mathcal{X}(c,a) \otimes \mathcal{X}(b,d) \leq \mathrm{hom}(c,d) \\\\
& \iff \\\\
& \mathcal{X}(a,b) \otimes \mathcal{X}(c,a) \otimes \mathcal{X}(b,d) \leq \mathcal{X}(c,d) \\\\
& \iff \\\\
& \mathcal{X}(c,a) \otimes \mathcal{X}(a,b) \otimes \mathcal{X}(b,d) \leq \mathcal{X}(c,d) \tag{Y}

So it suffices to show \\(\text{(Y)}\\). To do this, observe:

1. \\(- \otimes \mathcal{X}(x,y)\\) is monotone for all \\(x\\) and \\(y\\), since \\(\mathcal{V}\\) is a monoidal pre-order as defined in [Lecture 21](https://forum.azimuthproject.org/discussion/2082/lecture-21-chapter-2-monoidal-preorders)
2. \\(\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)\\); this is from condition (b) of the definition of a \\(\mathcal{V}\\)-enriched category in [Lecture 29](https://forum.azimuthproject.org/discussion/2121/lecture-29-chapter-2-enriched-categories)

From (2) we have

\[ \mathcal{X}(c,a) \otimes \mathcal{X}(a,b) \leq \mathcal{X}(c,b)\]

This with (1) means

\[ \mathcal{X}(c,a) \otimes \mathcal{X}(a,b) \otimes \mathcal{X}(b,d) \leq \mathcal{X}(c,b) \otimes \mathcal{X}(b,d) \tag{★} \]

But we also know from (2) that

\[ \mathcal{X}(c,b) \otimes \mathcal{X}(b,d) \leq \mathcal{X}(c,b)\]

This along with (★) and transitivity shows \\(\mathcal{X}(c,a) \otimes \mathcal{X}(a,b) \otimes \mathcal{X}(b,d) \leq \mathcal{X}(c,d)\\) as desired.
\\( \qquad \blacksquare \\)