>**Puzzle 225.** Suppose \$$\mathcal{X}\$$ is a \$$\mathcal{V}\$$-enriched categories. Show there are \$$\mathcal{V}\$$-enriched profunctors

>$\lambda_{\mathcal{X}} \colon \mathbf{1} \otimes \mathcal{X} \nrightarrow \mathcal{X}$

>and

>$\rho_{\mathcal{X}} \colon \mathcal{X} \otimes \mathbf{1} \nrightarrow \mathcal{X}$

>both of which have inverses.

Assume \$$a \leq x\$$ and \$$x' \leq a'\$$.

$(\mathbf{1} \otimes \mathcal{X})(\langle \cdot, a \rangle),\langle \cdot, x \rangle) \otimes \lambda_{\mathcal{X}}(\langle \cdot, x \rangle, \langle \cdot, x' \rangle) \otimes \mathcal{X}(x',a')$

$= \mathbf{1}(\cdot,\cdot) \otimes \mathcal{X}(a,x) \otimes \mathbf{1}(\cdot,\cdot) \otimes \mathcal{X}(x,x') \otimes \mathcal{X}(x',a')$

$= \mathbf{1}(\cdot,\cdot) \otimes \mathbf{1}(\cdot,\cdot) \otimes \mathcal{X}(a,x) \otimes \mathcal{X}(x,x') \otimes \mathcal{X}(x',a')$

$\leq \mathbf{1}(\cdot,\cdot) \otimes \mathcal{X}(a,a')$

$= \lambda_{\mathcal{X}}(\langle \cdot, a \rangle, \langle \cdot, a' \rangle)$

The symmetry is pretty obvious in this one so proof for inverses and \$$\rho\$$ is almost identical.