>**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.