I think there's a mistake in the proof that \\(\mathcal{X}^\text{op}\\) is a \\(\mathcal{V}\\)-enriched category:

> b) We also need to check that for all objects \\(x,x',x'' \\) of \\(\mathcal{X}^{\text{op}} \\) we have

>

> \[ \mathcal{X}^{\text{op}} (x,x') \otimes \mathcal{X}^{\text{op}}(x',x'') \le \mathcal{X}^{\text{op}}(x,x'') . \]

>

> Using the definitions, this just says that for all objects \\(x,x',x'' \\) of \\(\mathcal{X}\\) we have

>

> \[ \mathcal{X}(x'',x') \otimes \mathcal{X}(x',x) \le \mathcal{X}(x'',x) . \]

>

> This is true because \\(\mathcal{X}\\) is an enriched category. \\( \qquad \blacksquare \\).

>

> I just learned something: I thought we needed \\(\mathcal{V}\\) to be commutative here, but apparently we don't.

Surely the definitions tell us that

\[ \mathcal{X}^{\text{op}} (x,x') \otimes \mathcal{X}^{\text{op}}(x',x'') = \mathcal{X}(x',x) \otimes \mathcal{X}(x'',x') \]

So we _do_ need \\(\mathcal{V}\\) to be commutative to switch this to \\(\mathcal{X}(x'',x') \otimes \mathcal{X}(x',x)\\).

In the absence of commutativity/symmetry, all we can prove is that \\(\mathcal{X}^\text{op}\\) is a \\(\mathcal{V}^\text{op}\\)-enriched category, where \\(\mathcal{V}^\text{op}\\) is \\(\mathcal{V}\\) with the same partial order but the opposite monoidal multiplication, ie \\(u \otimes_{\mathcal{V}^\text{op}} v = v \otimes_\mathcal{V} u\\).

> b) We also need to check that for all objects \\(x,x',x'' \\) of \\(\mathcal{X}^{\text{op}} \\) we have

>

> \[ \mathcal{X}^{\text{op}} (x,x') \otimes \mathcal{X}^{\text{op}}(x',x'') \le \mathcal{X}^{\text{op}}(x,x'') . \]

>

> Using the definitions, this just says that for all objects \\(x,x',x'' \\) of \\(\mathcal{X}\\) we have

>

> \[ \mathcal{X}(x'',x') \otimes \mathcal{X}(x',x) \le \mathcal{X}(x'',x) . \]

>

> This is true because \\(\mathcal{X}\\) is an enriched category. \\( \qquad \blacksquare \\).

>

> I just learned something: I thought we needed \\(\mathcal{V}\\) to be commutative here, but apparently we don't.

Surely the definitions tell us that

\[ \mathcal{X}^{\text{op}} (x,x') \otimes \mathcal{X}^{\text{op}}(x',x'') = \mathcal{X}(x',x) \otimes \mathcal{X}(x'',x') \]

So we _do_ need \\(\mathcal{V}\\) to be commutative to switch this to \\(\mathcal{X}(x'',x') \otimes \mathcal{X}(x',x)\\).

In the absence of commutativity/symmetry, all we can prove is that \\(\mathcal{X}^\text{op}\\) is a \\(\mathcal{V}^\text{op}\\)-enriched category, where \\(\mathcal{V}^\text{op}\\) is \\(\mathcal{V}\\) with the same partial order but the opposite monoidal multiplication, ie \\(u \otimes_{\mathcal{V}^\text{op}} v = v \otimes_\mathcal{V} u\\).