> Sorry to be so fierce: I'm just playing the teacher, and one of the teacher's jobs is to get people to talk the official standard way, so they can communicate. I figured that if you, one of the best students, hadn't yet absorbed how to talk about enriched categories, I really need to emphasize how it's done.

Thank you for your kind words John.

I have been learning an awful lot from you and the other members on this forum.

I believe I have a handle on enriched categories now!

> If \\(\mathcal{V}\\) is discrete these inequalities are equations, so we get

>

> \[ I = \mathcal{X}(a,b) \otimes \mathcal{X}(b,a) \]

>

> which means that the subset

>

> \[ \{ \mathcal{X}(a,b) : a,b \in \mathrm{Ob}(\mathcal{X}) \} \subseteq \mathcal{V} \]

>

> is actually a group contained in the monoid \\(\mathcal{V}\\).

Not exactly.

The set \\(\\{ \mathcal{X}(a,b) : a,b \in \mathrm{Ob}(\mathcal{X}) \\}\\) may not be closed under \\(\otimes\\).

It *generates* a group, however.

Thank you for your kind words John.

I have been learning an awful lot from you and the other members on this forum.

I believe I have a handle on enriched categories now!

> If \\(\mathcal{V}\\) is discrete these inequalities are equations, so we get

>

> \[ I = \mathcal{X}(a,b) \otimes \mathcal{X}(b,a) \]

>

> which means that the subset

>

> \[ \{ \mathcal{X}(a,b) : a,b \in \mathrm{Ob}(\mathcal{X}) \} \subseteq \mathcal{V} \]

>

> is actually a group contained in the monoid \\(\mathcal{V}\\).

Not exactly.

The set \\(\\{ \mathcal{X}(a,b) : a,b \in \mathrm{Ob}(\mathcal{X}) \\}\\) may not be closed under \\(\otimes\\).

It *generates* a group, however.