Thanks a lot, [Matthew](https://forum.azimuthproject.org/discussion/comment/18702/#Comment_18702) and [Jonathan](https://forum.azimuthproject.org/discussion/comment/18703/#Comment_18703)! That makes a lot of sense to me. Two follow-up questions:

First, Matthew said
> Enriched categories substitute the sets in hom-sets with an arbitrary category. Regular categories can be see as Set-enriched categories, where Set is the category of sets.
> In order to make this happen, you need two thing: composition of arrows and identity morphisms.

Does this mean the reason why we need the two requirements when substituting the hom-set with an arbitrary category is because we need to make such extra effort to guarantee what we end up with after the substitution is still a category (I mean, still an "enriched category")? If so, what would the substitution yield if e.g. the arbitrary category we use is _nothing_ (such that we effectively leave no enrichment base at all, not even **Set**)?

Second, Jonathan said
>This angle also explains \$$I\leq\mathcal{X}(x,x)\$$: whatever \$$I\$$ is, \$$\mathcal{X}(x, x)\$$ better be _at least_ that much.

I managed to understand the adoption of "≤" in \$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \$$ via this angle, but am still not sure I fully grasp its necessity in \$$I\leq\mathcal{X}(x,x)\$$ - I (think I) see that \$$\mathcal{X}(x,x)\$$ need not be exactly \$$I\$$, but why must it be _at least_ that much? What would happen if \$$I>\mathcal{X}(x,x)\$$?