In [comment #88](https://forum.azimuthproject.org/discussion/comment/18708/#Comment_18708) Julio wrote:

> 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)\$$.

As [Simon pointed out](https://forum.azimuthproject.org/discussion/comment/18711/#Comment_18711), there's no "necessity" in math: we make up axioms that handle examples we're interested in, and give nice theorems. If we want to study examples that don't obey those axioms, we make up other axioms, and hope those also give nice theorems.

We get lots of nice examples of enriched categories, and lots of nice theorems about enriched categories, if we assume

$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)$

and

$I\leq\mathcal{X}(x,x) .$

You should look at all the examples and theorems I've stated — and others in the book — and see why these axioms are true in the examples, and how we use them in the theorems. You'll see these axioms work very well!

If we turn around _both) inequalities, and assume

$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\geq\mathcal{X}(x,z)$

and

$I\geq\mathcal{X}(x,x)$

everything works just as well! The reason is that every preorder has an "opposite", where we define inequality the opposite way. It's just an arbitrary convention whether we write \$$\le\$$ or \$$\ge\$$. It's like deciding to drive on the left or on the right of the road: neither one is better, but you have collisions if people don't agree on a choice!

If you want to explore some new territory, see what happens if you turn around just _one_ inequality. For example, assume

$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z)$

and

$I\geq\mathcal{X}(x,x) .$

I believe this territory will be an ugly swamp. But I haven't explored it, so I could be wrong!

Ultimately the answer to your question may be "wait a bit, and learn more!" For example, [Matthew](https://forum.azimuthproject.org/discussion/comment/18702/#Comment_18702) gave you a great answer. However, his answer only makes sense if you know about categories enriched in _monoidal categories_, not just in monoidal preorders. He wrote approximately this:

> In category theory, when we have two objects \$$X\$$ and \$$Y\$$ in a category we often care about the set of morphisms between them. This is the [*hom-set*](https://en.wikipedia.org/wiki/Morphism#Hom-set) \$$\mathrm{Hom}(X,Y)\$$.

> While this is a powerful generalization, it can be more general. Why do we demand the morphisms between \$$X\$$ and \$$Y\$$ be a set? Why not \$$\\{\mathtt{false}, \mathtt{true}\\}\$$, or a number, or a member of a group?

> _Enriched categories substitute the sets in hom-sets with objects in an arbitrary monoidal category_. Ordinary categories can be seen as **Set**-enriched categories, where **Set** is the category of sets.

> In order to make this happen, you need two things: composition of arrows and identity morphisms.

> \$$\mathcal{X}(x,y)\otimes\mathcal{X}(y,z)\leq\mathcal{X}(x,z) \$$ ensures morphisms are closed under composition.

> \$$I\leq\mathcal{X}(x,x) \$$ effectively ensures the presence of identity morphisms.

If this doesn't make sense now, it should later. You can hold on to this as a future goal.