[Xah wrote:](https://forum.azimuthproject.org/discussion/comment/16166/#Comment_16166)

> So, if S1 → S2, we can think of it as S1 ≥ S2.

Yes, you can do that. But notice that I'm not doing that! I'm saying that if that given a category with a morphism \\(f : S_1 \to S_2\\) , I can create a preorder with \\(S_1 \le S_2\\).

Why \\(\le\\) instead of \\(\ge\\)? It's an arbitrary choice:if you have a preorder where you say \\\(x \le_{\textrm{Xah}} y\\), I can make up a preorder where I say that \\( x \ge_{\textrm{John}} y \\). This is called the **opposite** preorder.

But there's a reason I'm making my choice! It's very nice to say that the inclusion of sets

{1,2,3} \\(\to\\) {1,2,3,4,5,6} is another way of saying \\( 3 \le 6 \\). So, everyone in category theory does it the way I'm doing it.

> So, if S1 → S2, we can think of it as S1 ≥ S2.

Yes, you can do that. But notice that I'm not doing that! I'm saying that if that given a category with a morphism \\(f : S_1 \to S_2\\) , I can create a preorder with \\(S_1 \le S_2\\).

Why \\(\le\\) instead of \\(\ge\\)? It's an arbitrary choice:if you have a preorder where you say \\\(x \le_{\textrm{Xah}} y\\), I can make up a preorder where I say that \\( x \ge_{\textrm{John}} y \\). This is called the **opposite** preorder.

But there's a reason I'm making my choice! It's very nice to say that the inclusion of sets

{1,2,3} \\(\to\\) {1,2,3,4,5,6} is another way of saying \\( 3 \le 6 \\). So, everyone in category theory does it the way I'm doing it.