/sci/ - Science & Math

>>15186607
f(x) <= g(x) for all x, because look.

With more abstract nonsense:

Posets are equivalent to (0,1)-categories, and (n,r)-categories form a (n+1,k+1)-category, so posets form a (1,2)-category.

A (n,r)-category is such that j-morphisms, if j > r, can only be equivalences, and if j > n, are unique up to equivalence if they exist.

The equivalent of natural transformations in the (1,2)-category of posets would be 2-morphisms, and since 2 > 1, they are unique up to equivalence, so we can expect a poset between any two compatible monotonic functions, just as we have a category between any two compatible functors.