Functors and hom class comparisons

Let $B,C$ be elements of a partially ordered set such that $B \subseteq C$. Then if we have an element $A \subseteq B$ we get $A \subseteq C$ or if we have an element $C \subseteq D$ then we can get $B \subseteq D$. This demonstrates that we can extend an interval in two ways: by making the predecessor smaller or making the successor larger. We will demonstrate an analagous concept exists in terms of the images of hom classes produced by functors.

Let $C$ and $D$ be categories, and $F : C \to D$ be a functor. Then we have the following I/O relationship pertaining to the morphism part of the functor: the hom class of the input morphism determines the hom class of the output morphism. \[ f : Mor_C(a,b) \to Mor_D(F(a), F(b)) \] Let $S$ be the class of all ordered pairs of objects $Ob(C)^2$. Then $f$ induces an equivalence relation $E$ on the ordered pairs of objects so that $E \subseteq (Ob(C)^2)^2$. Two ordered pairs of objects of objects in $C$ are equal if their hom classes are mapped to the same hom class in the output category by the action of the functor. This naturally leads to the hom class comparison preorder of a functor.

Definition. let $F: C \to D$ be a functor, $E$ be the equivalence relation induced by the functor on ordered pairs of objects, and $C$ an equivalence class in $E$. Then the hom class comparison preorder defined on $C$ is defined so that $(a,b) \leq (c,d)$ when $F(Mor_C(a,b)) \leq F(Mor_C(c,d))$.

It is immediate that the hom class comparison preorders of a functor are trivial for any functor which is object mono. The functor that maps any category to its underlying preorder by the hom class congruence, for example, is hom class comparison trivial because it is object mono. Therefore, from now on we can implicitly assume that $F$ is a faithful set-valued functor associated to a concrete category.

The sense that a structure preserving map is an inclusion can now be described functorially. Let $C$ be the category of preorders and monotone maps and suppose that $f : (A,\sqsubseteq) \to (B,\sqsubseteq)$ is a monotone map. Then if we make $(A,\sqsubseteq)$ a smaller preordering or we make $(B,\sqsubseteq)$ a larger preordering then the underlying function is still a monotone map, because it is still preserved as an image of the set-valued functor.

Proposition. let $f : (A,\sqsubseteq_A) \to (B,\sqsubseteq_B)$ be a monotone map then $f$ is still a monotone map for any smaller preorder then $\sqsubseteq_A$ or a larger preorder then $\sqsubseteq_B$.

A map $f$ is monotone iff $\sqsubseteq_A \subseteq map_{f,2}^{-1}(\sqsubseteq_B)$ this is an inclusion relation in a preorder so any transition which makes the predecessor smaller or the successor larger preserves the truth value of the statement. Inverse images are monotone so larger values of $\sqsubseteq_B$ produce larger values of $map_{f,2}^{-1}(\sqsubseteq_B)$ and smaller values of $\sqsubseteq_A$ produce smaller values of itself. $\square$

Recall that the family of all topological spaces on a set form a lattice. By the backwardness of continuous maps, we have that for any continuous map $f: S \to T$ any enlargement of the input topology on $S$ or any reduction of the output topology $T$ preserves the continuity of the function $f$. In particular, every map from a discrete topology or to an indiscrete topology is continuous.

This technique has been used to describe the inverse process: creating the largest/smallest topology on a given set that makes a family of continuous maps from a family of topological spaces into it continuous. The kernel topology is the smallest topology on the input of a common family, and the hull topology is the largest topology on the output of a common family. This sort of technique should be applicable to any category whose elements have structural lattices.