Posted Saturday April 1 2023.

If *C* is a category and *x*, *y* ∈ *C* are objects, then as long as *C* is locally small, you can form the *hom-set* *H**o**m*(*x*, *y*), which is literally the set of morphisms *x* → *y*. This defines a functor *C*^{op} × *C* → *S**e**t* called the *hom-functor*. (*C*^{op} is the opposite category of *C*.)

Enriched category theory generalizes this idea by allowing the hom-functor to return something other than a set, so long as it comes from some monoidal category *M*, so that the hom-functor goes *C*^{op} × *C* → *M*.

Consider the category “Bool” consisting of two objects {0, 1} and three morphisms {0 → 0, 1 → 1, 0 → 1}. We can define a monoidal product (aka an associative binary operation with identity) on objects 0 ⊗ 0 = 0 ⊗ 1 = 1 ⊗ 0 = 0 and 1 ⊗ 1 = 1. Altogether this makes Bool into a monoidal category with 2 objects.

If *C* a category enriched in Bool, with hom-functor *C*^{op} × *C* → *B**o**o**l*, then for every pair of objects *x*, *y* ∈ *C*, we have either *H**o**m*(*x*, *y*) = 0 or *H**o**m*(*x*, *y*) = 1. We can use this mapping to define a *relation* between objects by saying that *x* is related to *y* iff. *H**o**m*(*x*, *y*) = 1.

Theorem:The relationRonBooldefinedxRyiff.Hom(x,y) = 1 is a preorder onC.

To prove it, there are just two properties we need to show, for all *x*, *y*, *z* ∈ *C*:

- reflexivity:
*H**o**m*(*x*,*x*) = 1, - transititivity: if
*H**o**m*(*x*,*y*) = 1 and*H**o**m*(*y*,*z*) = 1 then*H**o**m*(*x*,*z*) = 1.

To show reflexivity, it is enough to look at the identity morphism *i**d*_{x} : 1 → *H**o**m*(*x*, *x*). Since its domain is 1, and the *only* arrow in Bool whose domain is 1 is the identity morphism on 1, this means that *i**d*_{x} must be the identity morphism on 1, whose codomain is also 1! Therefore *H**o**m*(*x*, *x*) = 1.

For transivitity, suppose *x**R**y* and *y**R**z*. We need to show *x**R**z*. We need to use the composition map *c* : *H**o**m*(*x*, *y*) ⊗ *H**o**m*(*y*, *z*) → *H**o**m*(*x*, *z*) which takes morphisms *f* : *x* → *y* from *H**o**m*(*x*, *y*) and *g* : *y* → *z* from *H**o**m*(*y*, *z*) and returns their composition *g* ∘ *f* : *x* → *z* in *H**o**m*(*x*, *z*). We have by assumption *H**o**m*(*x*, *y*) = *H**o**m*(*y*, *z*) = 1, and we know 1 ⊗ 1 = 1. Thus, the domain of *c* is 1. As before, the only morphism whose domain is 1 is the identity on 1, whose codomain is also 1. Thus the codomain of *c* must be *H**o**m*(*x*, *z*) = 1. Thus *x**R**z*.