## § Right Kan extensions as extending the domain of a functor

#### § First over functions (fake category fluff)

Given a function $g: C \rightarrow E$ and an embedding $j: C \rightarrow D$, then the Right Kan extension of $g$ along $j$, denoted $g/j$ is a new function $g/j: D \rightarrow E$. So we are extending the domain of $g$, along the extender $j$. Informally, we write the new function as $g/j$ because:
C---
|   \-g---*
j         |
|         |
v         v
D--g/j--->E

foo(j(c)) = g(c)
foo = (g/j)(c)
g/j(j(c)) = g(c)


#### § Next over preorders (real category stuff)

The kan extension provides us a bijection, for any function f: D -> E
C---
|   \-g---*
j         |
|         |
v         v
D--g/j--->E
D------f->E

hom(f.j, g) ~= hom(f, g/j)

That is, if we have some way to make congruent f.j with g, then we can "split" the congruence to have f.j congruent with (g/j).j. Cancelling j, we can have f congruent with g/j. Consider a preorder with a ordering ≤. Equip with a monoidal structure <>, which is a monotone map with a neutral element. (For example, integers with multiplication and ≤). The bijection of hom-sets is equivalent to saying
m*k <= n iff m <= n/k

(how? I have no idea; I gotta work this out!)

#### § Question: What happens in the context of vector spaces?

Since linear algebra is probably the nicest thing we have in math, I really want to understand what happens in the linear algebra case. I don't really understand how to make the correct version of a kan extension inside $Vect$, though. A kan extension seems to fundamentally be a 2 categorical construct, than a 1 categorical construct.