## ยง Yoneda from string concatenation

I'm trying to build intuition for Yoneda (again) this time from the perspective of strings and concatenation. The idea is that the identity function behaves like the empty string, and "free arrow composition" behaves like string concatenation. So can we understand yoneda from this model?
• First, let's think of Hom(X, X). What are the elements here? Well, for one,we need the identity arrow idX in Hom(X, X). Maybe we have other elementscalled a, b, c in Hom(X, X). So our picture of Hom(X, X) looks like this:
• what does it mean to have an element a in Hom(X, X)? It means that there'san arrow from X to X in the category. But this also means that we havea map from Hom(X, X) to Hom(X, X), given by composing with a! That is,we have a map - . a :: Hom(X, X) -> Hom(X, X).
• If we have such a map of "composition with a", then we need to know where thismap -.a maps all the elements of Hom(X, X). Thinking about this,we see that we need to add new elements to Hom(X, X), which are thecomposition of the current elements (idX, a, b, c) with a. Thisgives us the elementsidX.a = a, a.a = aa, b.a = ba, c.a = ca.
• Similarly, we need to know where these new elements aa, ba, ca map to,but let's hold off on that for now, for that simply demands an extrapolation ofimagination. Let's imagine having another object Y and an arrow g: X -> Y.This will give us a new hom-set Hom(X, Y) = Hom(X, X) . g
• In Hom(X, Y) we will have as elements all the arrows from X to Y. Let's say there's some arrow h: X -> Y. Then, we will find this arrow h in Hom(X, Y)as the image of idX under -.h . So really, for any arrow, we can findwhat element it maps to as long as know (a) idX and (b) -.h.
Now that we understand the "internal" structure, let's imagine we're representing this collection of objects and arrows by some other collection of objects and arrows. So we have a functor F that takes these sets to other sets, and takes these objects to other objects.