Union of Functors and a Functor on Union

Union of Functors and a Functor on Union

Intro

On 10/3/2016 Paul Phillips posted an interesting question, https://twitter.com/extempore2/status/783161511724273664?cn=cmVwbHk%3D&refsrc=email - how are \/[F[A], G[A]] and F[A\/B]related. Here are my musings on this. In 2011 I already wrote this: http://vpatryshev.blogspot.com/2011/05/monad-genome-part-1.html - pretty much close to the topic; but it's better to just go into details.

Basics

Let's fix some category C; assume that it has all the necessary limits or colimits.

Union is a special case of a coproduct; it's a functor C⨯C → C that is right adjoint to diagonal Δ:C → C⨯C. The diagonal maps an x to (x,x). 

This diagonal actually can be defined like this: given 1+1 → 1, where 1 is a unit, a terminal category, its exponential, C¹⁺¹ → C¹, is the same diagonal Δ.

Union of Two Endofunctors

An endofunctor on category C is a point in C^C; and a union of two endofunctors has the following signature: C^C⨯C^C → C^C which is the same as (C⨯C)^C → C^C, or C^(1+1)×C → C^C

In Scala its type can be written as (Bool => C => C) => C => C.  Here Bool is a cheap implementation of 1+1.

Since the union is calculated pointwise in this case, this functor, C^(1+1)×C → C^C, is a right adjoint to the diagonal  C^C → C^(1+1)×C.

Endofunctor following a Union

For the case where we first build a union and then apply a functor, we need three components: two objects and a functor. So we have C⨯C⨯C^C → C^C - here we map a triple (A,B,F) to F[A\/B].

Rewriting it, we have C^1+1+C → C^C- this is the signature of our operation. In other words, we have 

((Bool \/ C) => C) => C => C.