Re: [Haskell-cafe] adjoint of coproduct & diagonal

On Sat, 2008-08-02 at 23:40 -0700, Jason Dusek wrote:

Derek Elkins wrote:

...

[id,id] is the counit. [id,id] : C+C -> C Given a function f : A+B -> C there exists a unique function f* : (A,B) -> (C,C) that is a pair of functions h : A -> C and k : B -> C such that [id,id] . h+k = f.

This f is what I have labelled [f,g] (or f+g) in the diagram I linked to, right?

Yes. [f,g] is the right notation. The statement of the universal arrow, however, uses -any- arrow A+B -> C so I thought it best not to expose any of its structure. Anyway, showing that it has such structure is what proving uniqueness of f* does.

In that case h and k are just f and g, respectively, and the unique arrow that goes from A+B to C+C is f+g -- but that would make C+C just the same as C.

The unique arrow is f* : (A,B) -> (C,C), -not- an arrow A+B -> C+C. An arrow f : A+B -> C does -not- uniquely determine an arrow A+B -> C+C such that the universal arrow diagram commutes.