Derek Elkins derek.a.elkins at gmail.com
Sun Aug 3 03:26:33 EDT 2008

```On Sat, 2008-08-02 at 23:40 -0700, Jason Dusek wrote:
> Derek Elkins <derek.a.elkins at gmail.com> 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

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.

```