Jason Dusek jason.dusek at gmail.com
Sun Aug 3 03:52:14 EDT 2008

```Derek Elkins <derek.a.elkins at gmail.com> wrote:
> Jason Dusek wrote:
> > 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.

Yes, I have confused my meaning a bit. +(f*) is also unique,
and that was the arrow I was thinking of.

So, we have f* : (A, B) -> (C, C), comprising h and k, so I
am not sure sure how + will act on it. As I've seen with + so
far, it works like this on arrows:

+((A -> C), (B -> C))  |->  A+B -> C

So, for example, +(f, g) |-> [f,g] -- so what is f* made of?
This is what seems to be happening:

+((A, B) -> (C, C))  |->  A+B -> C+C

So how are h and k paired? Their pairing puts them CxC but
seemingly in a different way from the way I've paired f and g.

--
_jsn
```