RE: [Axiom-developer] Complex exponentiation and 0

[Page, Bill]

On Monday, June 21, 2004 6:39 PM William Sit
address@hidden wrote:

> 
> Bill Page wrote:
> >http://planetmath.org/?op=getobj&from=objects&name=CardinalArithmetic
> 
> Actually, if one accepts the convention that there is a 
> morphism from 0 to y in Cardinal numbers, then I don't see
> any reason not to accept the convention that there is a
> morphism from y to 0 (as there would be one in the opposite
> category, and therefore in the original category). This of
> course would create difficulty because one wants 0^y to
> be 0 for y \ne 0, not 1.
> 
> So it's all a matter of convenience for the "laws" to hold 
> "more generally".
>

No. The morphisms in the Cardinal numbers category are
functions. There is no function whose codomain is 0 unless
the domain is 0. See

  http://planetmath.org/encyclopedia/Function.html

There is exactly one function whose domain is 0 for each
codomain. So we do have x^0 = 1 and 0^y = 0 for y \ne 0.

Regards,
Bill Page.