Ralf Hemmecke <[EMAIL PROTECTED]> writes:

> Of course there must be one. \emptyset is an object in B and in a category
> there is always the 1 arrow from an object to itself.

OK, that's a good argument for thinking in caegories...

> >>> What is G[\sigma]? Ah, sure, it is a permutation in some S_m.
> >>> First question: what is m?
> >> The only good choice is m=\card G[0]. And to be functorial
> >> G[\sigma] must be the identity permuation of S_m.
> > I just wanted to look that up. OK, I did: one can say (although I
> > don't really have a good argument for that) that the sigma above is
> > the identity on the empty set...
> 
> No. G[\emptyset] is not necessarily empty. So the G transports the arrow
> 
> 1: \emptyset -> \emptyset to G[1]: G[\emptyset] -> G[\emptyset]. And the 
> latter
> is most certainly the identity on G[\emptyset], because G is a functor.

But that's exactly what I meant to say. (I didn't say that
G[\emptyset]=\emptyset, only that \sigma would be the identity on the empty
set)

> > Note that axiom says lcm(0,n)=0 for any n.
> 
> I don't think that is important since you suggested to take
> 
>    n := lcm([i for i in 1..#p | p.i ~= 0]);
> 
> and I would have to compute the lcm of the empty list. That is more
> interesting. :-)

I'd vote for zero. You see, I think that lcm(0,n)=0 for any n is somewhat
problematic, since 0 is a multiple of any number, so why don't we put
lcm(n,m)=0 for any n and m?
 
> >> So I would be happy if someone could confirm my thinking.  confirmed?
> 
> Thank you? ;-)

? ;-)

Martin


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