Vaughan Pratt wrote in part: ...
One notion of underlying object that makes sense for me in the context so far is the set E(1,X) of elements of an object X in a topos E. With this notion I have a prayer of making sense of your claim in the contex=
t
of the following four objects of a topos E with NNO N (and whatever els= e if anything is needed to ensure that objects 2-4 exist in E).
1. N^2.
2. The initial object among all objects X with an element 0: 1 --> X and maps s,t: X --> X satisfying st =3D ts.
3. As for 2 with ts =3D t in place of st =3D ts.
4. As for 3 with st =3D s in place of ts =3D t.
...
I could well imagine 1 and 2 turning out to be isomorphic, but I feel I would get a lot from seeing the proof of whichever way that goes.
I presume that 3 and 4 are isomorphic on the ground that s and t can be mapped to any two endomorphisms whence there must exist an isomorphism exchanging s and t. I imagine I would get less out of a rigorous proof of that unless there's a bug in the previous sentence. If there's no bug then this would seem to constitute a subtle difference from algebra=
,
where the signature labels the operations in a way that prevents such a= n interchange and puts 3 and 4 in distinct albeit equivalent varieties.
Whether or not 1 and 2 are isomorphic, they are in some sense both representations of the ordinary product of N with itself.
Where I have greater difficulty is imagining an isomorphism between 2 and 3 merely on the ground that they have the same underlying object. I'm really not following your logic here. 2 implements product concurrently whereas 3 and 4 implement it sequentially. How could they be isomorphic in every topos? (I can see that they'd have to be isomorphic in Set, and presumably certain other toposes which however I have no idea how to characterize.) I regard 3 and 4 as equivalent definitions of the ordinal w^2 (w =3D \omega) in a topos.
I=B4m pretty sure the objects (1)-(4) are isomorphic. Here=B4s a sketch o= f the proof: Let X be the object (3), i.e. the initial object with a morphism 0:1---->= X and morphisms s:X---->X, t:X---->X with st=3Ds. Let Y be the object (2), i.e. the initial object with a morphism 0:1---->Y and morphisms u:Y---->Y= , v:Y---->Y. (1)&(2): We clearly have maps 0 x 0: 1----> N^2, succ x 1_N and 1_N x succ : N^2---->N^2, inducing the map Y---->N^2. For an element a of Y, we can form maps f_{t,a}: N---->Y by 0 ]----> a, succ ]----> t, and f_{s,a} by 0 ]---->a, succ ]----> s. We can now form a map N x N ----> Y by f(a,b)=3Df_{t,{f_{s,0}(a)}}(b). We can check that these maps form an isomorphism. (2)&(3): Lemma: X is a monoid We construct the addition X x X---->X by its exponential transpose X---->X^X. This is the map induced by 1_X:1---->X^X, s^X:X^X---->X^X, and t^X:X^X---->X^X. We think of this monoid structure as ordinary ordinal addition. We now construct morphisms s=B4 and t=B4 X---->X, given by s=B4(x)=3Dx + = 1 and t=B4(x)=3D\omega + x. It is clear that s=B4 and t=B4 commute. This induce= s a morphism Y---->X. On Y, we can create the projection map p:Y---->N as Y is isomorphic to N = x N. We also have the inclusion i : N ----> N x N given by a ]----> (a,0). Now ipu and v satisfy ipuv=3Dipu (ipu is the map \_ + \omega. This induce= s a map X ----> Y, which will be the inverse of the map Y ----> X. I will check the details later to make sure. But I would be very surprise= d if there were any problems. Toby