Dusko Pavlovic wrote:
doesnt the 0-dim vector space carry the final (terminal) coalgebra? (it has just one vector, so there is not much choice for v_0 and v_1.)
Nope - v_0 and v_1 have to be linearly independent. (This is like the condition in Peter's result that the two distinguished points of the set be distinct, without which the result degenerates; and as in Peter's result, it can be regarded as a kind of flatness condition.) Incidentally, I was probably wrong to suspect that the answer is something to do with measure, as the question is posed over an arbitrary field. I now suspect that there's a similar question whose answer has to do with measure, but I won't attempt any further speculation here. Tom
Here's a question belonging to (10), to which I don't know the answer. Let C be the category whose objects are triples (V, v_0, v_1) where V is a vector space and v_0 and v_1 are linearly independent vectors in V, and whose maps preserve linear structure and the `basepoints'. There's a `wedge' functor C x C --> C defined by
(V, v_0, v_1) wedge (W, w_0, w_1) = ( (V + W)/~, v_0, w_1)
where (V + W)/~ is the direct sum with v_1 identified with w_0. (So dim(V wedge W) = dim V + dim W - 1.) There's then an endofunctor G of C given by self-wedging. Question: what, if any, is the terminal G-coalgebra?
(I suspect the answer is something to do with measure/integration - again see Peter's previous postings - but really have no idea.)
30-Nov-2004 08:03:55 -0400,1980;000000000000-00000000