categories: Re: Final Coalgebra Question

[<Tobias.Schroeder@sdm.de>]

Hi,
1. That's pretty easy. Since [X,_] maps the one-element
(final) set onto itself, i.e., it preserves the final
object, the final coalgebra of this functor is nothing
but the one-element-set itself. Of course one need not
use this abstract argument ("the forgetful functor
U:Set_F --> Set creates any limit which the functor
F:Set->Set preserves") but can give a direct proof.

2. That's much more complicated. I do not know any
elementary description. For one possible description
as a quotient of an automaton you may refer to

H. Peter Gumm, Tobias Schroeder :
Coalgebras of bounded type. Mathematical Structures in Computer 
Science, to appear

which you can find Peter Gumm's homepage

(http://www.mathematik.uni-marburg.de/~gumm/Papers/publ.html)

and the papers quoted there. ... If somebody could offer
a really nice description of that final coalgebra this
would be quite interesting I assume.

Hope that helps a bit

Tobias Schroeder
-----Original Message-----
From: N Ghani
To: categories@mta.ca
Cc: coalgebras@iti.cs.tu-bs.de
Sent: 17.01.02 12:23
Subject: coalgebras: Final Coalgebra Question


Can anyone help me with the following final coalgebra questions

1. Let X be a fixed Set. What is the final coalgebra of the functor
[X,_]:Set -> Set. If you wish to make X finite then that's fine by me.

2. Consider the functor [[_,2],2]:Set -> Set. This functor doesnt have
a final coalgebra for cardinality reasons. However one may define a
finitary variant of this functor as follows:

First let TX = [[X,2],2] if X is finitely presentable. Thus T:Set_fp
-> Set is a functor from the full subcategory of fintely presentable
objects of Set into Set. Next define T' to be the left Kan extension
of T along the inclusion Set_fp -> Set. In other words T'X is the
filtered colimit of all the TX_0 where X_0 is a finitely presentable
subobject of X

Now, T' is clearly finitary and from general nonsense we know that it
has a final coalgebra. But what is it concretely?

Thanks for any help you can offer

Neil