Sorry for the clutter. I forgot a counterexample. Yes, it is possible for (X*X)*X to have a final coalgebra but not X*X. On the discrete category with two objects, A and B, let the bifunctor be defined by: * | A B --+------ A | B A B | A A The unique cubical coalgebra is A but there is no square coalgebra. What I don't have is a counterexample with an _associative_ bifunctor.
* | A B --+------ A | B A B | A A
The unique cubical coalgebra is A but there is no square coalgebra.
What I don't have is a counterexample with an _associative_ bifunctor.
How about the positive integers with * as sum, with 1->3 as the only nonidentity arrow? The unique cubical coalgebra is 1 but there is no square coalgebra. Vaughan
How about the positive integers with * as sum, with 1->3 as the only nonidentity arrow? The unique cubical coalgebra is 1 but there is no square coalgebra.
Oops, that gets associativity at the cost of * no longer being a functor (pointed out to me by Peter Selinger). Ok, let me dig myself in deeper by making my example more complicated. Instead of 1->3, put an arrow from i to j whenever i <= j <= 2i. Now every i is a square coalgebra but no i is a cubical coalgebra. Now adjoin a new object oo (infinity), with x+oo = oo+x = oo for all x, and the identity at oo as the only new arrow. oo is both a square *and* a cubical coalgebra. Since it is disconnected from the other square coalgebras there can't be a final such. But oo is the only cubical coalgebra, with only one self-map, making it a final cubical coalgebra. Vaughan
Instead of 1->3, put an arrow from i to j whenever i <= j <= 2i.
I keep forgetting to make a list of what to remember. Peter Johnstone kindly put me out of my misery on this noncategory. Since I don't seem to be having much luck making the example more complicated, maybe making it simpler might work. The ring of integers mod 3 is a one-object monoidal category in the usual way, with multiplication as composition and addition as the monoid. Every arrow is clearly both a square coalgebra and a cubical coalgebra, i.e. we have three of each. Claim: There are no final square coalgebras, but 1 and 2 are final cubical coalgebras. Proof. Square coalgebra homomorphisms f from 2x to x (as square coalgebras) are those that satisfy xf = (2f)(2x). But 2x2 = 1 (mod 3) so every f solves this. Hence there are three square coalgebra homomorphisms from 2x to x, whence no x is a final square coalgebra. Cubical coalgebra homomorphisms f from y to x (as cubical coalgebras) must satisfy xf = (3f)y = 0. But for x other than 0, f = 0 is the only solution. So the two nonzero cubical coalgebras are final. The same example (unless I've forgottten yet another thing) solves the corresponding problem for initial algebras. Vaughan
participants (3)
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Dr. P.T. Johnstone -
Peter Freyd -
Vaughan Pratt