Talk:Free Boolean algebra

Disputed

I disputed myself here because I'm not quite sure of this fact--is this really sufficient to characterize free BAs? It's not obvious that this condition implies that any permutation of the generators extends to an automorphism, which is surely a property we want. Can't find any clear refere

Initial objects

I think the main definition is wrong. In any category, Initial objects if they exist are unique (up to isomorphism). The only initial object in the category of non-trivial boolean rings is Z₂. --CSTAR 19:36, 2 November 2005 (UTC)[reply]

Quite so, and that is my fault and not Trovatore's so I should fix it. There is a category of algebras in which the initial algebra is the initial object, namely the Eilenberg-Moore category, but that category has more structure than the plain category of algebras. The wording I introduced was confused and clear evidence of insufficient tea. --- Charles Stewart 19:47, 2 November 2005 (UTC)[reply]

I'm still not sure I understand this. What's the Eilenberg-Moore category? Anyway does this addition make any difference, because free object still is defined as an initial object in a category which does not depend on the generators in any way. --CSTAR 20:32, 2 November 2005 (UTC)[reply]

It's one of the two canonical categories generated by a monad (the other being the Kleisli category of resultions, which is sort of a dual to the Eilenberg-Moore category), where the algebra is defined in a different way to the regular way that is due to Lawvere:

The structure of the algebra is given by category and a monad T over that category, that is a functor equipped with two nat. trans. specifying its unit and product;
The Eilbenberg-Moore category over this algebra has its objects given by a pair of (i) an object A from the base category, and (ii) any of those arrows f:TA --> A which satisfy certain conditions that show they behave reasonably around the conditions on unit and product for the monad.

There's much more machinery on this definition of algebra than the regular definition of algebra, but this Lawverian conception of algebra is very powerful and admits a lot of nice generalisations of the regular notion. Alternatively it's all the worst sort of generalised abstract nonsense. I used Lambek&Scott's "Intro to higher-order categorical logic" to check my defs, though they call monads triples there.--- Charles Stewart 16:02, 3 November 2005 (UTC)[reply]

Postscript: I didn't explain where the regular homs on algebras will occur. The arrows of the E-M category will not contain all the automorphisms of the BA: I'm not sure where you will find these, but maybe they will be the nat. trans. from the monad onto itself. I'll ask someone: I didn't find this explained in L&S.--- Charles Stewart 16:10, 3 November 2005 (UTC)[reply]

Yikes. That makes the characterization I proposed seem, I dunno ... Jurassic :)--CSTAR 01:34, 4 November 2005 (UTC)[reply]

Well... It certainly frightens computer scientists who are told that these constructions are the right way to describe input/output for functional programming languages. BTW, Chapter VI of Mac Lane's CftWM is all about the monad - algebra connection; the history there contradicts some of what I said: it turns out the E-M construction is originited later but pretty much independently of Lawvere's work, and it was a later researcher, Pareigis, who spelt out the connection. --- Charles Stewart 19:20, 4 November 2005 (UTC)[reply]

Converse follows

It can easily be shown that the free Boolean generated by S, (π, B) pi; is injective. The universal proerty is the

This should follow by applying the universal property. More generally, the universal property implies that the induced mappings are functorial.--CSTAR 04:30, 3 November 2005 (UTC)[reply]