From: magidin@math.berkeley.edu (Arturo Magidin)
Subject: Re: Universal algebra question
Date: 4 May 2000 19:23:24 GMT
Newsgroups: sci.math
Summary: [missing]

In article <3911AE08.D0B70FEF@inrialpes.fr>,
Catalin Dima  <cdima@inrialpes.fr> wrote:
>
>Hi all,
>
>Did anyone hear of some "universal algebra style" modeling of
>conditional operations?
>
>I am thinking that partial algebras (i.e. with partial operations) would
>
>do the job, but I have only a shallow knowlegde on them... I am
>interested in answers to questions like: how are initial algebras like,
>what is an (appropriate notion of) variety etc. I remember from some
>postings on sci.math that partial algebras are not nicely behaving, but
>I don't remember why... I suspect it's about morphisms/homomorphic
>images, am I right?

Initial algebras are defined by the universal property; which
essentially tells you that initial (partial) algebras are the free
(partial) algebra on one element.

A variety of partial algebras is defined exactly the same was a a
variety of algebras: it is a class of (partial) algebras of a given
type, which is closed under products, subalgebras, and quotients.

The definition of subalgebras and quotients gives some trouble, since
the partial operations have to be "induced". If A is a partial
algebra, say with partial operation R, then R induces a partial
operation on the subset B: whenever the tuple of elements of B (b_i)
lies in the domain of R, then R(b_i) should lie in B; however, it is
possible that a subset have a structure "richer" than the algebra A
(say, by extending the partial operations) or "poorer" (by having the
partial operation even more partial), in which case it is not a
subalgebra.

With quotients you run into similar problems that have to be dealt
with accordingly. And with products, you simply define all operations
and partial operations coordinate-wise.

Aside from the difficulties with quotients and subalgebras, the other
problem is the sometimes annoying fact that if f:A->B is a morphism of
(partial) algebras of given type, it is NOT true that f(A) will always
be a subalgebra of , since the image of A may have a structure which
is "poorer" than the induced structyure on f(A). As Brian Davey says,
many beautiful, useful, and false "theorems" can be "proven" if one
forgets this fact.

======================================================================
"It's not denial. I'm just very selective about
 what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@math.berkeley.edu