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 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