Date: Sep 18, 2013 3:39 AM
Author: William Elliot
Subject: Re: Homomorphism of posets and lattices

On Wed, 18 Sep 2013, quasi wrote:
> William Elliot wrote:

> >What's your opinion about the following except from a manuscript?
> You should credit your source.
> Manuscript by who?


> Published when and where ?

Preprint; on his web site.

> >I consider the Definition 2.92 to be in error unless the domain
> >of the function is a linear order. For example, the identity
> >function from a two point antichain { a,b } to the chain a < b.
> >
> >Comments? Are not antichains preserved by order isomorphisms?

> Yes, of course.
> As your counterexample clearly shows, the author's definition
> 2.92 is flawed.

> >2,1,12 Homomorphism of posets and lattices
> >
> >Definition 2.90. A monotone function (also called order
> >homomorphism) from a poset A to a poset B is such a function f
> >that x <= y --> f(x) <= f(y).
> >
> >Definition 2.91. Order embedding is an injective monotone
> >function.
> >
> >Definition 2.92. Order isomorphism is an surjective order
> >embedding (= bijective monotone function).

> As noted, the above definition is flawed, but the fix is easy.
> Definition 2.92 (possible revision):
> An order isomorphism is a bijective function such that both it
> and its inverse are monotone functions.

Assume f and f^-1 are monotone and f(x) = f(y)
Then f(x) <= f(y) and f(y) <= f(x). Whence,
x = f^-1f(x) <= f^-1f(y) = y and similarily,
y <= x, so x = y.

Your definition is equivalent to a bimonotone surjection.
How would you revise order embedding? A bimonotone map?

My suggestions for the definitions, are in my recent thread "Order Embedding".

> >Order isomorphism preserves properties of posets, such as
> >order, joins and meets, etc.

> With the above suggested revision of definition 2.92, those
> properties _are_ preserved.

Even without knowing what a straight map
may be, what's your opinion of this excerpt?

Proposition 3.4. Let f be a monotone map from a meet-semilattice K
to some poset L. If
Aa,b in K: f(a meet b) = f(a) meet f(b)
then f is a straight map.