```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?Victor.> 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 Kto some poset L.  If	Aa,b in K:  f(a meet b) = f(a) meet f(b)then f is a straight map.
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