
Re: Homomorphism of posets and lattices
Posted:
Sep 18, 2013 3:39 AM


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 meetsemilattice 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.

