quasi
Posts:
9,921
Registered:
7/15/05


Re: Homomorphism of posets and lattices
Posted:
Sep 18, 2013 5:06 AM


William Elliot wrote: >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.
What you posted above is silly. By hypothesis, f is bijective, hence injective. Since f is injective,
f(x) = f(y) implies x = y.
No need to invoke monotonicity to get that implication.
>Your definition is equivalent to a bimonotone surjection.
How so?
>How would you revise order embedding?
I wouldn't revise it.
>A bimonotone map?
No.
quasi

