
Re: Homomorphism of posets and lattices
Posted:
Sep 18, 2013 5:29 AM


On Wed, 18 Sep 2013, quasi wrote: > >> 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. Oh yea, it'd have to be injective for an inverse to exist.
> >How would you revise order embedding? > > I wouldn't revise it.
Since order isomorphisms are order embeddings some revision appears likely. Recall, that if f:X > Y, is an embedding, then f:X > f(X) is an isomorphism.
So consider the monotone injection of the identity map from the antichain { a,b } into the chain a < b < c.
Were order embeddings simply monotone injections. then the antichain { a,b } would be order isomorphic to the chain a < b.


