```Date: Sep 18, 2013 5:29 AM
Author: William Elliot
Subject: Re: Homomorphism of posets and lattices

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 somerevision 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 mapfrom 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.----
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