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 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.

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