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Topic: Homomorphism of posets and lattices
Replies: 16   Last Post: Sep 20, 2013 6:22 AM

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William Elliot

Posts: 2,637
Registered: 1/8/12
Re: Homomorphism of posets and lattices
Posted: Sep 18, 2013 5:29 AM
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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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