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

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Posts: 221
Registered: 12/13/04
Re: Homomorphism of posets and lattices
Posted: Sep 20, 2013 6:22 AM
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quasi <quasi@null.set> wrote:
> quasi <quasi@null.set> wrote:
> >
> >Victor Porton wrote:

> >>
> >>I've corrected the definitions:
> >>
> >>Definition 1. A monotone function (also called order
> >>homomorphism) from a poset A to a poset B is such a function f
> >>that x<=y -> fx<=fy.
> >>
> >>Definition 2. Order embedding is an monotone function whose
> >>inverse is also monotone.

> >
> >Not quite right.
> >
> >To have an inverse, it would have to be bijective.
> >
> >Instead, try this (Wikipedia's version):
> >
> >Given posets X,Y, a function f:X -> Y is called an order
> >embedding if (x <= y iff f(x) <= f(y)).

> A better choice of letters:
> Given posets A,B, a function f:A -> B is called an order
> embedding if (x <= y iff f(x) <= f(y)).

> >>Obvious 3. Order embeddings are always injective.
> >
> >Yes.
> >

> >>Definition 4. Order isomorphism is a surjective order embedding.
> >
> >Yes, that's Wiki's definition.

> quasi

Also one could note that, once the notion of homomorphism is settled,
the contemporary way of defining isomorphism applies:

| An isomorphism f: A --> B is a homomorphism with a two sided inverse,
| i.e. a homomorphism g: B --> A with f o g = id_B and g o f = id_A.

Then the statement in the above "Definition 4" becomes a description:
Order isomorphism are exactly the surjective order embeddings.


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