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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: 11,254
Registered: 7/15/05
Re: Homomorphism of posets and lattices
Posted: Sep 18, 2013 5:06 AM
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William Elliot wrote:
>quasi wrote:
>> William Elliot wrote:

>>>What's your opinion about the following except from a

>> You should credit your source.
>> Manuscript by who?


>> Published when and where?
>Preprint; on his web site.

>>>I consider the Definition 2.92 to be in error unless the
>>>domain of the function is a linear order. For example,
>>>the identity function from a two point antichain { a,b }
>>>to the chain a < b.
>>>Comments? Are not antichains preserved by order isomorphisms?

>> Yes, of course.
>> As your counterexample clearly shows, the author's definition
>> 2.92 is flawed.

>> >2,1,12 Homomorphism of posets and lattices
>> >
>> >Definition 2.90. A monotone function (also called order
>> >homomorphism) from a poset A to a poset B is such a function f
>> >that x <= y --> f(x) <= f(y).
>> >
>> >Definition 2.91. Order embedding is an injective monotone
>> >function.
>> >
>> >Definition 2.92. Order isomorphism is an surjective order
>> >embedding (= bijective monotone function).

>> As noted, the above definition is flawed, but the fix is easy.
>> 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.

No need to invoke monotonicity to get that implication.

>Your definition is equivalent to a bimonotone surjection.

How so?

>How would you revise order embedding?

I wouldn't revise it.

>A bimonotone map?



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