Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
quasi

Posts: 10,396
Registered: 7/15/05
Re: Homomorphism of posets and lattices
Posted: Sep 18, 2013 5:06 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

William Elliot wrote:
>quasi wrote:
>> William Elliot wrote:
>>

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

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

>
>Victor.
>

>> 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?

No.

quasi



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.