Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question was firstly posted on Mathoverflow. Two answers are pretty interesting. The potential counter-example given in the second answer is really interesting, but it is not surely a counter-example.

Consider two partially ordered sets $A = {a< b,a< c}$, $B={x< z,y< z}$.

Their linear extensions (here we allow equality in linear extensions) for $A, B$ are $A_L={A_1={a< b< c}, A_2={a< b= c}, A_3={a< c< b}}$ and $B_L = { B_1 = {x< y< z}, B_2={y< x< z}, B_3 ={x= y< z}}$

We may define $f_1: A_1\to B_1$ by $f_1(a)=x, f_1(b)=y, f_1(c)=z$ and $f_2:A_3\to B_2$ by $f_2(a)=y, f_2(b)=x, f_2(c)=z$, but there are no bijective mappings from $A_2$ to $B_3$ s.t. $t< s\Longleftrightarrow f_3(t)< f_3(s)$ and $t=s \Longleftrightarrow f_3(t)=f_3(s)$. (And it's easy to see that no other pairing of the $A_i$ with the $B_j$ will allow such maps to be chosen.)

There are no order isomorphisms from $A$ to $B$, either.

Is the following conjecture true?

For any partially ordered sets $A,B$, if there exist isomorphisms from $A$'s linear extensions to $B$'s, then there exists an isomorphism from $A$ to $B$.

If not, could you give me a counter example?

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.