Suppose $\left(A,\leq_{A}\right)$ and $\left(B,\leq_{B}\right)$ are Posets and $f:A\to B$ is an order preserving bijection. I'm trying to show that $f^{-1}:B\to A$ is also order preserving. That is I want to show that given $b_{1},b_{2}\in B$ $b_{1}\leq_{B}b_{2}\Longrightarrow f^{-1}\left(b_{1}\right)\leq_{A}f^{-1}\left(b_{2}\right)$ Since $f$ is bijective there are uniquely defined $a_{1},a_{2}\in A$ such that $b_{1}=f\left(a_{1}\right)\Longrightarrow a_{1}=f^{-1}\left(b_{1}\right)$ $b_{2}=f\left(a_{2}\right)\Longrightarrow a_{2}=f^{-1}\left(b_{2}\right)$ If $a_{1},a_{2}$ are comparable and I assume $a_{2}\leq_{A}a_{1}$ I get a contradiction since $f$ is order preserving: $b_{2}=f\left(a_{2}\right)\leq_{B}\, f\left(a_{1}\right)=b_{1}$ So I deduce that if $a_{1},a_{2}$ are comparable then necessarily the required inequality holds. My question is whether it's possible that $a_{1},a_{2}$ are not comparable and the claim is actually false?
Inverse of order preserving transformation
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0This is a very good question, and it's very nice to see that you have posted your own attempts on the topic. – 2012-12-29
2 Answers
Let $A=\{0,1\}\times\Bbb N$, and define $\langle i,m\rangle\le_A\langle j,n\rangle$ iff $i=j$ and $m\le n$. Let $B=\Bbb N$ with the usual order. Finally, let
$f:A\to B:\langle i,n\rangle\mapsto 2n+i\;.$
Then $f$ is an order-preserving bijection, but $f^{-1}$ is not order-preserving for exactly the reason that you suggest: for any $n\in\Bbb N$, $2n<2n+1$ in $B$, but $f^{-1}(2n)=\langle 0,n\rangle$ and $f^{-1}(2n+1)=\langle 1,n\rangle$ are not comparable in $A$.
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0@Serpahimz: Thanks! – 2012-12-29
To drive Brian's point further, let's assume for simplicity that $f$ is the identity. This simply means that $\leq_A\subseteq\leq_B$ as sets of ordered pairs.
For example, if $(A,R)$ is any partially ordered set, and $(A,\leq_R)$ is a linearlization of $R$ then the identity is an order-preserving bijection, but its inverse need not be order-preserving.
Even further, if $A$ contains more than one point, then $(A,\mathrm{Id}_A)$ is a partial order (the discrete order) and the identity is an order-preserving map between $(A,R)$ for any partial order $R$. Its inverse, also the identity, is order-preserving if and only if $R=\mathrm{Id}_A$ as well.
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0@Serpahimz: It simply means that if $(A,R)$ is a partially ordered set then there is some $R'$ such that $R\subseteq R'$ and $(A,R')$ is a linearly ordered set. For finite sets this is provable in ZF, and for general posets this requires the axiom of choice, but this is strictly weaker than the axiom of choice. – 2012-12-29