Inverse of order preserving transformation

Question

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?

Answer 1

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 $A$.

Answer 2

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.