The reason of this question is an exercise in category theory:
Find a counterexample to the following statement: A monotone function $f:X\longrightarrow Y$ between posets $X$ and $Y$ which is a bijection on the underlying sets (an isomorphism in $\mathcal{S}et$) is necessarily an isomorphism in $\mathcal{P}arset$.
In short, this exercise asks me to give a bijective monotone function whose inverse is not monotone.
In this link: Are monotonic and bijective functions the same? the accepted answer gives an example that $f(x)$ is bijective but not monotone. I tried to define $f^{-1}(y)$ as the same as $f(x)$ defined in this post, and compute the $f$. Then we can have $f^{-1}(y)$ not monotone and $f$ bijective.
However, the problem is after the computation, I found that $f$ has the same expression as $f^{-1}$ in the same regions, which implies that $f$ is not monotone.
I also tried different constructions using linear function, it turned out that we can easily arrive either both $f$ and $f^{-1}$ are monotone or neither of them is monotone.
Is there any other possible construction I can follows?
By the way, in all the attempts I made, the construction is on $X=(\mathbb{R},\leq)$ and $Y=(\mathbb{R},\leq)$, so it is perhaps easier to choose other partially ordered set, for instance $Y:=(2^{\mathbb{R}},\subset)$, but then I don't know how to construct a bijective monotone function at all...(overall in the category theory, the monotone function can be respect to different partial orders, for instance, $x_{1}\leq_{X} x_{2}$ implies $f(x_{1})\leq_{Y} f(x_{2})$.)
Thank you!
Consider $X=Y=\{0,1\}$ with $ x \preceq_X y$ if and only if $x=y$ and $x \preceq_Y y$ if and only if $0 \preceq_Y 1$ or $x=y$.
The identity is a monotonic bijection between $(X,\preceq_X)$ and $(Y,\preceq_Y)$. However, the inverse is not monotonic.