Let be $(X,\preccurlyeq)$ and $(Y\curlyeqprec)$ partial ordered sets so that let be $f$ a function form $X$ to $Y$. So if $\rightthreetimes$ and $\leftthreetimes$ was an operation on $X$ and an operation on $Y$ then it is usual to say that $f$ is a groupoid homomorphism if for any $x_1$ and $x_2$ in $X$ the equality $$ f(x_1\rightthreetimes x_2)=f(x_1)\leftthreetimes f(x_2) $$ holds so that I would say that $f$ is an order homomorphism if the inequality $$ \tag{1}\label{1}x_1\preccurlyeq x_2 $$ implies the inequality $$ \tag{2}\label{2}f(x_1)\curlyeqprec f(x_2) $$
Now if $f$ was a bjiective groupoid homomorphism (namely a groupoid isomorphism) then it is not hard to show that even $f^{-1}$ is a groupoid homomorphism but I suspect that the same does not would be if $f$ was an order bjiective homomorphism (namely a groupoid isomorphism) since I know that actually many authors says that $f$ is an order isomorphism if it is a bijection such that the inequality \eqref{1} holds if and only if the inequality \eqref{2} holds so that I suspect it would be better say that $f$ is a order homomorphism whether \eqref{1} holds if and only if \eqref{2} holds: so by this I thought to put a specific question where I ask to prove or disprove whether the inverse application of an order isomorphism is an order isomorphism with respect the (first) definition I gave; moreover, I would like to know if any author actually defines an order homomorphism. Could someone help me, please?
The usual definition of an order homomorphism (also called "monotone", or "order preserving" map) is $(1)⟹(2),$ not $(1)⟺(2).$ Two references for this definition are given here on Wikipedia:
An order isomorphism is a bijection $f$ such that both $f$ and $f^{−1}$ are order homomorphisms.
An example of a bijective order homomorphism which is not an order isomorphism is $f={\rm id}_X,$ $X=Y=\{0,1\},$ $≼=\{(0,0),(1,1)\},$ $⋞=\{(0,0),(1,1),(0,1)\}.$