When reading the definition of an order Isomorphism again, it occurred to me - are there examples where a function between two ordered (for some orders) sets is bijective and is a homomorphism, but whose inverse is not a homomorphism?
I assume there are, otherwise the extra condition of the inverse having to be a homomorphism would be redundant, but I can't think of any.
By definition, if $X,Y$ are posets, a function $f$ is an order isomorphism from $X$ to $Y$ if
It's then automatic that $f^{-1}$ is an order isomorphism from $Y$ to $X$.
But if $f$ is only required to be a bijective homomorphism, then $f$ need not be an order isomorphism.
For a simple example, let $X=\{a,b,c\}$ with the partial order whose only strict comparison is $a < b$, and let $Y=\{a,b,c\}$ with the partial order whose strict comparisons are $a < b < c$.
Then if $f:X\to Y$ is given by $$f(a)=a,\;\;f(b)=c,\;\;f(c)=b$$ we get that $f$ is a bijective homomorphism, but not an isomorphism.
For an even simpler example, let $X,Y$ be sets such that $|X|=|Y| > 1$.
For $X$, let the partial order be such that no two distinct elements are comparable, and for $Y$, let the partial order be such that there exist two distinct elements which are comparable.
Then any function from $X$ to $Y$ is a homomorphism, but $X,Y$ are not isomorphic. Hence any bijective function from $X$ to $Y$ is a bijective homomorphism, but not an isomorphism.
On the other hand, if $X,Y$ are totally ordered, then a bijective homomorphism from $X$ to $Y$ is, in fact, an isomorphism, since for two distinct elements of $Y$, their comparison must be the same as the comparison of their element-wise inverse images. Thus, the "if and only if" requirement holds true.