strictly monotonic yet _non_ injective map

Question

If $X$ is a partially ordered set and $Y$ an ordered set, there may exist maps from $X$ to $Y$ that are :

• strictly monotonic
• non injective

Here are two examples :

• Consider $\mathcal{P}_f(\mathbb{N})$ the set all finite parts of $\mathbb{N}$, ordered by inclusion and let $$\varphi:\mathcal{P}_f(\mathbb{N})\to\mathbb{N},A\mapsto\text{card}(A)$$It is clear that, forall $A,B\in\mathcal{P}_f(\mathbb{N})$, we have $A\subsetneq B\implies \text{card}(A)<\text{card}(B)$ so that $\varphi$ is strictly increasing. At the same time, $\varphi$ is not injective because there are are obviously distinct finite subsets of $\mathbb{N}$ with the same cardinality.

• Let $\mathbb{P}$ be the set of prime numbers and $$\psi:\mathbb{N}-\{0,1\}\to\mathbb{N}-\{0\},n\mapsto\cases{1\text{ if }n\in\mathbb{P}\\n\text{ otherwise}}$$Let us equip $\mathbb{N}-\{0,1\}$ and $\mathbb{N}-\{0\}$ with the divisibility relation, denoted as usual by $\mid$. We can see that forall $m,n\in\mathbb{N}-\{0,1\}$ such that $m\,\Vert\,n$ (which means $m\mid n$ and $m\neq n$), we have $\psi(m)\,\Vert\,\psi(n)$, so that $\psi$ is strictly increasing. But $\psi$ is clearly not injective since all primes have the same image !

My questions :

• Given a poset $X$ (not totally ordered), does it necessarily exist a map $f:X\to X$ which is strictly monotonic yet non injective ?

• If so, can we describe some generic procedure to construct such a map ?

Any hints / help will be appreciated :)

Answer 1

Given a poset $X$ (not totally ordered), does it necessarily exist a map $f:X\to X$ which is strictly monotonic yet non injective?

Not necessarily. Let $\omega$ be the natural numbers in the usual order and let $X = \omega\;\|\;\omega^{\partial}$ denote the parallel sum of $\omega$ and its dual. (That is, put $\omega$ and $\omega^{\partial}$ side by side, with no comparabilities between them.) $X$ is not a total order, but any map $f\colon X\to X$ such that $x<y$ implies $f(x)<f(y)$ must map the copy of $\omega$ into itself injectively and the copy of $\omega^{\partial}$ into itself injectively, hence is injective.