The following question/task is from Taylor's, "Practical Foundations of Mathematics," page 175, ISBN 0 521 63107 6 hardback. I kind of answered it myself in typing this up, but I'll share it anyway so that others may benefit. This Approach0 search indicates that it is new to MSE.
The Details:
From page 103, ibid. . . .
Definition 2.6.1: A function $f:(X,<)\to(Y,\prec)$ between sets with binary relations is said to be strictly monotone if $$\forall x, x'.\, x'<x\implies f(x')\prec f(x).$$
(NB: The lack of $\in$ here is used throughout most of the book so far; I can't, for the life of me, figure out why - the answer is buried deep and forgotten long ago.)
The Question:
Give an example of a strictly monotone function which is not injective.
My Attempt:
I want to show that there exists a strictly monotone $f:(X,<)\to(Y,\prec)$ such that there exist $a,a'\in X$ with $f(a)=f(a')$ but $a\neq a'$.
One idea is to picture a v-shaped lattice on three nodes as $(X,<)$, where $a$ and $a'$ occupy the top nodes; essentially: there exists a $b\in X$ with $b<a$ and $b<a'$, but $a\neq a'$ and are otherwise not comparable. Then I'll take $(Y,\prec)$ as the lattice with two nodes, say $y,y'\in Y$, where $$y=f(b)\prec y'=f(a)=f(a').$$
Now for all $x,x'\in X$, if $x<x'$, then $f(x)\prec f(x')$, so $f$ is indeed strictly monotone and, by construction, not injective.
Is this correct?
Please help :)
Yes, this is a perfectly valid example. Good job!
Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.