Strictly monotonic function on an ordered set implies injectivity, when is the converse true? I'm looking for the "most general" case in which the following statement is true:

Let $\mathcal{F}_1$ and $\mathcal{F}_2$ be ordered sets and $f\colon \mathcal{F}_1\to\mathcal{F}_2$ an injective function, then $f$ is strictly monotonic.

I'm well aware that "the most general" isn't well defined. I'm wondering what are some (general) conditions that could be added so that the statement is true.

A couple of cases in which it's not true:
An injective real valued function of a real variable that is not monotonic on any interval.
$$\begin{align}\mathbb{R}&\to \mathbb{R}\\
x&\mapsto\begin{cases}x, &x \text{ is rational}\\
-x, & x \text{ is irrational}\end{cases}\end{align}$$
An injective continuous function that is not monotonic.
$$\begin{align}\mathbb{Q}&\to\mathbb{Q}\cup\sqrt{2}+\mathbb{Q}\\
x&\mapsto \begin{cases}x,& x< \frac{\sqrt{2}}{2},\\-x+\sqrt{2}, & x>\frac{\sqrt{2}}{2}.\end{cases}\end{align}$$
where $\mathbb{Q}$ are the rational numbers.
Edit: This question was about a function on an ordered field but, as suggested by Henry Davii on his answer, there is no need for $\mathcal{F}$ to be fields, they could just be sets. I've changed all the mentions of the word "field" on this question to "set".
 A: Actually since the $f:\mathcal{F}_1 \rightarrow \mathcal{F}_2$ is merely a function, we can think more generally as letting $\mathcal{F}_1$ and $\mathcal{F}_1$ be ordered sets, rather than ordered fields. I manage to contruct an equivalence between the condition that $f$ is strictly increasing and the condition that $f$ is continuous under some topology. This result can definitely be developed, but as a preliminary result I'll just leave it here today.
An ordered set $(S, \lt)$ has a natural topology called ordered topology, namely the open sets are exactly the sets $\{x| x < a\} $, $\emptyset$ and $S$. Denote this topology by $\mathcal{T}_1$.
Now I claim that $f$ is strictly increasing $\Leftrightarrow$ $f$ is continuous under the topology $\mathcal{T}_1$, which to me is a beautiful link between the order structure and topological structure. The proof is essentially done by checking all the possibilities.
"$\Leftarrow$": suppose $a,b\in \mathcal{F}_1$ and $a < b$. Let $E=\{y|y<f(b)\}$, since $f$ is continuous, $f^{-1}(E)$ is open (i.e. has the form $\{x|x<c\}$ for some $c\in \mathcal{F}_1$. We claim that $f^{-1}(E)=\{x|x<b\}$ (notice that this is not the same as requiring $c=b$).
If $x\in f^{-1}(E) - \{x|x<b\}$, then $x\ge b$. Since $f^{-1}(E)$ is open, $b\in f^{-1}(E)$ which implies $f(b) < f(b)$, contradiction. Hence $\{x|x<b \} \supset f^{-1}(E)$.
Conversely, since $b \notin f^{-1}(E)$, for all $x$ with $x\ge b$, we have $x\notin f^{-1}(E)$, hence $\{x|x<b \} \subset f^{-1}(E)$.
We conclude that if $a<b$, then $a \in \{x|x<b\} = f^{-1}(\{y|y<f(b)\})$, which implies $f(a) < f(b)$.
"$\Rightarrow$": suppose $f$ is strictly increasing, let $E = \{y|y < d\}$ for some $d\in \mathcal{F}_2$, then if $f^{-1}(E)$ is not open, there exists $a,b\in \mathcal{F}_1$ with $a<b, b\in f^{-1}(E), a\notin f^{-1}(E)$, but then $f(a)\ge d > f(b)$, contradictory to the assumption that $f$ is strictly increasing. Hence $f^{-1}(E)$ must be open, hence $f$ is continuous.         $\qquad \blacksquare$
