In working with the classification of stable vector bundles on $\mathbb{P}^2$, I've found that I need to answer a fairly basic question from analysis/point set topology. Here it is.

Suppose $f:\mathbb{Q}\to \mathbb{Q}$ is

- strictly increasing,
- not bounded above or below,
- a local homeomorphism (with the topology induced from $\mathbb{R}$), and
- extends to a continuous map (hence a bijection) $f':\mathbb{R}\to \mathbb{R}$.

Is $f$ a homeomorphism? (That is, is it surjective? Or, alternately, could there exist some irrational number $c$ such that $f(c)$ is rational?)

While I am dealing with a specific function, I state things in this generality because the function itself is fairly nasty and I'd rather not have to use its explicit definition more than I have to. My guess is that it is probably too much to hope for that this be true in this generality, so in case the general version is false here is a refined version of property (3) which incorporates a bit more about my present situation:

3'. there exists a partition $\mathbb{Q}=\bigcup_{\alpha \in A} I_\alpha$ of $\mathbb{Q}$ into countably many disjoint open intervals with irrational endpoints, such that $f$ is linear (with rational coefficients) on each interval.

The difficulty (at least for me) is that, viewing the intervals as intervals in the real numbers, their complement forms some kind of Cantor set.

Thanks!

(EDIT: Several counterexamples have shown the first formulation, with properties 1-4 are false. I imagine the formulation with 3' instead of 3 is also false, but it seems slightly less trivial to get a counterexample due to the condition on the irrationality of the endpoints. In particular, no two intervals can "match up" at an irrational number unless $f$ has the same slope on both intervals.)