A canonical injection $f:\alpha\to[0,1]$ for an arbitrary countable ordinal $\alpha$ Is there a canonical injection $f:\alpha\to[0,1]$ for an arbitrary countable ordinal $\alpha$?
As long as $\alpha$ is of the form $m\cdot\omega + n$ for $m,n<\omega$, this is fairly simple, but if powers are involved, this might be more difficult.
Edit: I am looking for an $f$ which is order-preserving.
 A: The answer is sorta negative.
Of course I'm saying "sorta" because I have to interpret the term "canonical" on my own. But there are a few very good reasons for this.


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*The axiom of choice plays a big role in here. It turns out that while every countable ordinal can be embedded in $[0,1]$ (in fact, in $\Bbb Q\cap[0,1]$), in order to find a sequence of injective functions $\langle f_\alpha\colon\alpha\to[0,1]\mid\alpha<\omega_1\rangle$ one has to resort to the axiom of choice. Even if you don't care about order preservation. Even if you replace $[0,1]$ by $\omega$.
It is consistent that the axiom of choice fails and there are no such sequences. Of course if something is "canonical" then we expect it to remain canonical when the axiom of choice is removed. But then we run into the above problem, the axiom is choice is needed for constructing any such sequence.

*Being countable is not absolute, different models of set theory may disagree on which ordinals are countable. More precisely, it is possible to have two models of set theory which are "very nice" and have the same ordinals, but they disagree on which ordinals are countable. 
Of course, this by itself is not sufficiently convincing to explain why there isn't a canonical embedding like that. But it can be used as a hint of something like that. Of course, one can require that the canonical embedding of $\alpha$ exists in every model where $\alpha$ is countable. But it doesn't quite cut it. 
The reason is that there is no "canonical" function which makes $\alpha$ countable. One could argue that such function is a generic function for the Levy collapse of $\alpha$ (whatever that means) which enumerates $\alpha$, but generic sets are probably the least canonical objects in the universe of set theory. So if a canonical function depends on how you enumerate $\alpha$, it's not canonical anymore.
Of course there are reasons of computability, as Zhen Lin notes in the comments, and there is the issue that you haven't given a concrete definition of canonical, and whether or not we are going to consider it in one fixed universe of set theory, and if so, what are the axioms it satisfies, and so on.
But the two reasons given above are sufficient, for me, to feel that there shouldn't be a canonical sequence such as that you are looking for.
