I am reading a paper where a side remark said that if a sequence $\langle g_\beta\colon\omega\to\beta\mid\beta<\omega_1\rangle$ is a sequence of surjections then $\omega_1$ is regular.
I have tried to proved the regularity of $\omega_1$ from this sequence by taking $\displaystyle\beta=\bigcup_{n<\omega}\beta_n$ for some countable ordinals and showing that $\beta$ is countable, i.e. to construct a surjection from $\omega$ onto $\beta$.
Let $p(k)=(r,s)$ the Cantor pairing function (or any other bijection of $\omega$ with $\omega^2$) then $F(n) = g_r(s) \text{ where } p(n)=(r,s)$
Then we have that for every $\alpha<\beta$ there is some $n<\omega$ such that $\alpha<\beta_n$, therefore $\alpha\in\operatorname{Rng}(g_n)$ therefore there is some $m<\omega$ such that $g_n(m)=\alpha$ and so $F(p^{-1}(n,m))=g_n(m)=\alpha$.
My question is (now that I have proved this fact) why this sequence of surjections cannot be created without some choice?
It seems to me that despite the fact that $\omega_1$ is singular, it is still true that for every $\alpha<\omega_1$ there exists a bijection with $\omega$. Can't there be some canonical choice of bijections?
For example, one can take $L^V$ where the GCH (and therefore choice) holds, define the sequence of surjections by taking the minimal in $<_L$ (the canonical ordering of $L$) for each ordinal.
Of course $\omega_1^L$ need not be $\omega_1^V$ but the argument why at some point all the bijections between countable ordinals and $\omega$ remain undefinable is unclear to me.