Is Axiom of Choice necessary with this particular quotient set? Learning rigorous set theory for the first time as a freshman in UNI here. I am working through some problems regarding equivalence relations, where one of such was to prove an isomorphism $\mathbb{R}^2/_\sim\cong [0,\infty)$, where $f:\mathbb{R}^2 \rightarrow \mathbb{R}$, $f(x,y) := \sqrt{x^2 + y^2}$ and $v \sim v' \Leftrightarrow f(v)=f(v')$.
Now the idea for the proof is pretty straight forward, as $f$ is essentially the Euclidean norm on $\mathbb{R}^2$, so each equivalence class is a circle with a certain radius, except for the class $[(0,0)]_\sim$, which is obviously just the origin. Therefore one would take a representative from an equivalence class $\xi \in \mathbb{R}^2/_\sim$, say $v \in \xi$ and then use $f$ to map $v$ to $[0,\infty)$, call this whole map $F:\mathbb{R}^2/_\sim \rightarrow [0,\infty)$, where $F([v]_\sim):=f(v)$. 
Obviously, $F$ is surjective, since for any $r \in [0,\infty)$ we construct the pair $w_r:=(r,0)$, therefore $F([w_r]_\sim) = r$, as $v \in [w_r]_\sim \iff f(v)=f(w_r)=r$. Moreover, $F$ is surely injective by the definition of the relation $\sim$, therefore $F$ is a bijection, hence an isomorphism too, which completes the proof.
The question that bugged me lies at taking the representative part. I asked myself, can I always specify a unique representative for any class in the quotient with reason(i.e. can I construct a map that will take a representative with what I have), or do I need AC to "provide" me with such a map? My answer was that yes I can, following the logic bellow:


*

*For every class $\xi \in \mathbb{R}^2/_\sim$ there exists $r \in [0,\infty)$ so that $(r,0) \in \xi$. 


Proof: Let $\xi \in \mathbb{R}^2/_\sim$. By definition there exists $v \in \mathbb{R}$ so that $\xi = [v]_\sim$, so $\xi$ is not empty. We notice that for any pair $u:=(a,b) \in \mathbb{R}^2$, $f(u) = \sqrt{a^2 + b^2} = \sqrt{\sqrt{(a^2 + b^2)^2} + 0^2} = f(f(u),0)$, which means that $u\sim(f(u),0)$! Therefore, as $\xi$ is not empty, there exists $r\in[0,\infty)$ so that $(r,0)\in \xi$ (We would otherwise get that $\xi$ is empty, a contradiction).


*Such an $r \in [0,\infty)$ is unique. 


Proof: Assume $r,r' \in [0,\infty)$ and that $(r',0) \in \xi:=[(r,0)]_\sim$. Then $f(r',0)=f(r,0)$, or $\sqrt{r'^2 + 0^2} = \sqrt{r^2 + 0^2}$, thus $r=r'$, for both are nonegative adn that completes the proof.


*Because of the above proven unique existence of such an $r\in[0,\infty)$, we can define a map $S:\mathbb{R}^2/_\sim \rightarrow \mathbb{R}^2$, where $S: \xi \mapsto (r,0), where r\in[0,\infty) and (r,0)\in \xi$. Therefore this map $S$ chooses a representative for a class from the quotient $\mathbb{R}^2/_\sim$.


Is this a sound and rigorous approach? Can it be done better? Is there a possibility that I am missing something here? I really want to understand the significance of the axiom of choice, so I try to really understand where I need to apply it and where not.
Hopefully this is a well fromed and good question. I am a novice in LaTeX and MD still, so take the outline/format with a grain of salt. Thank you for your help in advance!
 A: The axiom of choice tells you that for every indexed set of sets there exists a choice function. 
But if you have a particular indexed set of sets such as your set $\{f^{-1}(r) \mid r \in [0,\infty)\}$, and if you can yourself with your bare hands construct a choice function by writing down a concrete formula such as $\xi(r)=(r,0)$, then you have no need of the axiom of choice. 
You don't need a tool factory to build a lever.  
Nonetheless, there are abstract situations where you just don't have any concrete way of producing a choice function. I don't know what your level is, but one of the first places that one encounters this situation is proving the existence of a basis of an arbitrary vector space. For finite dimensional vector spaces, in a rigorous linear algebra course one learns how to employ the Gram-Schmidt process to construct a basis, based on a finite sequence of single choices --- single vectors --- which does not rely on the axiom of choice. But for infinite dimensional vector spaces, there is no recourse but to apply the axiom of choice. 
