Do I need axiom of choice to construct $sqrt x$?

I had a lecture concerning axiom of choice, and the teacher says that an equivalent property is that surjective maps are invertible at right. After, as an example, we had the function $\sqrt x$. The example goes as follow : $x\mapsto x^2$ is surjective $\mathbb R\to \mathbb R^+$. We are looking for function $f:\mathbb R^+\to \mathbb R$ s.t. $f(x)^2=x$. There are infinitely many such function, and there existence require axiom of choice. Strictly speaking, $x\mapsto \sqrt x$ is rather a representative of the class of the function $f$ s.t. $f(x)^2=x$ than a function

Question : For fix $x$, the equation $y^2=x$ has always $2$ solutions. Let denote $A_x=\{y\mid y^2=x\}$. Let $g(x)=\max A_x$. Then, $g(x)$ is really a function and not a representative. Moreover, I didn't need axiom of choice, right ? So why should we see an inverse as a representative class rather than as a function ?

Answer

We can prove, without the axiom of choice, that $A_x=\{y\in\Bbb R\mid y^2=x\}$ has at most $2$ elements for every $x\geq 0$.

Since $\Bbb R$ is linearly ordered, there is a "largest" and "smallest" of these two elements, so we can choose the largest. The axiom of choice is not needed.

Well, you might say, what happens when we want to consider $\Bbb C$ instead? Since there is no order on $\Bbb C$, then how do we do that? Well. There is no order which is compatible with the field, but there is an order on the set $\Bbb C$ (e.g. the lexicographic ordering) which we can fix and use to make our choices, of course there is some algebraic issue there with choosing a root for $-1$, but this is a single additional choice we need to make.

This works for any $\sqrt[n]x$ function as well, since there are at most $n$ roots in $\Bbb R$.

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To your other question, choosing a representative is in fact a function. It might not be spelled out in fancy formulas like $f(x)=x^2$ is given. But it is a function. That is the whole point of a representative, that it is unique for any two equivalent objects (in this case, any set of numbers with the same square).

I guess that whomever said "representative" to you was trying to point out that this is not somehow a unique choice, and to some extent we could also require the root to be the negative root. Although there are algebraic arguments against that (for example, you can't take $\sqrt{\sqrt 4}$ since $-\sqrt2$ does not have a real root).