TITLE: Does the definition of the fundamental group implicitly assume the Axiom of Choice?
QUESTION [8 upvotes]: Okay, I'm a little foggy around the axiom of choice, so help me out here.
The standard way the fundamental group of a connected space $X$ is defined is as follows. You consider the set of all loops based at some point $x$. You define an equivalence relation called homotopy on the elements of this set, and the equivalence relation partitions the set into equivalence classes. Now you impose a group structure on these classes by identifying identity and defining multiplication this way: to multiply to classes $A$ and $B$, you pick out an element $a$ from $A$, and an element $b$ from $B$, and multiply the loops $a$ and $b$ and call the product of $A$ and $B$ as the equivalence class of $ab$
$$A \circ B = [a\cdot b]$$
However, picking out $a$ from $A$ and $b$ from $B$ sounds a lot like the statement of axiom of choice, which talks about picking out representative elements from disjoint sets.
So what's the deal here. I'm getting something wrong about the axiom of choice (very likely) or fundamental groups do require the axiom of choice.
Much obliged for your time and help.

REPLY [5 votes]: This is a known issue, and the way we go around it is in fact by showing that the operation of the group does not depend on the choice of representatives. This allows us to make all sort of computations with only choose two elements, at a time.
And choosing two elements from two non-empty sets is hardly a choice at all. It is simply repeating two instances of "existential instantiation", where we replace an existential quantifier by an object.
The axiom of choice is needed when you have infinitely many choices to make at once. And even this is just a general rule of thumb.

You can find similar definitions when defining the real numbers as equivalence classes of Cauchy sequences, or when defining cardinal arithmetics, or just when you usually define any type of function on a set and show that you can define an equivalence relation that this function can be "pulled" to the quotient set.