TITLE: Axiom of Choice (for example in the Snake Lemma)
QUESTION [6 upvotes]: If we have to make a choice, but in the end it doesn't matter what choice we made, did we really make a choice to begin with?
More explicitly, somewhere in the standard diagram-chasing proof of the snake lemma for $R$-modules (see http://mathworld.wolfram.com/SnakeLemma.html) we use the fact we have a surjective map $A \twoheadrightarrow B$ and an element $b \in B$ to deduce that there is some element $a \in A$ which maps to $b \in B$. We use this element $a$ to define a map $\mathrm{Ker}(\gamma) \to \mathrm{Coker}(\alpha)$. It turns out of course, that the map we get is independent of the choice of $a$. 
Are we really using the axiom of choice here since the choice we make is irrelevant? I understand that there are proofs of the Snake Lemma in its various forms that avoid the usage of selecting an element but I am more interested in what happens here.

REPLY [2 votes]: The specific case in question was answered, however I wish to add on the general question.
Yes, there is a use of the axiom of choice even if the actual choice is irrelevant. Consider a Vitali set, for example, that is a choice of representatives for $\mathbb{R/Q}$. We wish to show that such set is non-measurable, and the fact it is non-measurable is indeed independent of the choice of representatives. However without the axiom of choice we may not be able to make a choice at all.