This may be a bit trivial an answer to this (old) question, but since a number of people have reacted with surprise when I mentioned this, and it seems to fall within the purview of what is being asked, let me state:
The Axiom of Choice is equivalent to the following statement:
If $X$ is a set, $\sim$ an equivalence relation on $X$, and $I$ a set, then the obvious map $\Phi\colon X^I/(\sim^I) \to (X/\sim)^I$$\Phi\colon X^I/(\sim^I) \to (X/{\sim})^I$ is a bijection.
(Here $X^I$ stands for the set of all $I$-indexed families of elements of $X$ and $\sim^I$ is the equivalence relation which holds between $(x_i)_{i\in I}$ and $(y_i)_{i\in I}$ in $X^I$ iff $x_i \sim y_i$ for all $i\in I$; and $\Phi$ takes the class of $(x_i)_{i\in I}$ to $([x_i])_{i\in I}$, where $[x]$ denotes the class of $x$ mod $\sim$.)
Note that, independently of AC, $\Phi$ is injective by definition of $\sim^I$.
(The equivalence with AC of the above is almost obvious. If AC holds, then we can lift $(\bar x)_{i\in I}$ in $(X/\sim)^I$$(X/{\sim})^I$ to a family $(x_i)_{\in I}$ in $X^I$ by choosing a representative for each class. Conversely, if the above holds and $(Z_i)_{i\in I}$ is a family of inhabited sets, let $X$ be their disjoint union, $\sim$ the equivalence relation for which $Z_i$ is the partition of $X$ so that $X/\sim$$X/{\sim}$ is $I$, and use the above to lift the “identity” element of $(X/\sim)^I = I^I$$(X/{\sim})^I = I^I$ to an element of $X^I$ to see that $\prod_{i\in I} Z_i$ is inhabited.)
Thus, in the above context, AC tells us that an injective map $\Phi$ is surjective, or, if we want, that every element in the target has a (necessarily unique) antecedent by $\Phi$.
