One of my favorite formulations of the Axiom of Choice is that for any nonempty family $A$ of nonempty sets, there is a choice function $F\colon A\to\cup A$ such that $F(X)\in X$ for each $X\in A$.
Using this, I was able to prove that there is also a function $f\colon\cup A\to A$ such that $x\in f(x)\in A$ for all $x\in\cup A$. I did this by considering any $x\in\cup A$, and letting $B_x=\{X\in A\mid x\in X\}$, and then letting $C=\{t\in\mathscr{P}(A)\mid \exists_{x\in\cup A}t=B_x\}$, i.e. $C$ is the set of all $B_x$. Now each $B_x$ is nonempty, so there is a choice function $F$ such that $F(B_x)\in B_x$ for each $B_x$. Defining $f(x)=F(B_x)$, I have $x\in f(x)\in A$ for every $x\in\cup A$. I suppose this is what I mean by a "backwards" choice function, even though it's not really one.
Is this equivalent to AC? That is, if for any set $A$, there exists a function $f\colon\cup A\to A$ such that $x\in f(x)\in A$ for all $x\in\cup A$, then AC holds? I couldn't immediately see a way to imply the Axiom of Choice, in any equivalent formulation assuming that the above is true for any set $A$. Is it just a one way implication? Thanks.