I have learned that the statements
"Every surjective function has right inverse" and the "Axiom of Choice" are equivalent each other. I could easily prove the $\Longleftarrow$ direction, but it's little tricky to do the reverse direction. The problematic part is that we can reduce the AC, that is,
$$\text{For any set } I, \text{ if } \forall i\in I, A_i\text{ are nonempty sets then there exist a choice function } \\
f:I\longrightarrow \bigcup_{i\in I} A_i \text{ such that } \forall i\in I, f(i)\in A_i$$
to a statement that
$$\text{For any set } I, \text{ if } \forall i\in I, A_i\text{ are nonempty pairwise disjoint sets then there exist a choice function }\\ f:I\longrightarrow \bigcup_{i\in I} A_i \text{ such that } \forall i\in I, f(i)\in A_i.$$
So that we can construct a surjection and make the right inverse. By reducing, I found that one uses the argument like this; surjection and axiom of choice.
But my question is, what if for some $i, j\in I, i\neq j, A_i=A_j$? Then we can't use this argument, because they make new collection which is no more disjoint.
So finally, I want to know what's wrong with my counterexample. If my counterexample is appropriate, then please give a perfect proof or idea of reducing statement. Thanks for reading my long question.
Best Answer
Let us call your second statement AC-disjoint. Clearly AC implies AC-disjoint. Conversely assume that we are given a family of nonempty sets $A_i$. Define $A'_i = A_i \times \{i\} \subset (\bigcup A_i) \times I$. These sets are pairwise disjoimt. By AC-disjoint there exists a choice function $f' : I \to \bigcup A'_i$ such that $f'(i) \in A'_i$. Then $$f = p \circ f' : I \to \bigcup A_i$$ with projection $p : (\bigcup A_i) \times I \to \bigcup A_i$ is the desired choice function.