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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.

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  • $\begingroup$ I don't understand your first question. Assuming that every surjective function has a right inverse we can show that $\sf AC$ holds for pairwise disjoint sets, and since this version of $\sf AC$ is equivalent to the usual one, you can prove thus the existence of a choice function for the case of existing $i,j \in I$ with $i \neq j$ and $A_i \neq A_j$. Perhaps you're asking for a proof which doesn't involve the pairwise disjoint set version of $\sf AC$? $\endgroup$
    – Rick
    Commented Jun 24, 2020 at 9:36
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    $\begingroup$ So I'm asking for a proof that why AC-disjoint is equivalent to AC. $\endgroup$
    – user785015
    Commented Jun 24, 2020 at 10:19

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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.

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  • $\begingroup$ Thanks! Actually I did this type of proof before but I wasn't quite sure. I maybe confused some concepts like union and cartesian product, ordered pair somehow. After your proof it became very clear now. $\endgroup$
    – user785015
    Commented Jun 24, 2020 at 10:56

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