# "Surjections have Right Inverse" to the "Axiom of Choice"

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.

• 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 $\sf AC$?
– Rick
Commented Jun 24, 2020 at 9:36
• So I'm asking for a proof that why AC-disjoint is equivalent to AC.
– user785015
Commented Jun 24, 2020 at 10:19

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.