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I will state the axiom of choice and give one equivallent formulation, i am interested in proving the $\Leftarrow$ way of the theorem. I will omit details.

Axiom of choice: For collection of nonempty sets $\mathcal{X}$ there exists a choice function $f:\mathcal{X}\rightarrow \bigcup\mathcal{X}$ such that for an set $A\in\mathcal{X}$ is $f(A)\in A$. In symbols $$(\forall x)(\emptyset \notin \mathcal{X}\Rightarrow(\exists f: \mathcal{X}\rightarrow \bigcup\mathcal{X})\wedge (\forall A)(A\in X\Rightarrow f(A)\in A)$$

Theorem: The following is equivalent to axiom of choice: For any surjective function $\varphi:X\rightarrow Y$ exists $\psi:Y\rightarrow X$ s.t. $\varphi \circ \psi = \operatorname{id}_Y$.

Proof. $\Rightarrow$: Assume AC holds and let $\varphi:X\rightarrow Y$ be surjective. Assume this particular system of nonempty sets $\mathcal{X}=\{\varphi^{-1}(y)\mid y\in Y\}$. (Yes, they are nonetmpyset, because $\varphi$ is a surjection). By AC there exists some $f:\mathcal{X}\rightarrow \bigcup\mathcal{X}$ s.t. $A\in \mathcal{X}$ implies $f(A)\in A$. Apparently $\bigcup\mathcal{X}=X$. Define $g:Y\rightarrow \mathcal{X}$ by $g:y\mapsto\varphi^{-1}(y)$ and finally $\psi:Y\rightarrow X$ by $\psi:y\mapsto f(g(y))$. Now it is straightforward to show that $\varphi\circ \psi = \operatorname{id}_Y$.

Could someone potentionally help me with the other way around? Assuming the theorem and proving axiom of choice? This is what I have so far:

$\Leftarrow$: Assume the theorem holds, and assume arbitrary system of nonempty sets $\mathcal{X}$. We wish to find a function $f:\mathcal{X}\rightarrow \bigcup\mathcal{X}$ satysfying $A\in \mathcal{X} \Rightarrow f(A)\in A$.

I am quite unsure where I should be finding that $f$. Or where to make use of the surjection. Will the fact that any function $h$ can be decomposed into two $s, i$ s.t. $i\circ s = h$ and $s$ is surjective and $i$ injective, help me in any way? Help please.

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marked as duplicate by Asaf Karagila axiom-of-choice Feb 4 at 8:12

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Define $\bigsqcup \mathcal{X} = \{ (A, a) \mid A \in \mathcal{X} \text{ and } a \in A \}$. This is just the disjoint union of the sets in $X$.

The function $\bigsqcup \mathcal{X} \to \mathcal{X}$ defined by $(A,a) \mapsto A$ is surjective, so by the assumption, there exists a function $f' : \mathcal{X} \to \bigsqcup{X}$ such that $f'(A) = (A,a)$ for some $a \in A$.

Now there is an evident function $\pi : \bigsqcup \mathcal{X} \to \bigcup \mathcal{X}$ defined by $\pi(A,a) = a$. The composite $$f = \pi \circ f' : \mathcal{X} \xrightarrow{f'} \textstyle\bigsqcup \mathcal{X} \xrightarrow{\pi} \bigcup \mathcal{X}$$ is then the choice function you desire.

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