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For the theorem, "Let A and B be sets and let $f : A \rightarrow B$ be a function. Then $f : A \rightarrow B$ is $surjective$ $iff$ there exists a function $g : B \rightarrow A$ such that $f \circ g = I_B$. "

The theorem implies Axiom of choice.

I proved the statement, but I want to know whether my proof is right or not.

Proof)

Suppose the theorem. Let A be a set and $Q_R$=$\text{{(R,r)|$r\in R$}}$ for all $R\in \mathscr P'(A)$ . Let $Q = \cup_{R\in\mathscr P'(A)} Q_R$.

Let $f: Q \rightarrow \mathscr P'(A)$ where $(R,r) \mapsto R$. Then, $\exists (R,r)\in Q$ for $\forall R \in \mathscr P'(A)$. So $f$ is surjective. By theorem, $\exists g: \mathscr P'(A) \rightarrow Q $ such that $f \circ g = I_B$. Then, $g(R) = (R,r)$ for some $ r\in R$.

So, let $\gamma : \mathscr P'(A) \rightarrow A$ such that $\gamma(R) = r$ where $\gamma(R) \in R$ for all $R \in \mathscr P'(A).$ Hence, Axiom of choice holds.

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    $\begingroup$ Yep, that does the trick. Good job! $\endgroup$ – Stefan Mesken May 26 '17 at 13:59
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Your proof is mostly fine. Note that you can define $Q$ more clearly as $$\{(R,r)\mid r\in R\subseteq A\},$$ or if you prefer a more explicit statement, $$\{(R,r)\mid R\subseteq A\text{ and }r\in R\}.$$

In the last line, however, you messed up a little bit. You just claimed that a choice function $\gamma$ exists. This requires proof, all that you know is that there is a function $g$ which is the inverse of $f$. You need to use that to define $\gamma$. In this case, $\gamma(R)=r$ if and only if $g(R)=(R,r)$.

Now you need to argue that $\gamma$ is a choice function indeed, which is not difficult from the way we picked $f$, and what we know about $g$.

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