[proofplan] We prove the [Axiom of Choice](/page/Axiom%20of%20Choice) directly from the assumed right-inverse property for surjections. Given an arbitrary set $A$ of nonempty sets, we build the set $E$ of membership pairs $(x,a)$ with $x \in A$ and $a \in x$. The first-coordinate projection $\pi: E \to A$ is surjective precisely because each $x \in A$ is nonempty, so the hypothesis supplies a section $s: A \to E$. Reading off the second coordinate of $s(x)$ gives an element of $x$, hence a choice function on $A$. [/proofplan] [step:Build the membership-pair surjection over the family $A$] Let $A$ be an arbitrary set whose elements are nonempty sets. Define the union set $U := \bigcup A$, and define \begin{align*} E := \{(x,a) \in A \times U : a \in x\}. \end{align*} Define the first-coordinate projection map \begin{align*} \pi: E &\to A \\ (x,a) &\mapsto x. \end{align*} This map is well-defined because every element of $E$ is, by definition, an ordered pair $(x,a)$ with $x \in A$. We now verify that $\pi$ is surjective. Let $x \in A$. Since every element of $A$ is nonempty, there exists $a \in x$. Since $x \in A$ and $a \in x$, the definition of $U = \bigcup A$ gives $a \in U$, and therefore $(x,a) \in E$. Hence $\pi(x,a) = x$. Since $x \in A$ was arbitrary, $\pi: E \to A$ is surjective. [guided] We want to turn the problem of choosing one element from each $x \in A$ into a problem about splitting a surjection. The natural surjection should remember which set an element came from. For that reason, define $U := \bigcup A$, so every element belonging to some member of $A$ lies in $U$, and define \begin{align*} E := \{(x,a) \in A \times U : a \in x\}. \end{align*} Thus an element of $E$ is not merely an element $a$ of the union; it is an element $a$ together with the particular set $x \in A$ from which it is being viewed. Define \begin{align*} \pi: E &\to A \\ (x,a) &\mapsto x. \end{align*} The map is well-defined because the first coordinate of every element of $E$ lies in $A$. Now check surjectivity. Take an arbitrary $x \in A$. The hypothesis on $A$ says that $x$ is nonempty, so there exists some $a \in x$. Since $x \in A$, the condition $a \in x$ implies $a \in \bigcup A = U$. Therefore $(x,a) \in E$, and applying the projection gives \begin{align*} \pi(x,a) = x. \end{align*} Every $x \in A$ is hit by $\pi$, so $\pi: E \to A$ is surjective. [/guided] [/step] [step:Apply the surjection splitting principle to obtain a section] By the assumed surjection splitting principle applied to the surjective map $\pi: E \to A$, there exists a map \begin{align*} s: A \to E \end{align*} such that \begin{align*} \pi \circ s = \operatorname{id}_A. \end{align*} Thus, for every $x \in A$, we have $\pi(s(x)) = x$. [/step] [step:Extract a choice function from the second coordinate of the section] For each $x \in A$, the value $s(x)$ lies in $E$, so there exist elements $y \in A$ and $a \in U$ such that $a \in y$ and $s(x) = (y,a)$. Since $\pi(s(x)) = x$ and $\pi(y,a) = y$, we get $y = x$. Hence for each $x \in A$, there is a unique element $a \in U$ such that $s(x) = (x,a)$. Define \begin{align*} c: A &\to U \\ x &\mapsto a, \end{align*} where $a$ is the unique element of $U$ satisfying $s(x) = (x,a)$. By the preceding paragraph, this definition is well-defined. Moreover, since $s(x) = (x,c(x))$ lies in $E$, the defining property of $E$ gives $c(x) \in x$ for every $x \in A$. Therefore $c: A \to \bigcup A$ is a choice function for the arbitrary set $A$ of nonempty sets. Since $A$ was arbitrary, the Axiom of Choice holds. [/step]