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  "body": "\n## Proof\n\nWe prove the two directions separately. Only the second direction uses the axiom of choice.\n\n1. Let $f \\colon A \\to B$ be a [[function|0XHDTE5YMZ]].\n2. First suppose that $f$ has a [[right inverse|0DDPZAG5ER]] $h \\colon B \\to A$. So $f(h(b)) = b$ for each $b \\in B$.\n3. Let $b \\in B$ and put $a = h(b) \\in A$. By step 2, $f(a) = b$. So $f$ is [[surjective|0XKH881BZ7]].\n4. Now suppose that $f$ is surjective. For $b \\in B$, let $F_b = f^{-1}(\\set{b}) = \\set{a \\in A \\mid f(a) = b}$ be the [[preimage|0FGCFY1AZX]] of $\\set{b}$.\n5. Since $f$ is surjective, each $F_b$ is nonempty. Let $X = \\set{F_b \\mid b \\in B}$. Then $X$ is a set of nonempty sets.\n6. By the [[axiom of choice|0HTGB13FE2]], there is a [[choice function|0YXEQWRGP0]] $c \\colon X \\to \\bigcup X$ with $c(S) \\in S$ for each $S \\in X$.\n7. Each $F_b$ is a subset of $A$, so $\\bigcup X \\subseteq A$. Define $h \\colon B \\to A$ by $h(b) = c(F_b)$.\n8. Let $b \\in B$. By step 6, $h(b) = c(F_b) \\in F_b$. By the definition of $F_b$, this says $f(h(b)) = b$.\n9. By step 8, $f \\circ h = \\mathrm{id}_B$, where $\\mathrm{id}_B$ is the [[identity function|0FSNQ0GNCC]] and $\\circ$ is [[composition|07HF9YCWPP]]. So $h$ is a right inverse of $f$.\n10. By steps 3 and 9, $f$ is surjective if and only if $f$ has a right inverse.\n\n## Idea\n\nA right inverse must pick, for each $b$, one point of $A$ that $f$ sends to $b$. Surjectivity says such points exist, and the axiom of choice lets us pick one for every $b$ at once. Conversely, the picked point $h(b)$ shows that $b$ is hit.\n",
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