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  "body": "\n## Proof\n\nWe prove the two directions separately, and we use no choice.\n\n1. Let $f \\colon A \\to B$ be a [[function|0XHDTE5YMZ]]. Here $\\circ$ is [[composition|07HF9YCWPP]] and $\\mathrm{id}$ is the [[identity function|0FSNQ0GNCC]].\n2. First suppose $f$ is [[bijective|0SNQEF7Z53]], that is, [[injective|0PVG1CRAJN]] and [[surjective|0XKH881BZ7]].\n3. Let $b \\in B$. Since $f$ is surjective, some $a \\in A$ has $f(a) = b$. If also $f(a') = b$, then $f(a) = f(a')$, so $a = a'$ by injectivity. So exactly one $a \\in A$ has $f(a) = b$.\n4. Define $g \\colon B \\to A$ by letting $g(b)$ be the unique $a$ of step 3. Then $g$ is a function, and $f(g(b)) = b$ for each $b \\in B$. So $f \\circ g = \\mathrm{id}_B$.\n5. Let $a \\in A$. By step 4, $f(g(f(a))) = f(a)$. Since $f$ is injective, $g(f(a)) = a$. So $g \\circ f = \\mathrm{id}_A$.\n6. Now suppose that $g \\colon B \\to A$ satisfies $g \\circ f = \\mathrm{id}_A$ and $f \\circ g = \\mathrm{id}_B$.\n7. Let $a, a' \\in A$ with $f(a) = f(a')$. Applying $g$ gives $g(f(a)) = g(f(a'))$. Since $g \\circ f = \\mathrm{id}_A$, this says $a = a'$. So $f$ is injective.\n8. Let $b \\in B$ and put $a = g(b)$. Since $f \\circ g = \\mathrm{id}_B$, we get $f(a) = f(g(b)) = b$. So $f$ is surjective.\n9. By steps 7 and 8, $f$ is bijective.\n10. By steps 4, 5 and 9, $f$ is bijective if and only if there is a function $g \\colon B \\to A$ with $g \\circ f = \\mathrm{id}_A$ and $f \\circ g = \\mathrm{id}_B$.\n\n## Idea\n\nFor a bijection, each $b$ has exactly one preimage, and sending $b$ to it undoes $f$ on both sides. Conversely, $g \\circ f = \\mathrm{id}_A$ forces $f$ to be injective, and $f \\circ g = \\mathrm{id}_B$ forces $f$ to be surjective.\n",
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