{
  "aliases": [
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  "body": "\n## Statement\n\nLet $f \\colon A \\to B$ and $g \\colon B \\to C$ be [[surjective functions|0XKH881BZ7]].\nThen the [[composition|07HF9YCWPP]] $g \\circ f \\colon A \\to C$ is surjective.\n\n## Remarks\n\nProof: let $c \\in C$. Since $g$ is surjective, $c = g(b)$ for some $b \\in B$. Since $f$ is surjective, $b = f(a)$ for some $a \\in A$. Then $c = (g \\circ f)(a)$.\nA partial converse holds: if $g \\circ f$ is surjective, then $g$ is surjective. The function $f$ need not be surjective.\nTogether with [[the result for injections|01FPXMTBX1]], this shows that a composition of [[bijections|0SNQEF7Z53]] is a bijection.\n\n## Explanations\n\n### L1 human\n\nTo reach a point $c$ of $C$, first find a point of $B$ that $g$ sends to $c$.\nThen find a point of $A$ that $f$ sends to that point of $B$.\n",
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    "name": "Composition of surjections is surjective",
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        "url": "https://en.wikipedia.org/wiki/Surjective_function"
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  "msc": [
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  "name": "Composition of surjections is surjective",
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