{
  "aliases": [
    "injective-comp"
  ],
  "backlinks": null,
  "body": "\n## Statement\n\nLet $f \\colon A \\to B$ and $g \\colon B \\to C$ be [[injective functions|0PVG1CRAJN]].\nThen the [[composition|07HF9YCWPP]] $g \\circ f \\colon A \\to C$ is injective.\n\n## Remarks\n\nProof: if $g(f(a)) = g(f(a'))$, then $f(a) = f(a')$ because $g$ is injective, and $a = a'$ because $f$ is injective.\nA partial converse holds: if $g \\circ f$ is injective, then $f$ is injective. The function $g$ need not be injective.\n\n## Explanations\n\n### L1 human\n\nThink of $f$ and $g$ as machines, and of $g \\circ f$ as the two machines in a chain.\nIf neither machine merges two inputs, the chain does not merge two inputs either.\n",
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    "aliases": [
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    "name": "Composition of injections is injective",
    "role": "proposition",
    "sources": [
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        "url": "https://en.wikipedia.org/wiki/Injective_function"
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  "msc": [
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  "name": "Composition of injections is injective",
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  "role": "proposition",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/01F/01FPXMTBX1.md",
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  "tag": "01FPXMTBX1",
  "uri": "m2://01FPXMTBX1",
  "url": "https://math2.tamnd.com/wiki/composition-of-injections-is-injective/",
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