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  "body": "\n## Statement\n\nLet $R$ be a [[ring|0894J980RK]], let $I$ be a two-sided ideal of $R$ and let $\\pi \\colon R \\to R/I$ be the quotient map.\nLet $f \\colon R \\to S$ be a [[ring homomorphism|0GGE0Z0JQ1]] with $I \\subseteq \\ker f$.\nThen there is exactly one ring homomorphism $\\bar{f} \\colon R/I \\to S$ such that $\\bar{f} \\circ \\pi = f$.\nIt is given by $\\bar{f}(a + I) = f(a)$.\n\n## Remarks\n\nThe map $\\bar{f}$ is well defined: if $a + I = b + I$, then $a - b \\in I \\subseteq \\ker f$, so $f(a) = f(b)$.\nThe image of $\\bar{f}$ is the image of $f$, and $\\ker \\bar{f} = (\\ker f)/I$.\nWith $I = \\ker f$, this gives the [[first isomorphism theorem|0ZMYX7GMH0]].\n\n## Explanations\n\n### L1 human\n\nA homomorphism that sends all of $I$ to zero does not see the difference between $a$ and $a + x$ for $x \\in I$.\nSo it already makes sense on the cosets of $I$.\nThis gives a homomorphism from the quotient ring.\n",
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