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  "body": "\n## Statement\n\nLet $\\mathcal{A}$ be an [[abelian category|Y9CZ3MTNF1]], such as the category of modules over a ring.\nConsider a [[commutative diagram|WV898P4X16]]\n$$\\begin{array}{ccccccc} A \u0026 \\to \u0026 B \u0026 \\to \u0026 C \u0026 \\to \u0026 D \\\\ \\downarrow{\\scriptstyle \\alpha} \u0026 \u0026 \\downarrow{\\scriptstyle \\beta} \u0026 \u0026 \\downarrow{\\scriptstyle \\gamma} \u0026 \u0026 \\downarrow{\\scriptstyle \\delta} \\\\ A' \u0026 \\to \u0026 B' \u0026 \\to \u0026 C' \u0026 \\to \u0026 D' \\end{array}$$\nin $\\mathcal{A}$ whose two rows are [[exact|JB7ZBFBA9C]].\n\n1. If $\\alpha$ is an [[epimorphism|RVXDSY04KG]] and $\\beta$ and $\\delta$ are [[monomorphisms|Y2NP1JFR86]], then $\\gamma$ is a monomorphism.\n2. If $\\delta$ is a monomorphism and $\\alpha$ and $\\gamma$ are epimorphisms, then $\\beta$ is an epimorphism.\n\n## Remarks\n\nPart 1 is a diagram chase.\nLet the row maps be $r \\colon A \\to B$, $s \\colon B \\to C$, $t \\colon C \\to D$ and $r', s', t'$ below.\nIf $\\gamma(x) = 0$ for $x \\in C$, then $\\delta(t(x)) = t'(\\gamma(x)) = 0$, so $t(x) = 0$ and $x = s(y)$ for some $y \\in B$.\nThen $s'(\\beta(y)) = 0$, so $\\beta(y) = r'(u')$ with $u' = \\alpha(u)$ for some $u \\in A$; now $\\beta(y - r(u)) = 0$, so $y = r(u)$ and $x = s(r(u)) = 0$.\nPart 2 is the dual statement, obtained by reversing all arrows.\nExample: take $A = A' = 0$, so that the rows are exact sequences $0 \\to B \\to C \\to D$ and $0 \\to B' \\to C' \\to D'$.\nIf $\\beta$ and $\\delta$ are monomorphisms, part 1 shows that $\\gamma$ is a monomorphism.\n\n## Explanations\n\n### L1 human\n\nThe four lemma is the five lemma split into two halves.\nOne half says when the third vertical map is one-to-one, and the other half says when the second vertical map is onto.\nEach half needs only four columns of the ladder.\n\n### L2 human\n\nThe two parts are proved by element chasing in modules, and hold in every abelian category by the [[Freyd–Mitchell embedding theorem|8V296XR2M3]] or by [[pseudo-elements|1WV2THA6NY]].\nApplying part 1 to the left four columns and part 2 to the right four columns of a five-column ladder gives the five lemma.\nThe four lemma is useful when only one of injectivity or surjectivity is needed, so fewer hypotheses have to be checked.\n",
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