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  "body": "\n## Statement\n\nLet $X$ be a topological space and let $A \\subseteq X$ be a subspace.\nLet $i \\colon A \\to X$ be the inclusion and $j \\colon (X, \\emptyset) \\to (X, A)$ the identity map seen as a map of pairs.\nFor $n \\ge 1$ let $\\partial \\colon H_n(X, A) \\to H_{n-1}(A)$ send the class of a relative cycle $c$ to the class of the cycle $\\partial c$ in $A$.\nThen the sequence of [[singular homology|M0Q5CFJ7ZH]] and [[relative homology|4BVWH0BN28]] groups\n$$\\cdots \\to H_n(A) \\xrightarrow{i_*} H_n(X) \\xrightarrow{j_*} H_n(X, A) \\xrightarrow{\\partial} H_{n-1}(A) \\xrightarrow{i_*} \\cdots \\to H_0(X) \\xrightarrow{j_*} H_0(X, A) \\to 0$$\nis [[exact|JB7ZBFBA9C]].\nIt is natural with respect to maps of pairs.\n\n## Remarks\n\nThe chain complexes form a short exact sequence $0 \\to C_\\bullet(A) \\to C_\\bullet(X) \\to C_\\bullet(X, A) \\to 0$.\nThe long exact sequence in homology of a short exact sequence of chain complexes gives the result, and its connecting map is the $\\partial$ above.\nIf $A$ is nonempty, the same sequence holds with [[reduced homology|NJQMWVK3QK]] $\\tilde H_n(A)$ and $\\tilde H_n(X)$ in place of $H_n(A)$ and $H_n(X)$.\n\nExample: for $(D^n, S^{n-1})$ with $n \\ge 1$, the disc has $\\tilde H_k(D^n) = 0$ for all $k$.\nExactness of the reduced sequence then forces $\\partial \\colon H_k(D^n, S^{n-1}) \\to \\tilde H_{k-1}(S^{n-1})$ to be an isomorphism for every $k$.\n\n## Explanations\n\n### L1 human\n\nThe homology of a space, of a subspace, and of the space relative to the subspace are tied together in one long chain of maps.\nAt each spot, what one map kills is exactly what the previous map produces.\nSo knowing two of the three kinds of groups often determines the third.\n\n### L2 human\n\nThis is the exactness axiom of the Eilenberg–Steenrod axioms.\nCombined with excision, it computes the homology of spheres by induction on dimension, and it gives the long exact sequences behind cellular homology.\nThe same algebra gives a long exact sequence for a triple $B \\subseteq A \\subseteq X$ and, dually, for singular cohomology with the arrows reversed.\n",
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