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  "body": "\n## Statement\n\nLet $M$ be a [[smooth manifold|F45VQ47GPV]].\nLet $\\omega$ be a closed $k$-form on $M$ and let $c$ be a smooth singular $k$-cycle in $M$.\nPut $\\langle [\\omega], [c] \\rangle = \\int_c \\omega$.\nLet $H^k_{\\mathrm{dR}}(M)$ be the [[de Rham cohomology|GX74C07T0G]] of $M$ and $H^k(M; \\R)$ its [[singular cohomology|6N7BC70CX0]] with real coefficients.\nFor every $k \\ge 0$, this pairing induces an isomorphism of real vector spaces\n$$H^k_{\\mathrm{dR}}(M) \\cong H^k(M; \\R).$$\n\n## Remarks\n\nThe map is well defined by the [[generalised Stokes theorem|0HPHKCGEQ0]].\nIt gives that the integral of an exact form over a cycle is $0$, and so is the integral of a closed form over a boundary.\nOne proof shows that both sides satisfy the Mayer–Vietoris sequence and agree on convex open subsets of $\\R^n$.\nOn such sets the Poincaré lemma gives the de Rham side.\nThen one inducts over a good cover of $M$.\nAnother proof says that both sides compute the sheaf cohomology of the constant sheaf $\\R$ on $M$.\nThe isomorphism also takes the wedge product of forms to the [[cup product|JX47MS1R1H]], so it is an isomorphism of graded rings.\n\nExample: on the circle $S^1$ with angle coordinate $\\theta$, the closed form $d\\theta$ is not exact.\nIts integral over the circle is $2\\pi \\neq 0$, so its class corresponds to a nonzero class in $H^1(S^1; \\R) \\cong \\R$.\n\nThe coefficients must be real (or contain $\\R$): differential forms do not see torsion, and $H^k(M; \\Z)$ can have torsion, as for the real projective plane.\n\n## Explanations\n\n### L1 human\n\nOn a smooth space there are two ways to count holes.\nOne uses calculus: differential forms whose derivative is zero but which are not derivatives themselves.\nThe other uses topology: cycles that are not boundaries.\nThe de Rham theorem says that the two counts agree, and the matching is given by integration.\n\n### L2 human\n\nThe theorem lets you compute topological invariants of a manifold with calculus, and conversely.\nFor example, it shows that the de Rham cohomology is a topological invariant, though it is defined with smooth structure.\nIt is the model for comparison theorems in other settings, such as the comparison between algebraic de Rham cohomology and singular cohomology of a complex variety.\nTogether with Poincaré duality and Hodge theory it is a basic tool of global analysis on manifolds.\n",
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