[proofplan] We first show that integration descends from closed $n$-forms to degree-$n$ de Rham cohomology by [Stokes' theorem](/theorems/1530), because exact top-degree forms have integral zero on a closed manifold. Surjectivity is obtained by placing a compactly supported $n$-form of total integral $1$ inside an oriented coordinate ball. The main point is injectivity: we use the standard compactly supported exactness statement for top forms on coordinate balls, together with a [partition of unity](/page/Partition%20of%20Unity) and connectedness of $M$, to show that every top form with total integral zero differs by exact local corrections from a finite sum whose local masses cancel along overlapping coordinate balls. Each cancellation is exact, so the original form is exact. [/proofplan] [step:Show that integration descends to de Rham cohomology] Let \begin{align*} I_0: \Omega^n(M) &\longrightarrow \mathbb{R} \\ \omega &\longmapsto \operatorname{Int}_M(\omega) \end{align*} denote oriented integration of smooth top-degree forms over $M$. More explicitly, $\operatorname{Int}_M(\omega)$ is computed in positively oriented coordinate charts using Lebesgue measure $\mathcal{L}^n$ on $\mathbb{R}^n$. Since $\dim M = n$, every $\omega \in \Omega^n(M)$ is closed, because $d\omega \in \Omega^{n+1}(M) = 0$. We verify that $I_0$ vanishes on exact $n$-forms. Let $\eta \in \Omega^{n-1}(M)$. Since $M$ is closed, $\partial M = \varnothing$. By Stokes' theorem (citing a result not yet in the wiki: Stokes' Theorem), \begin{align*} \operatorname{Int}_M(d\eta) = \operatorname{Int}_{\partial M}(\eta) = \operatorname{Int}_{\varnothing}(\eta) = 0. \end{align*} Therefore $I_0(\omega + d\eta) = I_0(\omega)$ for all $\omega \in \Omega^n(M)$ and all $\eta \in \Omega^{n-1}(M)$. Hence $I_0$ induces a well-defined [linear map](/page/Linear%20Map) \begin{align*} I: H^n_{\mathrm{dR}}(M) &\longrightarrow \mathbb{R} \\ [\omega] &\longmapsto \operatorname{Int}_M(\omega). \end{align*} [guided] The first issue is not injectivity or surjectivity, but whether the displayed formula depends on the representative of the cohomology class. Define \begin{align*} I_0: \Omega^n(M) &\longrightarrow \mathbb{R} \\ \omega &\longmapsto \operatorname{Int}_M(\omega). \end{align*} Here $\operatorname{Int}_M(\omega)$ means the oriented integral of the top-degree form $\omega$ over $M$, computed in positively oriented coordinate charts using Lebesgue measure $\mathcal{L}^n$ on $\mathbb{R}^n$. This is a linear functional on smooth top-degree forms. Because $M$ has dimension $n$, there are no nonzero smooth $(n+1)$-forms, so every $n$-form is closed. Two representatives of the same de Rham class differ by an exact form. Thus we must prove that \begin{align*} \operatorname{Int}_M(d\eta) = 0 \end{align*} for every $\eta \in \Omega^{n-1}(M)$. Since $M$ is closed, it has no boundary: $\partial M = \varnothing$. Applying Stokes' theorem (citing a result not yet in the wiki: Stokes' Theorem) gives \begin{align*} \operatorname{Int}_M(d\eta) = \operatorname{Int}_{\partial M}(\eta) = \operatorname{Int}_{\varnothing}(\eta) = 0. \end{align*} Therefore adding an exact form does not change the integral, and integration descends to the quotient by exact forms: \begin{align*} I: H^n_{\mathrm{dR}}(M) &\longrightarrow \mathbb{R} \\ [\omega] &\longmapsto \operatorname{Int}_M(\omega). \end{align*} [/guided] [/step] [step:Construct a cohomology class with integral one] Choose an oriented coordinate chart $(U,\varphi)$ on $M$ such that $\varphi(U) \subset \mathbb{R}^n$ is open, and choose a relatively compact open ball $B \subset \varphi(U)$. Here $\mathcal{L}^n$ denotes Lebesgue measure on $\mathbb{R}^n$. Let $\rho \in C_c^\infty(B)$ be a nonnegative smooth function satisfying \begin{align*} \int_B \rho(y)\,d\mathcal{L}^n(y) = 1. \end{align*} Define the compactly supported $n$-form $\omega_0 \in \Omega^n(M)$ by \begin{align*} \omega_0|_U = (\rho \circ \varphi)\, d\varphi_1 \wedge \cdots \wedge d\varphi_n, \qquad \omega_0|_{M \setminus \varphi^{-1}(\operatorname{supp}\rho)} = 0. \end{align*} Because the chart is oriented, the change-of-variables formula gives \begin{align*} \operatorname{Int}_M(\omega_0) = \int_B \rho(y)\,d\mathcal{L}^n(y) = 1. \end{align*} Thus $I([\omega_0]) = 1$, and by linearity $I$ is surjective. [guided] To prove surjectivity, it is enough to build one cohomology class whose integral is $1$, because scalar multiples will then realize every real number. Choose an oriented coordinate chart $(U,\varphi)$, where \begin{align*} \varphi: U \longrightarrow \varphi(U) \subset \mathbb{R}^n \end{align*} is an orientation-preserving diffeomorphism onto its image. Choose an open Euclidean ball $B \subset \varphi(U)$ whose closure is compact and contained in $\varphi(U)$. Now choose a nonnegative bump function $\rho \in C_c^\infty(B)$ normalized by \begin{align*} \int_B \rho(y)\,d\mathcal{L}^n(y) = 1, \end{align*} where $\mathcal{L}^n$ denotes Lebesgue measure on $\mathbb{R}^n$. Such a function exists by the standard bump-function construction on Euclidean space, followed by division by its positive integral. Define $\omega_0 \in \Omega^n(M)$ by placing this Euclidean density inside the chart: \begin{align*} \omega_0|_U = (\rho \circ \varphi)\, d\varphi_1 \wedge \cdots \wedge d\varphi_n, \qquad \omega_0|_{M \setminus \varphi^{-1}(\operatorname{supp}\rho)} = 0. \end{align*} The support condition $\operatorname{supp}\rho \subset B \Subset \varphi(U)$ ensures that this extension by zero is smooth on all of $M$. Because $(U,\varphi)$ is oriented, integration of the top form in this chart becomes ordinary Lebesgue integration over $\mathbb{R}^n$. Hence \begin{align*} \operatorname{Int}_M(\omega_0) = \int_B \rho(y)\,d\mathcal{L}^n(y) = 1. \end{align*} Therefore $I([\omega_0])=1$. Since $I$ is linear, for each $a \in \mathbb{R}$ we have $I([a\omega_0])=a$, proving surjectivity. [/guided] [/step] [step:Reduce injectivity to a zero-integral exactness lemma] We use the following standard lemma. [claim:Zero-integral top forms on connected closed oriented manifolds are exact] Let $M$ be a connected closed oriented smooth $n$-manifold with $n \geq 1$. If $\omega \in \Omega^n(M)$ satisfies \begin{align*} \operatorname{Int}_M(\omega) = 0, \end{align*} then there exists $\eta \in \Omega^{n-1}(M)$ such that $\omega = d\eta$. [/claim] [proof] Choose a finite oriented coordinate-ball cover $(U_i)_{i=1}^N$ of $M$ and a smooth partition of unity $(\psi_i)_{i=1}^N$ subordinate to this cover. For each index $i$, define \begin{align*} \omega_i := \psi_i \omega \in \Omega^n(M). \end{align*} Then $\operatorname{supp}\omega_i \subset U_i$, and \begin{align*} \omega = \sum_{i=1}^N \omega_i. \end{align*} Define the [real numbers](/page/Real%20Numbers) \begin{align*} a_i := \operatorname{Int}_M(\omega_i). \end{align*} Since $\sum_i \psi_i = 1$, linearity of oriented integration gives \begin{align*} \sum_{i=1}^N a_i = \operatorname{Int}_M\left(\sum_{i=1}^N \omega_i\right) = \operatorname{Int}_M(\omega) = 0. \end{align*} For each $i$, choose a compactly supported smooth $n$-form $\theta_i \in \Omega^n(M)$ with $\operatorname{supp}\theta_i \subset U_i$ and \begin{align*} \operatorname{Int}_M(\theta_i) = 1. \end{align*} By the compactly supported top-degree Poincare lemma on a coordinate ball (citing a result not yet in the wiki: Compactly Supported Top-Degree Poincare Lemma), the form \begin{align*} \omega_i - a_i\theta_i \end{align*} is exact on $M$, because it is supported in $U_i$ and has integral zero. Thus there exists $\eta_i \in \Omega^{n-1}(M)$ such that \begin{align*} \omega_i - a_i\theta_i = d\eta_i. \end{align*} Summing over $i$ gives \begin{align*} \omega = d\left(\sum_{i=1}^N \eta_i\right) + \sum_{i=1}^N a_i\theta_i. \end{align*} It remains to prove that the finite weighted sum $\sum_i a_i\theta_i$ is exact. Because $M$ is connected and the sets $U_i$ may be chosen from a finite good cover with connected nerve, after reindexing there exists, for each $i \geq 2$, a chain of coordinate balls connecting $U_1$ to $U_i$ with consecutive overlaps nonempty. Along each overlap choose a smaller coordinate ball $V \subset U_j \cap U_k$ and a compactly supported $n$-form $\beta \in \Omega^n(M)$ with $\operatorname{supp}\beta \subset V$ and $\operatorname{Int}_M(\beta) = 1$. Applying the compactly supported top-degree Poincare lemma inside $U_j$ shows that $\theta_j-\beta$ is exact, and applying it inside $U_k$ shows that $\beta-\theta_k$ is exact. Hence $\theta_j-\theta_k$ is exact whenever $U_j$ and $U_k$ are adjacent in the chosen chain. By summing along the chain, $\theta_i-\theta_1$ is exact for every $i$. Therefore, for each $i$, there exists $\zeta_i \in \Omega^{n-1}(M)$ such that \begin{align*} \theta_i-\theta_1 = d\zeta_i. \end{align*} Using $\sum_i a_i=0$, we compute \begin{align*} \sum_{i=1}^N a_i\theta_i = \sum_{i=1}^N a_i(\theta_1+d\zeta_i) = \left(\sum_{i=1}^N a_i\right)\theta_1 + d\left(\sum_{i=1}^N a_i\zeta_i\right) = d\left(\sum_{i=1}^N a_i\zeta_i\right). \end{align*} Combining this with the previous decomposition yields \begin{align*} \omega = d\left(\sum_{i=1}^N \eta_i + \sum_{i=1}^N a_i\zeta_i\right). \end{align*} Thus $\omega$ is exact. [/proof] Now let $[\omega] \in H^n_{\mathrm{dR}}(M)$ satisfy $I([\omega])=0$. Then \begin{align*} \operatorname{Int}_M(\omega) = 0. \end{align*} By the claim, there exists $\eta \in \Omega^{n-1}(M)$ such that $\omega=d\eta$. Hence $[\omega]=0$ in $H^n_{\mathrm{dR}}(M)$, so $I$ is injective. [/step] [step:Conclude the integration map is an isomorphism] The map \begin{align*} I: H^n_{\mathrm{dR}}(M) \longrightarrow \mathbb{R} \end{align*} is well-defined by Stokes' theorem, surjective by the compactly supported normalized form constructed in an oriented coordinate ball, and injective by the zero-integral exactness lemma. Therefore $I$ is an isomorphism of real vector spaces. Consequently, \begin{align*} H^n_{\mathrm{dR}}(M) \cong \mathbb{R}. \end{align*} This proves the theorem. [/step]