[proofplan] The map $I_k: \Omega^k(M) \to C^k_{\mathrm{sm}}(M;\mathbb{R})$ defined by integration over smooth simplices is, by [Stokes' Theorem](/theorems/1530), a cochain map: it satisfies the [cochain Stokes identity](/theorems/3594) $\delta \circ I_k = I_{k+1} \circ d$. From this single identity, both well-definedness statements follow algebraically. Closed forms map to singular cocycles and exact forms map to singular coboundaries, so $I_k$ descends to a [linear map](/page/Linear%20Map) on cohomology. The pairing statement is the dual computation: for a closed form $\omega$ and a boundary $\partial b$, the integral $\int_{\partial b}\omega$ equals $\int_b d\omega = 0$. [/proofplan] [step:Record the cochain Stokes identity satisfied by the integration map] Let $\partial: C_{k+1}^{\mathrm{sm}}(M;\mathbb{R}) \to C_k^{\mathrm{sm}}(M;\mathbb{R})$ denote the singular boundary operator, defined on a smooth $(k+1)$-simplex $\sigma: \Delta^{k+1} \to M$ by $\partial \sigma = \sum_{i=0}^{k+1} (-1)^i \sigma \circ F_i$, where $F_i: \Delta^k \to \Delta^{k+1}$ is the $i$-th face inclusion, and extended $\mathbb{R}$-linearly. Let $\delta: C^k_{\mathrm{sm}}(M;\mathbb{R}) \to C^{k+1}_{\mathrm{sm}}(M;\mathbb{R})$ denote the dual coboundary, defined by $(\delta \varphi)(\tau) := \varphi(\partial \tau)$ for $\varphi \in C^k_{\mathrm{sm}}(M;\mathbb{R})$ and $\tau \in C_{k+1}^{\mathrm{sm}}(M;\mathbb{R})$. We claim that for every $\omega \in \Omega^k(M)$, \begin{align*} \delta\, I_k(\omega) = I_{k+1}(d\omega) \quad \text{in } C^{k+1}_{\mathrm{sm}}(M;\mathbb{R}). \end{align*} Fix a smooth singular $(k+1)$-simplex $\tau: \Delta^{k+1} \to M$. By [Stokes' Theorem](/theorems/1530) applied to the smooth $(k+1)$-manifold-with-corners $\Delta^{k+1}$ and the smooth $k$-form $\tau^*\omega \in \Omega^k(\Delta^{k+1})$, we have \begin{align*} \int_{\Delta^{k+1}} d(\tau^*\omega)\, d\mathcal{L}^{k+1} = \int_{\partial \Delta^{k+1}} \tau^*\omega \, d\mathcal{H}^k, \end{align*} where $\partial \Delta^{k+1}$ carries the induced boundary orientation. The boundary of $\Delta^{k+1}$ decomposes as $\partial \Delta^{k+1} = \bigcup_{i=0}^{k+1} F_i(\Delta^k)$, with the $i$-th face contributing the sign $(-1)^i$ relative to the standard orientation on $\Delta^k$. Naturality of the [exterior derivative](/theorems/1525) gives $d(\tau^* \omega) = \tau^*(d\omega)$. Substituting, \begin{align*} \int_{\Delta^{k+1}} \tau^*(d\omega)\, d\mathcal{L}^{k+1} = \sum_{i=0}^{k+1} (-1)^i \int_{\Delta^k} (\tau \circ F_i)^* \omega \, d\mathcal{L}^k. \end{align*} By definition of $I_{k+1}$, the left-hand side equals $I_{k+1}(d\omega)(\tau)$. By definition of $\partial \tau$ and $I_k$, the right-hand side equals $I_k(\omega)(\partial \tau) = (\delta\, I_k(\omega))(\tau)$. Since $\tau$ was arbitrary, $\delta\, I_k(\omega) = I_{k+1}(d\omega)$ as claimed. [guided] The whole argument that follows rests on a single algebraic identity: $I_\bullet$ is a chain map from the de Rham complex $(\Omega^\bullet(M), d)$ to the smooth singular cochain complex $(C^\bullet_{\mathrm{sm}}(M;\mathbb{R}), \delta)$. Once we have this, well-definedness on cohomology is purely formal — exact maps to exact, closed maps to closed. So we first prove the chain-map identity. Let $\partial: C_{k+1}^{\mathrm{sm}}(M;\mathbb{R}) \to C_k^{\mathrm{sm}}(M;\mathbb{R})$ denote the singular boundary, defined on a smooth $(k+1)$-simplex $\sigma: \Delta^{k+1} \to M$ by $\partial \sigma = \sum_{i=0}^{k+1} (-1)^i \sigma \circ F_i$, where $F_i: \Delta^k \to \Delta^{k+1}$ is the $i$-th face inclusion (the affine map identifying $\Delta^k$ with the face opposite the $i$-th vertex of $\Delta^{k+1}$). Extend $\partial$ $\mathbb{R}$-linearly. The dual coboundary $\delta: C^k_{\mathrm{sm}} \to C^{k+1}_{\mathrm{sm}}$ is defined by $(\delta \varphi)(\tau) = \varphi(\partial \tau)$. We want to show: for every $\omega \in \Omega^k(M)$ and every smooth $(k+1)$-simplex $\tau$, \begin{align*} (\delta\, I_k(\omega))(\tau) = I_{k+1}(d\omega)(\tau). \end{align*} Why is this true? It is exactly [Stokes' theorem](/theorems/1530) on the standard simplex $\Delta^{k+1}$, applied to the pulled-back form $\tau^*\omega$. Fix a smooth simplex $\tau: \Delta^{k+1} \to M$. The form $\tau^*\omega$ is a smooth $k$-form on $\Delta^{k+1}$, and $\Delta^{k+1}$ is a smooth $(k+1)$-manifold with corners. [Stokes' Theorem](/theorems/1530) gives \begin{align*} \int_{\Delta^{k+1}} d(\tau^*\omega)\, d\mathcal{L}^{k+1} = \int_{\partial \Delta^{k+1}} \tau^*\omega \, d\mathcal{H}^k. \end{align*} Two facts simplify each side. First, *naturality of the [exterior derivative](/theorems/1525)*: for any smooth map $\tau$ and any smooth form $\omega$, one has $d(\tau^*\omega) = \tau^*(d\omega)$ (this is a coordinate-free identity, verified locally from the chain rule). So the left-hand integrand is $\tau^*(d\omega)$, and the left-hand side becomes $\int_{\Delta^{k+1}} \tau^*(d\omega)\, d\mathcal{L}^{k+1} = I_{k+1}(d\omega)(\tau)$ by definition of $I_{k+1}$. Second, the boundary $\partial \Delta^{k+1}$ decomposes as the union of its $(k+2)$ codimension-one faces $F_i(\Delta^k)$, $i = 0, \dots, k+1$. The induced boundary orientation on the $i$-th face differs from the standard orientation on $\Delta^k$ (transported via $F_i$) by the sign $(-1)^i$. Hence \begin{align*} \int_{\partial \Delta^{k+1}} \tau^*\omega \, d\mathcal{H}^k = \sum_{i=0}^{k+1} (-1)^i \int_{\Delta^k} F_i^*(\tau^*\omega)\, d\mathcal{L}^k = \sum_{i=0}^{k+1} (-1)^i \int_{\Delta^k} (\tau \circ F_i)^*\omega \, d\mathcal{L}^k, \end{align*} using $F_i^* \tau^* = (\tau \circ F_i)^*$ (contravariance of pullback). By definition of $I_k$ and $\partial\tau$, this last sum is \begin{align*} \sum_{i=0}^{k+1} (-1)^i I_k(\omega)(\tau \circ F_i) = I_k(\omega)(\partial \tau) = (\delta\, I_k(\omega))(\tau). \end{align*} Combining the two sides of Stokes gives $(\delta\, I_k(\omega))(\tau) = I_{k+1}(d\omega)(\tau)$. Since $\tau$ was arbitrary, $\delta \, I_k(\omega) = I_{k+1}(d\omega)$. This is the *[cochain Stokes identity](/theorems/3594)*. The signs in the singular boundary $\partial$ were rigged precisely to make this hold — that is the conceptual content of the alternating-sum definition. [/guided] [/step] [step:Closed forms map to singular cocycles] Let $\omega \in \Omega^k(M)$ be closed, i.e., $d\omega = 0$. By the [cochain Stokes identity](/theorems/3594) established above, \begin{align*} \delta\, I_k(\omega) = I_{k+1}(d\omega) = I_{k+1}(0) = 0, \end{align*} using $\mathbb{R}$-linearity of $I_{k+1}$. Hence $I_k(\omega) \in \ker \delta \subseteq C^k_{\mathrm{sm}}(M;\mathbb{R})$ is a smooth singular $k$-cocycle, so the class $[I_k(\omega)] \in H^k_{\mathrm{sing}}(M;\mathbb{R})$ is defined. [/step] [step:Exact forms map to singular coboundaries] Let $\omega \in \Omega^k(M)$ be exact, so $\omega = d\eta$ for some $\eta \in \Omega^{k-1}(M)$ (this presupposes $k \ge 1$; for $k = 0$ there are no exact forms and the step is vacuous). Applying the [cochain Stokes identity](/theorems/3594) from the previous step with $\omega$ replaced by $\eta$, \begin{align*} I_k(\omega) = I_k(d\eta) = \delta\, I_{k-1}(\eta). \end{align*} Hence $I_k(\omega) \in \operatorname{im} \delta$ is a smooth singular $k$-coboundary, and its cohomology class $[I_k(\omega)] = 0$ in $H^k_{\mathrm{sing}}(M;\mathbb{R})$. [/step] [step:Conclude that $I$ descends to a well-defined $\mathbb{R}$-linear map on cohomology] Combining the previous two steps, $I_k$ restricts to a map $Z^k_{\mathrm{dR}}(M) \to Z^k_{\mathrm{sing}}(M;\mathbb{R})$ between closed forms and singular cocycles, and sends the subspace $B^k_{\mathrm{dR}}(M)$ of exact forms into the subspace $B^k_{\mathrm{sing}}(M;\mathbb{R})$ of coboundaries. By the universal property of quotient vector spaces, the composition \begin{align*} Z^k_{\mathrm{dR}}(M) \xrightarrow{I_k} Z^k_{\mathrm{sing}}(M;\mathbb{R}) \twoheadrightarrow H^k_{\mathrm{sing}}(M;\mathbb{R}) \end{align*} vanishes on $B^k_{\mathrm{dR}}(M)$ and therefore factors uniquely through the quotient $H^k_{\mathrm{dR}}(M) = Z^k_{\mathrm{dR}}(M)/B^k_{\mathrm{dR}}(M)$, producing a well-defined map \begin{align*} I: H^k_{\mathrm{dR}}(M) &\to H^k_{\mathrm{sing}}(M;\mathbb{R}) \\ [\omega] &\mapsto [I_k(\omega)]. \end{align*} Linearity of $I$ follows from $\mathbb{R}$-linearity of $I_k$ (immediate from linearity of the integral and of pullback in the form argument) together with $\mathbb{R}$-linearity of the two quotient projections. [guided] We have shown: - (Step 2) $I_k$ sends closed forms to cocycles: $I_k(Z^k_{\mathrm{dR}}) \subseteq Z^k_{\mathrm{sing}}$. - (Step 3) $I_k$ sends exact forms to coboundaries: $I_k(B^k_{\mathrm{dR}}) \subseteq B^k_{\mathrm{sing}}$. The construction of the induced map on cohomology is now formal. The quotient $H^k_{\mathrm{dR}}(M) = Z^k_{\mathrm{dR}}(M)/B^k_{\mathrm{dR}}(M)$ has the *universal property*: any $\mathbb{R}$-[linear map](/page/Linear%20Map) $f: Z^k_{\mathrm{dR}}(M) \to V$ (for $V$ any real [vector space](/page/Vector%20Space)) which vanishes on $B^k_{\mathrm{dR}}(M)$ factors uniquely through a linear $\bar f: H^k_{\mathrm{dR}}(M) \to V$. Apply this to \begin{align*} f := \pi \circ I_k\big|_{Z^k_{\mathrm{dR}}(M)}: Z^k_{\mathrm{dR}}(M) \to H^k_{\mathrm{sing}}(M;\mathbb{R}), \end{align*} where $\pi: Z^k_{\mathrm{sing}}(M;\mathbb{R}) \twoheadrightarrow H^k_{\mathrm{sing}}(M;\mathbb{R})$ is the quotient projection. By Step 3, $f$ vanishes on $B^k_{\mathrm{dR}}(M)$ (exact forms hit $B^k_{\mathrm{sing}}$, which $\pi$ kills). So we get the unique factorisation $I: H^k_{\mathrm{dR}}(M) \to H^k_{\mathrm{sing}}(M;\mathbb{R})$ with $I([\omega]) = [I_k(\omega)]$. For linearity: $I_k$ is $\mathbb{R}$-linear because $\omega \mapsto \int_{\Delta^k} \sigma^*\omega \, d\mathcal{L}^k$ is linear in $\omega$ (pullback is linear in the form, and the integral is linear). The quotient projections are linear by construction. The composition of linear maps factoring through a quotient is linear. So $I$ is $\mathbb{R}$-linear. [/guided] [/step] [step:Verify the pairing depends only on the homology class of the cycle] Let $\omega \in Z^k_{\mathrm{dR}}(M)$ be a closed $k$-form and let $c \in C_k^{\mathrm{sm}}(M;\mathbb{R})$ be a smooth singular $k$-cycle, i.e., $\partial c = 0$. The pairing value $\int_c \omega := I_k(\omega)(c)$ is the evaluation of the cocycle $I_k(\omega) \in Z^k_{\mathrm{sing}}(M;\mathbb{R})$ (cocycle by Step 2) on the cycle $c \in Z_k^{\mathrm{sm}}(M;\mathbb{R})$. Replace $c$ by another representative $c + \partial b$ of the same homology class, where $b \in C_{k+1}^{\mathrm{sm}}(M;\mathbb{R})$. Using $\mathbb{R}$-linearity of $I_k(\omega)$ and the [cochain Stokes identity](/theorems/3594) from Step 1, \begin{align*} \int_{c + \partial b} \omega - \int_c \omega &= I_k(\omega)(\partial b) = (\delta\, I_k(\omega))(b) = I_{k+1}(d\omega)(b) = I_{k+1}(0)(b) = 0, \end{align*} where the penultimate equality uses $d\omega = 0$. Hence $\int_c \omega$ depends only on $[c] \in H_k^{\mathrm{sing}}(M;\mathbb{R})$. Replace $\omega$ by another representative $\omega + d\eta$ of the same de Rham class, where $\eta \in \Omega^{k-1}(M)$. By $\mathbb{R}$-linearity of $I_k$ and the [cochain Stokes identity](/theorems/3594), \begin{align*} \int_c (\omega + d\eta) - \int_c \omega = I_k(d\eta)(c) = (\delta\, I_{k-1}(\eta))(c) = I_{k-1}(\eta)(\partial c) = I_{k-1}(\eta)(0) = 0, \end{align*} using $\partial c = 0$. Hence the pairing depends only on $[\omega] \in H^k_{\mathrm{dR}}(M)$. Combining both invariances, the pairing \begin{align*} H^k_{\mathrm{dR}}(M) \times H_k^{\mathrm{sing}}(M;\mathbb{R}) &\to \mathbb{R} \\ ([\omega], [c]) &\mapsto \int_c \omega \end{align*} is well defined, and by construction it satisfies $I([\omega])([c]) = \int_c \omega$. This completes the proof. [/step]