[proofplan] We verify the identity by evaluating both cochains on an arbitrary smooth singular $(k+1)$-simplex $\sigma:\Delta^{k+1}\to M$. Unfolding the singular coboundary expresses $(\delta I_k(\omega))(\sigma)$ as a signed sum of integrals over the faces of $\Delta^{k+1}$, which equals the integral of $\sigma^*\omega$ over the oriented topological boundary $\partial\Delta^{k+1}$. [Stokes' theorem](/theorems/1530) on the compact oriented manifold with corners $\Delta^{k+1}$ converts this boundary integral into the integral of $d(\sigma^*\omega)$ over $\Delta^{k+1}$. Naturality of the [exterior derivative](/theorems/1525) under pullback rewrites the integrand as $\sigma^*(d\omega)$, which is precisely $I_{k+1}(d\omega)(\sigma)$. [/proofplan] [step:Reduce the identity to a pointwise check on smooth singular simplices] A smooth singular cochain in $C^{k+1}_\infty(M;\mathbb{R})$ is determined by its values on smooth singular $(k+1)$-simplices, since these freely generate $C_{k+1}^\infty(M)$. Therefore, to prove the cochain identity $\delta I_k(\omega) = I_{k+1}(d\omega)$ it suffices to fix an arbitrary smooth singular $(k+1)$-simplex \begin{align*} \sigma : \Delta^{k+1} \to M \end{align*} and verify that $(\delta I_k(\omega))(\sigma) = I_{k+1}(d\omega)(\sigma)$. We regard $\Delta^{k+1} \subset \mathbb{R}^{k+2}$ as a compact oriented smooth manifold with corners of dimension $k+1$, equipped with the orientation induced from the standard orientation on the affine hyperplane $\{\sum_i t_i = 1\}$ in $\mathbb{R}^{k+2}$. [/step] [step:Unfold the left-hand side as an integral over the boundary of $\Delta^{k+1}$] The singular coboundary is the $\mathbb{R}$-linear transpose of the singular boundary: for $c \in C^k_\infty(M;\mathbb{R})$ and a smooth singular $(k+1)$-simplex $\tau$, \begin{align*} (\delta c)(\tau) := c(\partial \tau), \end{align*} where the smooth singular boundary is \begin{align*} \partial \tau = \sum_{i=0}^{k+1} (-1)^i \, \tau \circ F_i^{k+1}, \qquad F_i^{k+1} : \Delta^k \to \Delta^{k+1} \end{align*} and $F_i^{k+1}$ is the $i$-th face inclusion (the affine embedding that omits the $i$-th vertex of $\Delta^{k+1}$). Applying this with $c = I_k(\omega)$, $\tau = \sigma$ and using the definition of $I_k$: \begin{align*} (\delta I_k(\omega))(\sigma) &= I_k(\omega)(\partial \sigma) \\ &= \sum_{i=0}^{k+1} (-1)^i \int_{\Delta^k} (\sigma \circ F_i^{k+1})^* \omega. \end{align*} By contravariant functoriality of pullback for smooth maps, $(\sigma \circ F_i^{k+1})^* \omega = (F_i^{k+1})^*(\sigma^*\omega)$, so \begin{align*} (\delta I_k(\omega))(\sigma) = \sum_{i=0}^{k+1} (-1)^i \int_{\Delta^k} (F_i^{k+1})^*(\sigma^*\omega). \end{align*} The right-hand side is, by the very definition of integration of a smooth $k$-form over a singular boundary chain on the compact oriented manifold-with-corners $\Delta^{k+1}$, the integral of $\sigma^*\omega$ over the oriented topological boundary $\partial \Delta^{k+1}$: \begin{align*} (\delta I_k(\omega))(\sigma) = \int_{\partial \Delta^{k+1}} \sigma^*\omega. \end{align*} [guided] We need to convert $(\delta I_k(\omega))(\sigma)$ into a single integral over $\partial \Delta^{k+1}$. Three ingredients combine. **(i) Definition of the singular coboundary.** The map $\delta : C^k_\infty(M;\mathbb{R}) \to C^{k+1}_\infty(M;\mathbb{R})$ is the $\mathbb{R}$-linear transpose of the smooth singular boundary $\partial : C_{k+1}^\infty(M) \to C_k^\infty(M)$: \begin{align*} (\delta c)(\sigma) := c(\partial \sigma). \end{align*} **(ii) The smooth singular boundary.** On a smooth singular $(k+1)$-simplex, \begin{align*} \partial \sigma = \sum_{i=0}^{k+1} (-1)^i \, \sigma \circ F_i^{k+1}, \end{align*} where $F_i^{k+1} : \Delta^k \to \Delta^{k+1}$ is the affine map that places the vertices $e_0, \dots, e_k$ of $\Delta^k$ at the vertices of $\Delta^{k+1}$ in order, skipping the $i$-th vertex. The signs $(-1)^i$ encode the orientation induced on each face by the outward normal convention. **(iii) Definition of $I_k$ and its linear extension.** By definition $I_k(\omega)(\tau) = \int_{\Delta^k} \tau^*\omega$ for a single smooth singular $k$-simplex $\tau$. The map $I_k(\omega) \in C^k_\infty(M;\mathbb{R})$ is $\mathbb{R}$-linear on $C_k^\infty(M)$, so on the formal sum $\partial \sigma$ it gives the signed sum. Combining (i)-(iii): \begin{align*} (\delta I_k(\omega))(\sigma) = I_k(\omega)(\partial \sigma) = \sum_{i=0}^{k+1} (-1)^i \int_{\Delta^k} (\sigma \circ F_i^{k+1})^* \omega. \end{align*} Why pull pullback inside? The pullback functor satisfies $(g \circ f)^* = f^* \circ g^*$ for smooth maps $f, g$ (contravariant functoriality). With $f = F_i^{k+1}$ and $g = \sigma$: \begin{align*} (\sigma \circ F_i^{k+1})^* \omega = (F_i^{k+1})^*(\sigma^*\omega). \end{align*} This is the right rewrite because it isolates the global form $\sigma^*\omega \in \Omega^k(\Delta^{k+1})$ — a single form living on a single space — and expresses each face integral as the integral of *this* form pulled back to $\Delta^k$ along the $i$-th face inclusion. We now have: \begin{align*} (\delta I_k(\omega))(\sigma) = \sum_{i=0}^{k+1} (-1)^i \int_{\Delta^k} (F_i^{k+1})^*(\sigma^*\omega). \end{align*} This signed sum is, by definition, the integral of $\sigma^*\omega$ over the singular boundary chain $\partial \Delta^{k+1} = \sum_{i=0}^{k+1}(-1)^i \, F_i^{k+1}$, which coincides with the integral over the topological boundary $\partial \Delta^{k+1}$ equipped with the induced orientation (the alternating signs reproduce the induced orientation on each codimension-one face): \begin{align*} (\delta I_k(\omega))(\sigma) = \int_{\partial \Delta^{k+1}} \sigma^*\omega. \end{align*} [/guided] [/step] [step:Apply Stokes' theorem on the standard simplex $\Delta^{k+1}$] We invoke [Stokes' Theorem](/theorems/1530). Its hypotheses require: (a) a compact oriented smooth $(k+1)$-dimensional manifold with corners, and (b) a smooth $k$-form on a neighbourhood of this manifold. We verify both: - $\Delta^{k+1}$ is compact (closed and bounded in $\mathbb{R}^{k+2}$), oriented (by the induced orientation from the standard orientation of $\mathbb{R}^{k+2}$ restricted to the affine hyperplane $\{\sum_i t_i = 1\}$), and is a smooth $(k+1)$-manifold with corners. - The form $\sigma^*\omega$ is smooth on $\Delta^{k+1}$: since $\sigma : \Delta^{k+1} \to M$ is smooth (extending to a smooth map on a neighbourhood) and $\omega \in \Omega^k(M)$ is smooth, the pullback is smooth. [Stokes' theorem](/theorems/1530) therefore yields \begin{align*} \int_{\partial \Delta^{k+1}} \sigma^*\omega = \int_{\Delta^{k+1}} d(\sigma^*\omega). \end{align*} [guided] This is the heart of the argument. [Stokes' Theorem](/theorems/1530) states that for a compact oriented smooth $(k+1)$-manifold with corners $N$ and a smooth $k$-form $\eta$ on $N$, \begin{align*} \int_{\partial N} \eta = \int_N d\eta, \end{align*} where $\partial N$ carries the induced (Stokes) orientation. Two hypotheses must be checked. *Compactness and orientability of $N = \Delta^{k+1}$.* The standard simplex $\Delta^{k+1} \subset \mathbb{R}^{k+2}$ is closed and bounded in the ambient Euclidean space, hence compact. It is a smooth $(k+1)$-manifold with corners — its interior is an open subset of the affine hyperplane $H = \{\sum t_i = 1\}$, and each boundary stratum (face, edge, vertex) is modelled on the standard quadrant. We orient $\Delta^{k+1}$ by the orientation of $H$ inherited from the standard orientation of $\mathbb{R}^{k+2}$ via the outward normal $(1,\dots,1)/\sqrt{k+2}$ to $H$. *Smoothness of $\eta = \sigma^*\omega$.* By hypothesis $\sigma$ is a smooth singular simplex — that is, $\sigma$ extends to a smooth map on an open neighbourhood of $\Delta^{k+1}$ in $H$, and $\omega \in \Omega^k(M)$ is smooth. The pullback of a smooth form by a smooth map is smooth. Both conditions hold, so [Stokes' theorem](/theorems/1530) applies and gives the displayed identity. The induced orientation on $\partial \Delta^{k+1}$ in [Stokes' theorem](/theorems/1530) coincides with the one encoded by the alternating-sign convention $\partial \Delta^{k+1} = \sum_i (-1)^i F_i^{k+1}$ from Step 2 — this is exactly why the singular boundary operator carries those signs. [/guided] [/step] [step:Apply naturality of the exterior derivative under pullback] The [exterior derivative](/theorems/1525) is natural with respect to smooth maps: for every smooth map $f : N \to M$ between smooth manifolds and every $\omega \in \Omega^k(M)$, \begin{align*} d(f^*\omega) = f^*(d\omega) \qquad \text{in } \Omega^{k+1}(N). \end{align*} This is a defining property of the [exterior derivative](/theorems/1525) on smooth manifolds (see [Exterior Derivative](/theorems/1525)). Applying it with $f = \sigma$ and the form $\omega \in \Omega^k(M)$: \begin{align*} d(\sigma^*\omega) = \sigma^*(d\omega) \qquad \text{in } \Omega^{k+1}(\Delta^{k+1}). \end{align*} Integrating over $\Delta^{k+1}$: \begin{align*} \int_{\Delta^{k+1}} d(\sigma^*\omega) = \int_{\Delta^{k+1}} \sigma^*(d\omega). \end{align*} [/step] [step:Identify the right-hand side and conclude] By the definition of the de Rham cochain map $I_{k+1}$ applied to the form $d\omega \in \Omega^{k+1}(M)$ and the smooth singular $(k+1)$-simplex $\sigma$: \begin{align*} I_{k+1}(d\omega)(\sigma) = \int_{\Delta^{k+1}} \sigma^*(d\omega). \end{align*} Chaining the equalities established in Steps 2-4: \begin{align*} (\delta I_k(\omega))(\sigma) &= \int_{\partial \Delta^{k+1}} \sigma^*\omega &&\text{(Step 2)} \\ &= \int_{\Delta^{k+1}} d(\sigma^*\omega) &&\text{(Stokes' theorem, Step 3)} \\ &= \int_{\Delta^{k+1}} \sigma^*(d\omega) &&\text{(naturality of }d\text{, Step 4)} \\ &= I_{k+1}(d\omega)(\sigma). \end{align*} Because $\sigma$ was an arbitrary smooth singular $(k+1)$-simplex and a smooth singular cochain is determined by its values on such simplices (Step 1), we conclude \begin{align*} \delta I_k(\omega) = I_{k+1}(d\omega) \qquad \text{in } C^{k+1}_\infty(M;\mathbb{R}), \end{align*} which is the desired identity. [/step]