[proofplan] The identity $\partial(\partial M) = 0$ at the level of de Rham currents is the dual statement of $d \circ d = 0$ on differential forms. The proof has three ingredients: (i) [Stokes' Theorem](/theorems/1530) applied to the compactly supported $(n-1)$-form $d\eta$, which converts the integral of $d(d\eta)$ over $M$ into a boundary integral of $\iota^*(d\eta)$; (ii) the identity $d \circ d = 0$ on the [Exterior Derivative](/theorems/1525), which annihilates the bulk integral; and (iii) the naturality of the [Exterior Derivative](/theorems/1525) under smooth pullback, $\iota^* \circ d = d \circ \iota^*$, which rewrites the boundary integrand as $d(\iota^* \eta)$. Combining these gives $\int_{\partial M} d(\iota^* \eta) = 0$. [/proofplan] [step:Set up the pullback to the boundary and verify the form has the right degree] Equip $\partial M$ with the induced boundary orientation. Because $M$ has dimension $n$, the boundary $\partial M$ is a smooth $(n-1)$-manifold without boundary, and the inclusion \begin{align*} \iota: \partial M &\hookrightarrow M \end{align*} is a smooth embedding. Pullback along $\iota$ gives a degree-preserving [linear map](/page/Linear%20Map) \begin{align*} \iota^*: \Omega^k(M) &\to \Omega^k(\partial M), \qquad k \in \{0, 1, \dots, n\}, \end{align*} which sends compactly supported forms to compactly supported forms. Since $\eta \in \Omega_c^{n-2}(M)$ and $n \ge 2$, the degree $n - 2 \ge 0$ is valid, and $\iota^* \eta \in \Omega_c^{n-2}(\partial M)$. The expression $d(\iota^* \eta)$ is therefore a compactly supported $(n-1)$-form on the $(n-1)$-manifold $\partial M$, and the integral $\int_{\partial M} d(\iota^* \eta)$ is well defined. [guided] Before invoking [Stokes' theorem](/theorems/1530) we record the structure of the data. The manifold $M$ has dimension $n \ge 2$ and is oriented; its boundary $\partial M$ inherits the *induced* (or *Stokes*) orientation — the unique orientation on $\partial M$ such that [Stokes' theorem](/theorems/1530) holds without sign corrections. The boundary $\partial M$ is itself a smooth manifold without boundary of dimension $n - 1$, and the inclusion \begin{align*} \iota: \partial M &\hookrightarrow M \end{align*} is a smooth embedding. Why does the hypothesis $n \ge 2$ appear? It guarantees that the degree $n - 2$ of $\eta$ is non-negative, so $\Omega_c^{n-2}(M)$ is a non-trivial space and $\eta$ is a genuine differential form rather than a formal symbol. For $n = 1$ the statement would be vacuous (there is no $(-1)$-form) and the geometric content — "the boundary of a boundary is empty" — becomes the trivial statement that the boundary of a $0$-manifold is empty. Pullback along the embedding $\iota$ is a degree-preserving graded algebra homomorphism \begin{align*} \iota^*: \Omega^\bullet(M) &\to \Omega^\bullet(\partial M) \end{align*} that sends compactly supported forms to compactly supported forms (the support of $\iota^* \alpha$ is contained in $\iota^{-1}(\operatorname{supp} \alpha)$, which is compact whenever $\operatorname{supp} \alpha$ is). In particular, $\iota^* \eta \in \Omega_c^{n-2}(\partial M)$, and its [exterior derivative](/theorems/1525) $d(\iota^* \eta)$ is a compactly supported $(n-1)$-form on $\partial M$. Because $\dim \partial M = n - 1$, the integral $\int_{\partial M} d(\iota^* \eta)$ pairs a top-degree compactly supported form against the orientation, so it converges and is well defined. [/guided] [/step] [step:Apply Stokes' Theorem to the $(n-1)$-form $d\eta$] Since $\eta \in \Omega_c^{n-2}(M)$ and the [exterior derivative](/theorems/1525) $d$ preserves compact support, the form \begin{align*} d\eta &\in \Omega_c^{n-1}(M) \end{align*} has compact support and degree $n - 1 = \dim M - 1$. The hypotheses of [Stokes' Theorem](/theorems/1530) are met: $M$ is an oriented smooth $n$-manifold with boundary, and $d\eta$ is a compactly supported smooth $(n-1)$-form on $M$. Applying [Stokes' theorem](/theorems/1530) to $d\eta$ yields \begin{align*} \int_M d(d\eta) &= \int_{\partial M} \iota^*(d\eta). \end{align*} [guided] The generalised [Stokes' Theorem](/theorems/1530) states: if $N$ is an oriented smooth $m$-manifold with boundary and $\omega \in \Omega_c^{m-1}(N)$, then \begin{align*} \int_N d\omega &= \int_{\partial N} \iota^* \omega, \end{align*} where $\partial N$ carries the induced orientation. We apply this with $N = M$, $m = n$, and $\omega = d\eta$. The required hypotheses are: (i) *$M$ is an oriented smooth $n$-manifold with boundary.* This is given. (ii) *$d\eta \in \Omega_c^{n-1}(M)$.* The [exterior derivative](/theorems/1525) is local — its support satisfies $\operatorname{supp}(d\eta) \subseteq \operatorname{supp}(\eta)$ — so the compact support of $\eta$ transfers to $d\eta$. The degree $n - 1$ is correct: $d$ raises degree by one and $\eta$ has degree $n - 2$. With these verified, Stokes gives \begin{align*} \int_M d(d\eta) &= \int_{\partial M} \iota^*(d\eta). \end{align*} The strategy is now to argue that the left-hand side vanishes (because $d^2 = 0$) and to rewrite the right-hand side as the integral we care about (because pullback commutes with $d$). [/guided] [/step] [step:Annihilate the bulk integral using $d \circ d = 0$] The [Exterior Derivative](/theorems/1525) satisfies the cochain identity $d \circ d = 0$ on $\Omega^\bullet(M)$. Applied to $\eta$, this gives $d(d\eta) = 0$ as a section of $\Lambda^n T^*M$, hence pointwise as an $n$-form on $M$. Therefore \begin{align*} \int_M d(d\eta) &= \int_M 0 \, d\mathrm{vol} = 0, \end{align*} and the Stokes identity from the previous step reduces to \begin{align*} 0 &= \int_{\partial M} \iota^*(d\eta). \end{align*} [guided] The [Exterior Derivative](/theorems/1525) on a smooth manifold $M$ is the unique graded derivation $d: \Omega^\bullet(M) \to \Omega^{\bullet + 1}(M)$ extending $df(X) = X(f)$ on functions and satisfying $d \circ d = 0$. The identity $d^2 = 0$ is the cornerstone of de Rham cohomology — it is exactly the statement that $(\Omega^\bullet(M), d)$ is a cochain complex. Applying $d^2 = 0$ to $\eta \in \Omega_c^{n-2}(M)$ gives $d(d\eta) = 0$ as a smooth section of $\Lambda^n T^*M \to M$; in particular, $d(d\eta)$ vanishes pointwise. Integrating the zero $n$-form against the orientation of $M$ gives zero: \begin{align*} \int_M d(d\eta) &= 0. \end{align*} Combined with Stokes from the previous step: \begin{align*} \int_{\partial M} \iota^*(d\eta) &= \int_M d(d\eta) = 0. \end{align*} We are one rewrite away from the stated identity: the integrand $\iota^*(d\eta)$ must be recognised as $d(\iota^* \eta)$. [/guided] [/step] [step:Rewrite the boundary integrand using naturality of $d$ under pullback] The [Exterior Derivative](/theorems/1525) is natural with respect to smooth maps: for any smooth map $F: N_1 \to N_2$ between smooth manifolds and any form $\alpha \in \Omega^k(N_2)$, \begin{align*} F^*(d\alpha) &= d(F^* \alpha). \end{align*} Applying this to the smooth inclusion $\iota: \partial M \to M$ and to the $(n-2)$-form $\eta$ gives \begin{align*} \iota^*(d\eta) &= d(\iota^* \eta) \in \Omega_c^{n-1}(\partial M). \end{align*} Substituting into the identity from the previous step: \begin{align*} \int_{\partial M} d(\iota^* \eta) &= \int_{\partial M} \iota^*(d\eta) = 0. \end{align*} [guided] The naturality identity $F^*(d\alpha) = d(F^*\alpha)$ — that pullback commutes with the [Exterior Derivative](/theorems/1525) — holds for every smooth map $F: N_1 \to N_2$ and every smooth differential form $\alpha$ on $N_2$. It is the statement that $d$ is determined by its behaviour on functions and is compatible with the chain-rule structure of pullback. We apply this with $F = \iota: \partial M \to M$ (which is smooth, being an embedding) and $\alpha = \eta \in \Omega^{n-2}(M)$, obtaining \begin{align*} \iota^*(d\eta) &= d(\iota^* \eta). \end{align*} Both sides are smooth $(n-1)$-forms on $\partial M$ with support contained in $\iota^{-1}(\operatorname{supp} \eta)$, hence compactly supported. Substituting this rewrite into the conclusion of the previous step: \begin{align*} \int_{\partial M} d(\iota^* \eta) &= \int_{\partial M} \iota^*(d\eta) \\ &= \int_M d(d\eta) && \text{(Stokes' Theorem, Step 2)} \\ &= 0 && (d^2 = 0, \text{ Step 3}). \end{align*} This is the stated identity. [/guided] [/step] [step:Interpret the result as $\partial(\partial M) = 0$] The vanishing \begin{align*} \int_{\partial M} d(\iota^* \eta) &= 0 \quad \text{for all } \eta \in \Omega_c^{n-2}(M) \end{align*} is the de Rham–dual formulation of the topological identity $\partial(\partial M) = 0$. Viewing $\partial M$ as the de Rham current $T_{\partial M} \in \mathcal{D}_{n-1}'(M)$ defined by $T_{\partial M}(\alpha) := \int_{\partial M} \iota^* \alpha$ for $\alpha \in \Omega_c^{n-1}(M)$, the boundary operator $\partial$ on currents is the adjoint of $d$: \begin{align*} (\partial T_{\partial M})(\eta) &:= T_{\partial M}(d\eta) = \int_{\partial M} \iota^*(d\eta) = \int_{\partial M} d(\iota^* \eta) = 0 \end{align*} for every $\eta \in \Omega_c^{n-2}(M)$. Thus $\partial T_{\partial M} = 0$ as a current, which is the precise statement that the boundary of the boundary is null. This completes the proof. [/step]