[proofplan] The proof proceeds by localisation. We cover $\operatorname{supp}\omega$ by finitely many oriented coordinate charts modelled on either $\mathbb{R}^n$ (interior charts) or the closed upper half-space $\mathbb{H}^n = \{x \in \mathbb{R}^n : x_n \geq 0\}$ (boundary charts), choose a smooth [partition of unity](/page/Partition%20of%20Unity) subordinate to this cover, and split $\omega$ into a finite sum of pieces each supported in a single chart. Naturality of the [exterior derivative](/theorems/1525) and the change-of-variables formula reduce the identity to two model computations on $\mathbb{R}^n$ and $\mathbb{H}^n$. In both models, [Fubini's theorem](/theorems/2961) and the one-dimensional [Fundamental Theorem of Calculus](/theorems/632) reduce $\int d\eta$ to a boundary integral; in the interior case every term vanishes by compact support, while in the boundary case exactly one term survives and matches $\int_{\partial \mathbb{H}^n} \iota^*\eta$ with the correct sign prescribed by the boundary orientation convention. [/proofplan]

[step:Localise via a partition of unity to charts on $\mathbb{R}^n$ or $\mathbb{H}^n$]Set $\mathbb{H}^n := \{x = (x_1, \dots, x_n) \in \mathbb{R}^n : x_n \geq 0\}$, so $\partial \mathbb{H}^n = \{x_n = 0\}$. Since $K := \operatorname{supp}\omega \subset M$ is compact, cover $K$ by finitely many oriented smooth charts $(U_\alpha, \varphi_\alpha)_{\alpha=1}^{N}$ where each \begin{align*} \varphi_\alpha: U_\alpha &\to \widehat{U}_\alpha \end{align*} is an orientation-preserving diffeomorphism onto an open subset $\widehat{U}_\alpha$ of either $\mathbb{R}^n$ (when $U_\alpha \cap \partial M = \varnothing$, an *interior chart*) or $\mathbb{H}^n$ (when $U_\alpha \cap \partial M \neq \varnothing$, a *boundary chart*). By [Existence of Smooth Partitions of Unity](/theorems/57), there exists a smooth [partition of unity](/page/Partition%20of%20Unity) $(\psi_\alpha)_{\alpha=1}^{N}$ subordinate to $(U_\alpha)$ with $\sum_\alpha \psi_\alpha \equiv 1$ on a neighbourhood of $K$. Define \begin{align*} \omega_\alpha &:= \psi_\alpha\, \omega \in \Omega^{n-1}_c(M), \qquad \operatorname{supp}\omega_\alpha \subset U_\alpha. \end{align*} Then $\omega = \sum_{\alpha=1}^{N} \omega_\alpha$ on $M$. By linearity of $d$, integration, and $\iota^*$, it suffices to prove \begin{align*} \int_M d\omega_\alpha &= \int_{\partial M} \iota^* \omega_\alpha \qquad (\alpha = 1, \dots, N). \end{align*}[/step]

[guided]The strategy is the standard manifold-level localisation. We have an $(n-1)$-form $\omega$ on a possibly complicated manifold, and we want to convert a manifold integral into a coordinate integral so that one-variable calculus applies. The mechanism is a [partition of unity](/page/Partition%20of%20Unity) that lets us split $\omega$ into pieces, each supported inside a single chart. Two technical points deserve care. First, the charts must be **oriented**: $\varphi_\alpha$ pushes the orientation of $U_\alpha$ to the standard orientation on $\widehat{U}_\alpha \subset \mathbb{R}^n$ (or on $\widehat{U}_\alpha \subset \mathbb{H}^n$, which inherits the standard orientation from $\mathbb{R}^n$). This is what allows us to identify $\int_{U_\alpha} \eta$ with the ordinary [Lebesgue integral](/page/Lebesgue%20Integral) of the coefficient function. Second, we must distinguish interior charts ($\widehat{U}_\alpha$ open in $\mathbb{R}^n$) from boundary charts ($\widehat{U}_\alpha$ open in $\mathbb{H}^n$ and meeting $\partial \mathbb{H}^n$): the boundary integral on the right-hand side is non-trivial only for boundary charts. Compactness of $K = \operatorname{supp}\omega$ is what makes the cover finite, so the sum $\omega = \sum_\alpha \psi_\alpha \omega$ is a finite sum and all manipulations remain rigorous. The [partition of unity](/page/Partition%20of%20Unity) provided by [Theorem 57](/theorems/57) is smooth and subordinate to $(U_\alpha)$, so each $\omega_\alpha = \psi_\alpha \omega$ is smooth, compactly supported, and lies inside the corresponding chart domain. Linearity then reduces the global identity to the per-chart identity. Note that we are *not* yet claiming the per-chart identities are independent — the [partition of unity](/page/Partition%20of%20Unity) sums to $1$ only on a neighbourhood of $K$, but outside $K$ the form $\omega$ already vanishes, so this is enough.[/guided]

[step:Push each piece forward to a model form on $\mathbb{R}^n$ or $\mathbb{H}^n$]Fix $\alpha$ and drop the subscript: write $\varphi: U \to \widehat U$ for the chart and $\omega_\alpha = \psi \omega$ as $\sigma \in \Omega^{n-1}_c(U)$ with $\operatorname{supp}\sigma \subset U$. Define \begin{align*} \eta := (\varphi^{-1})^* \sigma \in \Omega^{n-1}(\widehat U), \end{align*} the pullback of $\sigma$ under the diffeomorphism $\varphi^{-1}: \widehat U \to U$. Since $\operatorname{supp}\eta = \varphi(\operatorname{supp}\sigma)$ is compact and contained in $\widehat U$, we may extend $\eta$ by zero to a smooth compactly supported $(n-1)$-form on the entire model space — either $\mathbb{R}^n$ (interior chart) or $\mathbb{H}^n$ (boundary chart). Denote the extension still by $\eta$. By naturality of the [exterior derivative](/theorems/1525) ($d \circ (\varphi^{-1})^* = (\varphi^{-1})^* \circ d$, see [Exterior Derivative](/theorems/1525)) and the definition of [Integration of Differential Forms](/theorems/1529) via orientation-preserving charts, \begin{align*} \int_M d\sigma &= \int_U d\sigma = \int_{\widehat U} (\varphi^{-1})^* d\sigma = \int_{\widehat U} d\big((\varphi^{-1})^*\sigma\big) = \int_{\widehat U} d\eta. \end{align*} Similarly, writing $\iota_M: \partial M \hookrightarrow M$ and $\iota_{\mathbb{H}}: \partial \mathbb{H}^n \hookrightarrow \mathbb{H}^n$, the restriction of $\varphi$ to $U \cap \partial M$ is an orientation-preserving (boundary orientation to boundary orientation) diffeomorphism onto $\widehat U \cap \partial \mathbb{H}^n$, and naturality of pullback under composition gives \begin{align*} \int_{\partial M} \iota_M^* \sigma &= \int_{\widehat U \cap \partial \mathbb{H}^n} \iota_{\mathbb{H}}^* \eta = \int_{\partial \mathbb{H}^n} \iota_{\mathbb{H}}^* \eta. \end{align*} (For an interior chart, $U \cap \partial M = \varnothing$, so the left side is $0$; correspondingly, $\eta \in \Omega^{n-1}_c(\mathbb{R}^n)$ has no boundary contribution.) Thus the per-chart identity reduces to proving, for $\eta \in \Omega^{n-1}_c(\mathbb{R}^n)$ or $\eta \in \Omega^{n-1}_c(\mathbb{H}^n)$: \begin{align*} \int_{\mathbb{R}^n} d\eta &= 0 \qquad \text{(interior model)}, \\ \int_{\mathbb{H}^n} d\eta &= \int_{\partial \mathbb{H}^n} \iota_{\mathbb{H}}^* \eta \qquad \text{(boundary model)}. \end{align*}[/step]

[guided]The point of this step is to translate the manifold identity into a flat-space identity that ordinary multivariable calculus can handle. Two pieces of structure transport correctly under the chart map $\varphi$: **The [exterior derivative](/theorems/1525) is natural.** For any smooth map $F$ and form $\alpha$, $F^*(d\alpha) = d(F^*\alpha)$ — see [Exterior Derivative](/theorems/1525). Applied with $F = \varphi^{-1}$ this turns $(\varphi^{-1})^* d\sigma$ into $d\eta$, exactly what we want under the integral. **Integration is defined by pullback.** By the definition of [Integration of Differential Forms](/theorems/1529) on an oriented manifold, $\int_U \alpha$ is computed in any oriented chart as the [Lebesgue integral](/page/Lebesgue%20Integral) of the coefficient function of $\varphi^*$ (or equivalently $(\varphi^{-1})_*$) of $\alpha$. Since our chart is orientation-preserving, no sign appears. Why extend $\eta$ by zero to the whole model space? Because the model identities are cleanest when stated on all of $\mathbb{R}^n$ or all of $\mathbb{H}^n$ — we then have no domain-boundary terms beyond those from $\partial \mathbb{H}^n$ itself, and Fubini applies on the full Euclidean space. The extension is smooth because $\operatorname{supp}\eta$ is compactly contained in the [open set](/page/Open%20Set) $\widehat U$, so a neighbourhood of $\partial \widehat U$ inside the model space sees $\eta \equiv 0$. For the boundary integral: the chart $\varphi$ sends $U \cap \partial M$ diffeomorphically onto $\widehat U \cap \partial \mathbb{H}^n$. A standard fact about boundary orientations (which is essentially the *definition* of the induced boundary orientation in this context) is that $\varphi|_{U \cap \partial M}$ is orientation-preserving when both sides carry the boundary orientation induced from their respective ambient orientations. This is the consistency that makes the global identity well-posed — different charts produce consistent contributions to the boundary integral. The interior model identity $\int_{\mathbb{R}^n} d\eta = 0$ is the case where the chart does not touch $\partial M$; the boundary integral on the right vanishes automatically because $\operatorname{supp}\iota_M^* \sigma = \varnothing$.[/guided]

[step:Compute $d\eta$ in standard coordinates]In coordinates $(x_1, \dots, x_n)$ on the model space, every $(n-1)$-form on a subset of $\mathbb{R}^n$ has the unique expansion \begin{align*} \eta &= \sum_{i=1}^{n} f_i \, dx_1 \wedge \dots \wedge \widehat{dx_i} \wedge \dots \wedge dx_n, \end{align*} where $f_i: \mathbb{R}^n \to \mathbb{R}$ (or $f_i: \mathbb{H}^n \to \mathbb{R}$) is smooth and compactly supported, and $\widehat{dx_i}$ indicates omission of the $i$-th factor. A direct computation of the [exterior derivative](/theorems/1525) gives \begin{align*} d\eta = \sum_{i=1}^{n} df_i \wedge dx_1 \wedge \dots \wedge \widehat{dx_i} \wedge \dots \wedge dx_n = \sum_{i=1}^{n} (-1)^{i-1}\, \frac{\partial f_i}{\partial x_i}\, dx_1 \wedge \dots \wedge dx_n, \end{align*} where we used $df_i = \sum_j (\partial_{x_j} f_i)\, dx_j$ and that $dx_j \wedge dx_1 \wedge \dots \wedge \widehat{dx_i} \wedge \dots \wedge dx_n$ vanishes unless $j = i$, in which case bringing $dx_i$ into position $i$ contributes the sign $(-1)^{i-1}$. Consequently \begin{align*} \int_{\text{model}} d\eta &= \sum_{i=1}^{n} (-1)^{i-1} \int_{\text{model}} \frac{\partial f_i}{\partial x_i}\, d\mathcal L^n(x), \end{align*} where "model" is $\mathbb{R}^n$ in the interior case and $\mathbb{H}^n$ in the boundary case.[/step]

[guided]This is the coordinate computation that converts the abstract identity into a sum of one-variable problems. The form $\eta$ has degree $n-1$ in an $n$-dimensional space, so its basis consists of the $n$ wedge products obtained by omitting exactly one of $dx_1, \dots, dx_n$. The sign computation merits unpacking. We need to evaluate \begin{align*} df_i \wedge dx_1 \wedge \cdots \wedge \widehat{dx_i} \wedge \cdots \wedge dx_n. \end{align*} Expand $df_i = \sum_j \partial_j f_i\, dx_j$. The term with $dx_j$ for $j \neq i$ contains $dx_j$ twice (once in $df_i$, once in the basis), so it vanishes by antisymmetry. Only $j = i$ survives: \begin{align*} (\partial_i f_i)\, dx_i \wedge dx_1 \wedge \cdots \wedge \widehat{dx_i} \wedge \cdots \wedge dx_n. \end{align*} To compare with the standard volume form $dx_1 \wedge \cdots \wedge dx_n$, slide $dx_i$ past $dx_1, \dots, dx_{i-1}$ — that is $i-1$ transpositions, each contributing a sign $-1$. Hence the coefficient is $(-1)^{i-1}$, giving the formula above.[/guided]

[step:Apply Fubini and the Fundamental Theorem of Calculus to each summand]Fix $i \in \{1, \dots, n\}$. Since $f_i$ is smooth and compactly supported, there exists $R > 0$ with $\operatorname{supp} f_i \subset [-R, R]^n$. Apply [Fubini's Theorem](/theorems/513) to the integrable function $\partial_{x_i} f_i$ — integrability holds because $\partial_{x_i} f_i$ is continuous and compactly supported, hence bounded with bounded support. Integrating $x_i$ last: \begin{align*} \int_{\text{model}} \frac{\partial f_i}{\partial x_i}\, d\mathcal L^n(x) &= \int_{x' \in \mathbb{R}^{n-1}_{\hat i}} \left( \int_{a}^{b} \frac{\partial f_i}{\partial x_i}(x)\, d\mathcal L^1(x_i) \right) d\mathcal L^{n-1}(x'), \end{align*} where $x' = (x_1, \dots, \widehat{x_i}, \dots, x_n) \in \mathbb{R}^{n-1}$, and the inner integration limits are $(a, b) = (-\infty, \infty)$ on $\mathbb{R}^n$, and $(a, b) = (-\infty, \infty)$ for $i < n$ on $\mathbb{H}^n$, and $(a, b) = (0, \infty)$ for $i = n$ on $\mathbb{H}^n$. The map $x_i \mapsto f_i(x', x_i)$ is smooth and compactly supported in $x_i$, so by the [Fundamental Theorem of Calculus](/theorems/632): **Case I:** $(a, b) = (-\infty, \infty)$. Then \begin{align*} \int_{-\infty}^{\infty} \frac{\partial f_i}{\partial x_i}(x', x_i)\, d\mathcal L^1(x_i) &= \lim_{R \to \infty} \bigl( f_i(x', R) - f_i(x', -R) \bigr) = 0, \end{align*} since $f_i$ vanishes outside a compact set. **Case II:** $(a, b) = (0, \infty)$, which occurs only when the model is $\mathbb{H}^n$ and $i = n$. Then \begin{align*} \int_{0}^{\infty} \frac{\partial f_n}{\partial x_n}(x', x_n)\, d\mathcal L^1(x_n) &= \lim_{R \to \infty} f_n(x', R) - f_n(x', 0) = -f_n(x', 0), \end{align*} again using compact support to discard the limit at $+\infty$.[/step]

[guided]This is the analytic heart of the proof. Two ingredients combine: Fubini reduces the $n$-dimensional integral to an iterated integral, and then FTC reduces the inner one-variable integral of a derivative to a difference of boundary values. **Verifying Fubini's hypotheses.** [Theorem 513](/theorems/513) (Fubini) applies to any integrable function on a product [measure space](/page/Measure%20Space). We have $\partial_{x_i} f_i \in C^\infty_c(\text{model})$: smoothness gives continuity, compact support gives boundedness with bounded support, hence $\partial_{x_i} f_i \in L^1$. So $\int |\partial_{x_i} f_i| d\mathcal L^n < \infty$ and Fubini lets us iterate in any order, in particular taking $x_i$ as the innermost variable. **Why we order $x_i$ last.** Because we want to evaluate the $x_i$-derivative of $f_i$. The remaining $n-1$ variables collectively become $x'$, ranging over the appropriate copy of $\mathbb{R}^{n-1}$. **Why FTC applies in the unbounded form.** Each $f_i$ has compact support in $\mathbb{R}^n$ (or $\mathbb{H}^n$), so for any fixed $x'$ there exists $R = R(x')$ with $f_i(x', x_i) = 0$ for $|x_i| > R$. Hence \begin{align*} \int_{-\infty}^{\infty} \partial_{x_i} f_i(x', x_i)\, d\mathcal L^1(x_i) = \int_{-R}^{R} \partial_{x_i} f_i(x', x_i)\, d\mathcal L^1(x_i), \end{align*} and the latter is a proper integral on a bounded interval where [FTC](/theorems/632) gives $f_i(x', R) - f_i(x', -R) = 0 - 0 = 0$. **The crucial asymmetry in Case II.** When integrating $x_n$ over $[0, \infty)$ on $\mathbb{H}^n$, the lower endpoint $0$ is *not* outside the support — it lies on $\partial \mathbb{H}^n$, where $f_n$ can be nonzero. Thus $f_n(x', 0)$ does not vanish in general, and we pick up the boundary contribution $-f_n(x', 0)$. This is exactly the surviving term that will match the boundary integral in the next step. **Why $i < n$ vanishes even on $\mathbb{H}^n$.** When $i < n$, the variable $x_i$ ranges over all of $\mathbb{R}$ (only $x_n$ is restricted to be $\geq 0$ in $\mathbb{H}^n$), so Case I applies and the term contributes zero.[/guided]

[step:Conclude the interior model identity] In the interior case, the model is $\mathbb{R}^n$ and every variable $x_i$ ranges over $\mathbb{R}$, so Case I of the previous step applies for every $i$. Therefore \begin{align*} \int_{\mathbb{R}^n} d\eta &= \sum_{i=1}^{n} (-1)^{i-1} \int_{\mathbb{R}^n} \frac{\partial f_i}{\partial x_i}\, d\mathcal L^n(x) = \sum_{i=1}^{n} (-1)^{i-1} \cdot 0 = 0. \end{align*} This proves the interior model identity. Since interior charts contribute nothing to $\int_{\partial M} \iota^*\omega_\alpha$ either, the per-chart identity holds for every interior chart. [/step]

[step:Compute the boundary pullback $\iota_{\mathbb{H}}^* \eta$]For the boundary model identity, take $\eta \in \Omega^{n-1}_c(\mathbb{H}^n)$. The inclusion $\iota_{\mathbb{H}}: \partial \mathbb{H}^n \hookrightarrow \mathbb{H}^n$ has $\iota_{\mathbb{H}}^* dx_n = 0$ (since $x_n$ is constant on $\partial \mathbb{H}^n$) and $\iota_{\mathbb{H}}^* dx_j = dx_j$ for $j < n$. Hence, in the expansion $\eta = \sum_i f_i\, dx_1 \wedge \cdots \widehat{dx_i} \cdots \wedge dx_n$, every term with $i < n$ contains the factor $dx_n$ and pulls back to zero, while the term with $i = n$ contains no $dx_n$ and pulls back to itself: \begin{align*} \iota_{\mathbb{H}}^* \eta &= f_n(x', 0)\, dx_1 \wedge \dots \wedge dx_{n-1}, \end{align*} viewed as an $(n-1)$-form on $\partial \mathbb{H}^n \cong \mathbb{R}^{n-1}$ with coordinates $x' = (x_1, \dots, x_{n-1})$. Now we must integrate this with respect to the **boundary orientation** of $\partial \mathbb{H}^n$. The convention (the "outward-normal-first" rule) is: an ordered basis $(v_1, \dots, v_{n-1})$ of $T_p(\partial \mathbb{H}^n)$ is positively oriented for the boundary orientation iff $(N_p, v_1, \dots, v_{n-1})$ is positively oriented for $\mathbb{H}^n$, where $N_p$ is the outward-pointing normal. On $\partial \mathbb{H}^n$, the outward normal is $-\partial_{x_n}$, so we test whether $(-\partial_{x_n}, \partial_{x_1}, \dots, \partial_{x_{n-1}})$ matches the standard orientation $(\partial_{x_1}, \dots, \partial_{x_n})$ of $\mathbb{H}^n$. Moving $-\partial_{x_n}$ from the first to the last slot requires $n-1$ transpositions, giving the sign $(-1)^{n-1}$, and the substitution $-\partial_{x_n} \leadsto \partial_{x_n}$ contributes an additional sign $-1$; the net sign is $(-1)^{n-1} \cdot (-1) = (-1)^n$. Hence the boundary orientation on $\partial \mathbb{H}^n \cong \mathbb{R}^{n-1}$ equals $(-1)^n$ times the standard orientation. Therefore \begin{align*} \int_{\partial \mathbb{H}^n} \iota_{\mathbb{H}}^* \eta &= (-1)^n \int_{\mathbb{R}^{n-1}} f_n(x', 0)\, d\mathcal L^{n-1}(x'). \end{align*}[/step]

[guided]Two distinct facts are being combined here, and it is important to separate them. **Fact 1: Pullback kills $dx_n$-terms.** Geometrically, $\iota_{\mathbb{H}}$ is the inclusion of the hyperplane $\{x_n = 0\}$. For any $1$-form $\alpha = \sum_j g_j dx_j$ on $\mathbb{H}^n$ and any tangent vector $v \in T_p(\partial \mathbb{H}^n)$, the pullback is $(\iota_{\mathbb{H}}^* \alpha)_p(v) = \alpha_p(d\iota_{\mathbb{H}}(v)) = \alpha_p(v)$ (since $d\iota_{\mathbb{H}}$ is the inclusion of $T_p(\partial \mathbb{H}^n)$ into $T_p\mathbb{H}^n$). Now $T_p(\partial \mathbb{H}^n) = \operatorname{span}(\partial_{x_1}, \dots, \partial_{x_{n-1}})$, so $dx_n(v) = 0$ for $v$ in this subspace. Hence $\iota_{\mathbb{H}}^* dx_n = 0$ and $\iota_{\mathbb{H}}^* dx_j = dx_j|_{\partial \mathbb{H}^n}$ for $j < n$. Of the $n$ basis $(n-1)$-forms in the expansion of $\eta$, exactly the one with $i = n$ — that is, the one missing $dx_n$ — survives. **Fact 2: The boundary orientation has sign $(-1)^n$.** This is where the convention enters. Different textbooks use different sign conventions for the boundary orientation; the one fixed in the statement of the theorem ("outward-normal-first") is standard (e.g. Lee, *Introduction to Smooth Manifolds*, Chapter 15). The verification is a small linear-algebra computation: we ask whether $(-\partial_{x_n}, \partial_{x_1}, \dots, \partial_{x_{n-1}})$ has the same orientation as $(\partial_{x_1}, \dots, \partial_{x_n})$. The permutation moving the first vector to the last position is the cyclic permutation $(1\, 2\, \cdots\, n)$, which is a product of $n-1$ transpositions, hence has sign $(-1)^{n-1}$. The vector $-\partial_{x_n}$ then differs from $\partial_{x_n}$ by a single sign flip, contributing an additional $-1$. Net sign: $(-1)^n$. **Consequence for the integral.** When integrating an $(n-1)$-form $g(x') dx_1 \wedge \cdots \wedge dx_{n-1}$ on $\partial \mathbb{H}^n$ with respect to the boundary orientation, we treat $dx_1 \wedge \cdots \wedge dx_{n-1}$ as the *standard* volume form on $\mathbb{R}^{n-1}$ and then multiply by the orientation sign $(-1)^n$. This gives \begin{align*} \int_{\partial \mathbb{H}^n} g(x')\, dx_1 \wedge \cdots \wedge dx_{n-1} &= (-1)^n \int_{\mathbb{R}^{n-1}} g(x')\, d\mathcal L^{n-1}(x'). \end{align*} The signs $(-1)^n$ here and $(-1)^n$ from Case II of the previous step are *not* a coincidence — they are precisely calibrated by the boundary orientation convention so that the final identity holds without an extra sign.[/guided]

[step:Conclude the boundary model identity and assemble the global result]Combining the computation of $\int_{\mathbb{H}^n} d\eta$ with Cases I and II: \begin{align*} \int_{\mathbb{H}^n} d\eta &= \sum_{i=1}^{n-1} (-1)^{i-1} \underbrace{\int_{\mathbb{H}^n} \frac{\partial f_i}{\partial x_i}\, d\mathcal L^n}_{= 0 \text{ by Case I}} \;+\; (-1)^{n-1} \int_{\mathbb{H}^n} \frac{\partial f_n}{\partial x_n}\, d\mathcal L^n \\ &= (-1)^{n-1} \int_{\mathbb{R}^{n-1}} \bigl( -f_n(x', 0) \bigr) d\mathcal L^{n-1}(x') \qquad \text{by Case II and Fubini} \\ &= (-1)^n \int_{\mathbb{R}^{n-1}} f_n(x', 0)\, d\mathcal L^{n-1}(x') \\ &= \int_{\partial \mathbb{H}^n} \iota_{\mathbb{H}}^* \eta \qquad \text{by the previous step}. \end{align*} This proves the boundary model identity. Together with the interior model identity from Step 5, we have established \begin{align*} \int_M d\omega_\alpha &= \int_{\partial M} \iota^* \omega_\alpha \end{align*} for every $\alpha = 1, \dots, N$. Summing over $\alpha$ and using linearity together with $\sum_\alpha \omega_\alpha = \omega$ on $M$ (and hence on $\partial M$): \begin{align*} \int_M d\omega &= \int_M d\left( \sum_{\alpha=1}^{N} \omega_\alpha \right) = \sum_{\alpha=1}^{N} \int_M d\omega_\alpha = \sum_{\alpha=1}^{N} \int_{\partial M} \iota^* \omega_\alpha = \int_{\partial M} \iota^*\omega. \end{align*} This is the [Generalised Stokes' Theorem](/theorems/1530).[/step]

[guided]The boundary-model computation is the place where the orientation sign $(-1)^n$ produced by FTC (the boundary value $-f_n(x', 0)$, with its minus sign coming from $[f]_0^\infty$) cancels against the orientation sign $(-1)^n$ produced by the boundary orientation convention. Numerically: \begin{align*} \underbrace{(-1)^{n-1}}_{\text{from } d\eta \text{ sign}} \cdot \underbrace{(-1)}_{\text{from FTC: } -f_n(x',0)} &= (-1)^n = \underbrace{(-1)^n}_{\text{from boundary orientation}}. \end{align*} Both sides give the same coefficient in front of $\int_{\mathbb{R}^{n-1}} f_n(x', 0)\, d\mathcal L^{n-1}$, so the identity holds *without* an extra sign — and this is precisely *why* the outward-normal-first convention is the right one to pair with the standard expansion of $d\eta$. The global assembly is then routine. We have proved a per-chart identity for each $\omega_\alpha = \psi_\alpha \omega$. Summing over $\alpha$ produces the global identity because (i) integration is linear, (ii) $d$ is linear, (iii) $\iota^*$ is linear, and (iv) $\sum_\alpha \psi_\alpha = 1$ on $\operatorname{supp}\omega$. Outside $\operatorname{supp}\omega$, the form $\omega$ is identically zero, so the [partition of unity](/page/Partition%20of%20Unity) only needs to sum to $1$ on the support, which is exactly what subordinate partitions give. If $\partial M = \varnothing$, all charts are interior charts and the right-hand side vanishes; the theorem then reduces to $\int_M d\omega = 0$ for compactly supported $\omega$ on a closed (boundaryless) oriented manifold. This is the special case sometimes stated separately and underlies de Rham cohomology pairings.[/guided]