[proofplan] The integral of a top-degree form is defined via positively oriented charts together with a subordinate [partition of unity](/page/Partition%20of%20Unity). We construct, from any positive oriented atlas $\mathcal{A}$ for $M$, a positive oriented atlas $\widetilde{\mathcal{A}}$ for $-M$ obtained by post-composing each chart with the reflection $r(x_1,\dots,x_n) = (-x_1,x_2,\dots,x_n)$. We then compute, on each partition-of-unity piece, the pullback of $\omega$ to Euclidean space under the reflected chart; the antisymmetry of the wedge product produces a sign $-1$ relative to the original chart, and the Lebesgue change of variables under $r$ (which has unit Jacobian determinant) returns the integration domain to that used for $\int_M \omega$. Summing over the finitely many pieces that meet $\operatorname{supp}\omega$ yields the identity. [/proofplan] [step:Fix a positive oriented atlas for $M$ and recall the definition of $\int_M$] Let $\mathcal{A} = \{(U_\alpha, \varphi_\alpha)\}_{\alpha \in A}$ be a smooth atlas for $M$ that represents the orientation of $M$: that is, every transition map $\varphi_\beta \circ \varphi_\alpha^{-1}$ is a diffeomorphism between open subsets of $\mathbb{R}^n$ whose Jacobian determinant is strictly positive at every point. Let $\{\rho_\alpha\}_{\alpha \in A}$ be a smooth [partition of unity](/page/Partition%20of%20Unity) subordinate to $\{U_\alpha\}$; such a partition exists by [Existence of Smooth Partitions of Unity](/theorems/57), since $M$ is paracompact Hausdorff. By the definition of the integral of a top-degree form on an oriented manifold (see [Integration of Differential Forms](/theorems/1529)), for $\omega \in \Omega^n_c(M)$, \begin{align*} \int_M \omega = \sum_{\alpha \in A} \int_M \rho_\alpha\, \omega, \end{align*} where for each $\alpha$ the form $\rho_\alpha \omega$ is supported inside the chart domain $U_\alpha$, and \begin{align*} \int_M \rho_\alpha \omega := \int_{\varphi_\alpha(U_\alpha)} g_\alpha(x)\, d\mathcal{L}^n(x), \end{align*} in which $g_\alpha : \varphi_\alpha(U_\alpha) \to \mathbb{R}$ is the unique smooth function such that the Euclidean pullback satisfies \begin{align*} (\varphi_\alpha^{-1})^* (\rho_\alpha \omega) = g_\alpha\, dx_1 \wedge \cdots \wedge dx_n \end{align*} on $\varphi_\alpha(U_\alpha)$. Because $\rho_\alpha \omega$ has compact support inside $U_\alpha$, the function $g_\alpha$ has compact support inside $\varphi_\alpha(U_\alpha)$, and only finitely many indices $\alpha$ contribute to a sum involving $\omega$ (since $\operatorname{supp}\omega$ is compact and the cover $\{U_\alpha\}$ is locally finite). [guided] The integral on an oriented manifold is defined chart-by-chart, then patched together with a [partition of unity](/page/Partition%20of%20Unity). The two ingredients we need to keep straight are: 1. **What "positive chart" means.** A chart $(U,\varphi)$ is positive for the orientation if, after passing to Euclidean coordinates, the local frame $\partial_{x_1},\dots,\partial_{x_n}$ agrees with the chosen orientation on each tangent space $T_pM$. Equivalently, an atlas $\mathcal{A}$ represents the orientation if and only if every transition $\varphi_\beta \circ \varphi_\alpha^{-1}$ has positive Jacobian determinant. 2. **How the integral is built.** Given $\omega \in \Omega^n_c(M)$, pick a [partition of unity](/page/Partition%20of%20Unity) $\{\rho_\alpha\}$ subordinate to $\{U_\alpha\}$. Each $\rho_\alpha \omega$ is a compactly supported top form inside a single chart $U_\alpha$, so we can transport it to $\mathbb{R}^n$ via the pullback by $\varphi_\alpha^{-1}$, write the pulled-back form as $g_\alpha\, dx_1 \wedge \cdots \wedge dx_n$, and Lebesgue-integrate $g_\alpha$ over $\varphi_\alpha(U_\alpha)$. The total integral $\int_M \omega$ is then the sum of these chart integrals. A [partition of unity](/page/Partition%20of%20Unity) exists because a smooth manifold is paracompact Hausdorff, and on such a space [Existence of Smooth Partitions of Unity](/theorems/57) produces a smooth partition $\{\rho_\alpha\}$ subordinate to any open cover. The sum $\sum_\alpha \int_M \rho_\alpha \omega$ has only finitely many nonzero terms: $\operatorname{supp}\omega$ is compact, the cover $\{U_\alpha\}$ is locally finite, so only finitely many supports $\operatorname{supp}\rho_\alpha$ meet $\operatorname{supp}\omega$. The independence of $\int_M \omega$ from the choice of positive atlas and subordinate [partition of unity](/page/Partition%20of%20Unity) is part of the content of [Integration of Differential Forms](/theorems/1529); we will use this independence below for $-M$. [/guided] [/step] [step:Build a positive oriented atlas for $-M$ by reflecting the first coordinate] Define the reflection \begin{align*} r: \mathbb{R}^n &\to \mathbb{R}^n, \\ (x_1, x_2, \dots, x_n) &\mapsto (-x_1, x_2, \dots, x_n). \end{align*} The map $r$ is a smooth involution ($r \circ r = \operatorname{id}_{\mathbb{R}^n}$) with Jacobian matrix $J r = \operatorname{diag}(-1, 1, \dots, 1)$, hence $\det J r = -1$. For each $\alpha \in A$, define \begin{align*} \widetilde{\varphi}_\alpha &:= r \circ \varphi_\alpha : U_\alpha \to \widetilde{V}_\alpha := r(\varphi_\alpha(U_\alpha)). \end{align*} Each $\widetilde{\varphi}_\alpha$ is a diffeomorphism onto its image because both $r$ and $\varphi_\alpha$ are. We claim $\widetilde{\mathcal{A}} := \{(U_\alpha, \widetilde{\varphi}_\alpha)\}_{\alpha \in A}$ is a positive oriented atlas for $-M$. A transition map in $\widetilde{\mathcal{A}}$ is \begin{align*} \widetilde{\varphi}_\beta \circ \widetilde{\varphi}_\alpha^{-1} = (r \circ \varphi_\beta) \circ (\varphi_\alpha^{-1} \circ r^{-1}) = r \circ (\varphi_\beta \circ \varphi_\alpha^{-1}) \circ r. \end{align*} By the chain rule, its Jacobian determinant equals $(\det J r) \cdot \det(J(\varphi_\beta \circ \varphi_\alpha^{-1})) \cdot (\det J r) = (-1)\cdot(\text{positive})\cdot(-1) > 0$, so $\widetilde{\mathcal{A}}$ is a smooth oriented atlas. Moreover, the Jacobian determinant of each $\widetilde{\varphi}_\alpha \circ \varphi_\alpha^{-1} = r$ is $-1 < 0$, so $\widetilde{\mathcal{A}}$ induces the orientation opposite to $\mathcal{A}$, i.e., the orientation of $-M$. The same [partition of unity](/page/Partition%20of%20Unity) $\{\rho_\alpha\}$ remains subordinate to $\{U_\alpha\}$, since the underlying cover is unchanged. [guided] We need a concrete positive atlas for $-M$ so that the definition of $\int_{-M}\omega$ can be unfolded in computable terms. The trick is the reflection $r(x_1,\dots,x_n) = (-x_1,x_2,\dots,x_n)$. Composing each positive chart $\varphi_\alpha$ for $M$ with $r$ produces a new chart $\widetilde{\varphi}_\alpha = r \circ \varphi_\alpha$ on the same [open set](/page/Open%20Set) $U_\alpha$. Why does this reverse orientation? On overlaps, the new transition is \begin{align*} \widetilde{\varphi}_\beta \circ \widetilde{\varphi}_\alpha^{-1} = r \circ (\varphi_\beta \circ \varphi_\alpha^{-1}) \circ r, \end{align*} using $r^{-1} = r$. The chain rule gives Jacobian determinant \begin{align*} \det J(\widetilde{\varphi}_\beta \circ \widetilde{\varphi}_\alpha^{-1}) = (\det J r)^2 \cdot \det J(\varphi_\beta \circ \varphi_\alpha^{-1}) = (-1)^2 \cdot (\text{positive}) > 0. \end{align*} So $\widetilde{\mathcal{A}}$ is internally consistent — its transitions are all orientation-preserving as maps of $\mathbb{R}^n$. It is therefore an oriented atlas. Which orientation does it pick out? Compare $\widetilde{\varphi}_\alpha$ to $\varphi_\alpha$ via the transition \begin{align*} \widetilde{\varphi}_\alpha \circ \varphi_\alpha^{-1} = r, \end{align*} whose Jacobian determinant is $-1 < 0$. So $\widetilde{\mathcal{A}}$ is incompatible with $\mathcal{A}$ as an oriented atlas — they pick out opposite orientations on $M$. Since $\mathcal{A}$ represents $M$'s orientation, $\widetilde{\mathcal{A}}$ represents the orientation of $-M$. We may therefore use $\widetilde{\mathcal{A}}$ together with the same [partition of unity](/page/Partition%20of%20Unity) $\{\rho_\alpha\}$ to evaluate $\int_{-M}\omega$. [/guided] [/step] [step:Compute the pullback of $\rho_\alpha \omega$ under the reflected chart] Fix $\alpha \in A$ and let $g_\alpha: \varphi_\alpha(U_\alpha) \to \mathbb{R}$ be the smooth compactly supported function from Step 1 satisfying \begin{align*} (\varphi_\alpha^{-1})^* (\rho_\alpha \omega) = g_\alpha\, dx_1 \wedge \cdots \wedge dx_n \end{align*} on $\varphi_\alpha(U_\alpha) \subseteq \mathbb{R}^n$. For the reflected chart $\widetilde{\varphi}_\alpha = r \circ \varphi_\alpha$, we have $\widetilde{\varphi}_\alpha^{-1} = \varphi_\alpha^{-1} \circ r^{-1} = \varphi_\alpha^{-1} \circ r$. By functoriality of pullback, \begin{align*} (\widetilde{\varphi}_\alpha^{-1})^* (\rho_\alpha \omega) = r^* \big((\varphi_\alpha^{-1})^* (\rho_\alpha \omega)\big) = r^* \big(g_\alpha\, dx_1 \wedge \cdots \wedge dx_n\big), \end{align*} viewed on $\widetilde{V}_\alpha = r(\varphi_\alpha(U_\alpha))$. We evaluate this pullback. Writing the coordinate functions on $\mathbb{R}^n$ as $x_1, \dots, x_n$, the pullback satisfies $r^* x_1 = -x_1$ and $r^* x_i = x_i$ for $i \ge 2$, hence $r^* (dx_1) = d(r^* x_1) = d(-x_1) = -dx_1$ and $r^*(dx_i) = dx_i$ for $i \ge 2$. Pullback commutes with wedge, so \begin{align*} r^*(dx_1 \wedge dx_2 \wedge \cdots \wedge dx_n) = (-dx_1) \wedge dx_2 \wedge \cdots \wedge dx_n = -\, dx_1 \wedge dx_2 \wedge \cdots \wedge dx_n. \end{align*} Combined with $r^* g_\alpha = g_\alpha \circ r$, we obtain \begin{align*} (\widetilde{\varphi}_\alpha^{-1})^* (\rho_\alpha \omega) = -\, (g_\alpha \circ r)\, dx_1 \wedge \cdots \wedge dx_n \end{align*} on $\widetilde{V}_\alpha$. [guided] We want to evaluate $\int_{-M} \rho_\alpha \omega$, which by the definition of integration on $-M$ (using the positive atlas $\widetilde{\mathcal{A}}$ from Step 2) requires expressing the Euclidean pullback $(\widetilde{\varphi}_\alpha^{-1})^* (\rho_\alpha \omega)$ as a multiple of the standard volume form on $\widetilde{V}_\alpha$. The key observation is that the new chart's inverse factors: \begin{align*} \widetilde{\varphi}_\alpha^{-1} = (r \circ \varphi_\alpha)^{-1} = \varphi_\alpha^{-1} \circ r^{-1} = \varphi_\alpha^{-1} \circ r, \end{align*} where we used $r^{-1} = r$ from Step 2. Pullback is contravariantly functorial: $(f \circ g)^* = g^* \circ f^*$. Hence \begin{align*} (\widetilde{\varphi}_\alpha^{-1})^* = (\varphi_\alpha^{-1} \circ r)^* = r^* \circ (\varphi_\alpha^{-1})^*. \end{align*} This lets us reuse the data $g_\alpha = (\varphi_\alpha^{-1})^* (\rho_\alpha \omega) / (dx_1 \wedge \cdots \wedge dx_n)$ from the original chart and simply apply $r^*$ to it. Now compute $r^*(dx_1 \wedge \cdots \wedge dx_n)$. Pullback of a $1$-form satisfies $r^*(df) = d(f \circ r)$. The coordinate function $x_1$ pulls back to $x_1 \circ r = -x_1$, so $r^* dx_1 = d(-x_1) = -dx_1$. The other coordinates $x_i$ ($i \ge 2$) are fixed by $r$, so $r^* dx_i = dx_i$. Pullback distributes over the wedge product, so a single sign $-1$ appears: \begin{align*} r^*(dx_1 \wedge \cdots \wedge dx_n) = (-dx_1) \wedge dx_2 \wedge \cdots \wedge dx_n = -\, dx_1 \wedge \cdots \wedge dx_n. \end{align*} The function $g_\alpha$ pulls back by composition: $r^* g_\alpha = g_\alpha \circ r$. Multiplying, \begin{align*} (\widetilde{\varphi}_\alpha^{-1})^*(\rho_\alpha \omega) = -\,(g_\alpha \circ r)\, dx_1 \wedge \cdots \wedge dx_n, \end{align*} which is what we needed to extract. The sign here is **the geometric content of orientation reversal**: it is not bookkeeping, but the antisymmetry of the wedge product responding to the reflection of a single basis vector. [/guided] [/step] [step:Apply the Lebesgue change of variables under $r$ to restore the integration domain] By Step 3 and the definition of integration on $-M$ using the positive atlas $\widetilde{\mathcal{A}}$, \begin{align*} \int_{-M} \rho_\alpha \omega = \int_{\widetilde{V}_\alpha} \big(-\, g_\alpha \circ r\big)(x)\, d\mathcal{L}^n(x) = -\int_{r(\varphi_\alpha(U_\alpha))} g_\alpha(r(x))\, d\mathcal{L}^n(x). \end{align*} Apply the Lebesgue [Change of Variables (general)](/theorems/22) with the diffeomorphism $r: \mathbb{R}^n \to \mathbb{R}^n$. We verify its hypotheses: $r$ is a $C^1$ diffeomorphism (it is linear with $\det Jr = -1 \ne 0$); the integrand $g_\alpha \circ r$ is continuous with compact support inside $\widetilde{V}_\alpha = r(\varphi_\alpha(U_\alpha))$ (since $g_\alpha$ is so on $\varphi_\alpha(U_\alpha)$ and $r$ is a homeomorphism), hence Lebesgue-integrable. Substituting $y = r(x)$, so $x = r^{-1}(y) = r(y)$ and $d\mathcal{L}^n(x) = |\det Jr|\, d\mathcal{L}^n(y) = d\mathcal{L}^n(y)$, \begin{align*} \int_{r(\varphi_\alpha(U_\alpha))} g_\alpha(r(x))\, d\mathcal{L}^n(x) = \int_{\varphi_\alpha(U_\alpha)} g_\alpha(y)\, d\mathcal{L}^n(y) = \int_M \rho_\alpha \omega, \end{align*} where the last equality is the definition from Step 1. Therefore \begin{align*} \int_{-M} \rho_\alpha \omega = -\int_M \rho_\alpha \omega. \end{align*} [guided] We now have an expression for $\int_{-M}\rho_\alpha\omega$ as a [Lebesgue integral](/page/Lebesgue%20Integral) over $\widetilde{V}_\alpha = r(\varphi_\alpha(U_\alpha))$ — but the integral $\int_M \rho_\alpha\omega$ uses the domain $\varphi_\alpha(U_\alpha)$. We need to transport the integration variable back across $r$ to make the two domains match. This is exactly the Lebesgue [Change of Variables (general)](/theorems/22). To apply it, we verify the hypotheses: - **$r$ is a $C^1$ diffeomorphism $\mathbb{R}^n \to \mathbb{R}^n$.** Indeed $r$ is linear and $\det Jr = -1 \neq 0$, so $r$ is a linear isomorphism, hence a $C^\infty$ diffeomorphism. - **The integrand is Lebesgue-integrable.** The function $g_\alpha$ is smooth and compactly supported inside $\varphi_\alpha(U_\alpha)$ (Step 1), so $g_\alpha \circ r$ is smooth and compactly supported inside $\widetilde{V}_\alpha$, which suffices for integrability. - **Compute the absolute Jacobian.** $|\det Jr| = |-1| = 1$. This is why the reflection contributes **only one sign**, captured back in Step 3 via the wedge product: the Lebesgue measure does not "see" the orientation reversal. The substitution $y = r(x)$ (equivalently $x = r(y)$, since $r$ is an involution) maps $\widetilde{V}_\alpha = r(\varphi_\alpha(U_\alpha))$ bijectively onto $\varphi_\alpha(U_\alpha)$, with $d\mathcal{L}^n(x) = |\det Jr|\, d\mathcal{L}^n(y) = d\mathcal{L}^n(y)$. The integrand transforms by $g_\alpha(r(x)) = g_\alpha(y)$. Therefore \begin{align*} \int_{r(\varphi_\alpha(U_\alpha))} g_\alpha(r(x))\, d\mathcal{L}^n(x) = \int_{\varphi_\alpha(U_\alpha)} g_\alpha(y)\, d\mathcal{L}^n(y). \end{align*} The right-hand side is by definition $\int_M \rho_\alpha\omega$ (Step 1). Pulling the leftover sign through, we conclude \begin{align*} \int_{-M}\rho_\alpha\omega = -\int_M\rho_\alpha\omega. \end{align*} [/guided] [/step] [step:Sum over the partition of unity to conclude] Since $\operatorname{supp}\omega$ is compact and $\{U_\alpha\}$ is locally finite, the set $A_0 := \{\alpha \in A : \operatorname{supp}\rho_\alpha \cap \operatorname{supp}\omega \ne \varnothing\}$ is finite, and for $\alpha \notin A_0$ both $\int_M \rho_\alpha \omega$ and $\int_{-M} \rho_\alpha \omega$ vanish. Summing the per-chart identity from Step 4 over $\alpha \in A_0$, \begin{align*} \int_{-M} \omega = \sum_{\alpha \in A_0} \int_{-M} \rho_\alpha \omega = \sum_{\alpha \in A_0} \Big(-\int_M \rho_\alpha \omega\Big) = -\sum_{\alpha \in A_0} \int_M \rho_\alpha \omega = -\int_M \omega. \end{align*} This is the claimed identity, completing the proof. [/step]