[proofplan] The integral of $\omega$ over a parametrisation $\Phi$ is defined as the [Lebesgue integral](/page/Lebesgue%20Integral) over the parameter domain of the coefficient of the pullback $\Phi^*\omega$ against the standard volume form. We rewrite $\int_{\Phi \circ \psi}\omega$ using functoriality of the pullback as $\int_{W'} \psi^*(\Phi^*\omega)$, expand $\Phi^*\omega = f\, du_1 \wedge \cdots \wedge du_k$, and use the coordinate pullback formula to show that $\psi^*(\Phi^*\omega) = (f \circ \psi) \det(J\psi)\, dv_1 \wedge \cdots \wedge dv_k$. The classical change-of-variables theorem for Lebesgue integrals then converts this back to $\int_W f\, d\mathcal{L}^k$, with the sign of the integral controlled by the sign of $\det J\psi$ — since $|\det J\psi|$ appears in the change of variables formula but $\det J\psi$ appears in the pullback formula, a connectedness argument on $W'$ pins the sign down to $\pm 1$ globally. [/proofplan] [step:Reduce the integral over $\Phi \circ \psi$ to a pullback integral on $W'$ via functoriality] By definition, \begin{align*} \int_{\Phi \circ \psi}\omega \;=\; \int_{W'} (\Phi \circ \psi)^*\omega. \end{align*} By the [Functoriality of the Pullback of Differential Forms](/theorems/3568), the pullback of $\omega$ along the composition $\Phi \circ \psi$ equals the composition of pullbacks: \begin{align*} (\Phi \circ \psi)^*\omega \;=\; \psi^*(\Phi^*\omega) \qquad \text{in } \Omega^k(W'). \end{align*} Hence \begin{align*} \int_{\Phi \circ \psi}\omega \;=\; \int_{W'} \psi^*(\Phi^*\omega). \end{align*} [guided] Our goal is to compare two integrals: $\int_{W'}(\Phi\circ\psi)^*\omega$ and $\int_W \Phi^*\omega$. The first lives on $W'$ and the second on $W$, so we need a bridge between them. That bridge is the pullback. The [Functoriality of the Pullback of Differential Forms](/theorems/3568) states that for smooth maps $\psi: W' \to W$ and $\Phi: W \to U$ and any form $\omega \in \Omega^k(U)$, \begin{align*} (\Phi \circ \psi)^*\omega \;=\; \psi^*(\Phi^*\omega). \end{align*} This is exactly what we need: it lets us pass through the intermediate space $W$, where $\Phi^*\omega$ lives as a concrete $k$-form, and then use the simpler map $\psi$ to transport it to $W'$. So \begin{align*} \int_{\Phi \circ \psi}\omega \;=\; \int_{W'}(\Phi\circ\psi)^*\omega \;=\; \int_{W'}\psi^*(\Phi^*\omega). \end{align*} Now the entire problem is reduced to comparing $\int_{W'}\psi^*\alpha$ with $\int_W \alpha$ for the specific $k$-form $\alpha := \Phi^*\omega \in \Omega^k(W)$. The hypothesis that $\Phi^*\omega$ has compact support in $W$, combined with $\psi$ being a diffeomorphism, ensures the same for $\psi^*(\Phi^*\omega)$ on $W'$ (the support of $\psi^*\alpha$ is $\psi^{-1}(\operatorname{supp}\alpha)$, which is compact since $\psi$ is a homeomorphism), so both integrals are well-defined. [/guided] [/step] [step:Express $\Phi^*\omega$ in the standard top-form basis on $W$] Since $W \subseteq \mathbb{R}^k$ is open and $\Phi^*\omega \in \Omega^k(W)$ is a $k$-form on a $k$-dimensional domain, by the [Coordinate Basis for Differential Forms on an Open Subset of $\mathbb{R}^n$](/theorems/3562) applied with ambient dimension $k$, there exists a unique smooth function \begin{align*} f: W &\to \mathbb{R}, & u &\mapsto f(u), \end{align*} such that \begin{align*} \Phi^*\omega \;=\; f\, du_1 \wedge \cdots \wedge du_k \qquad \text{on } W, \end{align*} where $(u_1, \dots, u_k)$ are the standard coordinates on $W$. The support of $f$ coincides with the support of $\Phi^*\omega$, and is therefore compact in $W$ by hypothesis. By the very definition of integration of a $k$-form over an open subset of $\mathbb{R}^k$, \begin{align*} \int_\Phi \omega \;=\; \int_W \Phi^*\omega \;=\; \int_W f(u)\, d\mathcal{L}^k(u). \end{align*} [/step] [step:Compute $\psi^*(\Phi^*\omega)$ using the determinant formula for the pullback of a top form] Let $(v_1, \dots, v_k)$ be the standard coordinates on $W'$, and write the components of $\psi$ as \begin{align*} \psi: W' &\to W, & v &\mapsto (\psi_1(v), \dots, \psi_k(v)). \end{align*} By the [Coordinate Formula for the Pullback of a Differential Form](/theorems/3570), the pullback of $du_1 \wedge \cdots \wedge du_k$ under $\psi$ is \begin{align*} \psi^*(du_1 \wedge \cdots \wedge du_k) \;=\; \det(J\psi_v)\, dv_1 \wedge \cdots \wedge dv_k, \end{align*} where $J\psi_v \in \mathbb{R}^{k \times k}$ is the Jacobian matrix of $\psi$ at $v$, with entries $(J\psi_v)_{ij} = \partial_{v_j}\psi_i(v)$. Combined with the multiplicativity of pullback over scalar coefficients, we obtain \begin{align*} \psi^*(\Phi^*\omega) \;=\; \psi^*(f\, du_1 \wedge \cdots \wedge du_k) \;=\; (f \circ \psi)(v)\, \det(J\psi_v)\, dv_1 \wedge \cdots \wedge dv_k. \end{align*} Therefore \begin{align*} \int_{W'} \psi^*(\Phi^*\omega) \;=\; \int_{W'} f(\psi(v))\, \det(J\psi_v)\, d\mathcal{L}^k(v). \tag{$\ast$} \end{align*} [guided] The pullback under $\psi$ of the standard top form $du_1 \wedge \cdots \wedge du_k$ on $W$ is governed by the antisymmetry of the wedge product. The [Coordinate Formula for the Pullback of a Differential Form](/theorems/3570) is the precise statement: for $\psi: W' \to W$ with $W', W \subseteq \mathbb{R}^k$ open of the same dimension, we have \begin{align*} \psi^*(du_1 \wedge \cdots \wedge du_k) \;=\; \det(J\psi_v)\, dv_1 \wedge \cdots \wedge dv_k. \end{align*} The determinant — not its absolute value — appears here. This is the **algebraic** Jacobian, with sign, and it is the source of the orientation distinction in the final answer. Pullback commutes with multiplication by scalar-valued functions in the natural way: $\psi^*(f \alpha) = (f \circ \psi)\,\psi^*\alpha$ for any form $\alpha$. Hence \begin{align*} \psi^*(\Phi^*\omega) \;=\; (f \circ \psi)\cdot \psi^*(du_1 \wedge \cdots \wedge du_k) \;=\; (f \circ \psi) \det(J\psi)\, dv_1 \wedge \cdots \wedge dv_k. \end{align*} Integrating this $k$-form over $W'$ — which by definition means integrating its coefficient against $\mathcal{L}^k$ — yields the right-hand side of $(\ast)$. [/guided] [/step] [step:Determine the sign of $\det J\psi$ via connectedness of the components of $W'$] We claim that on each connected component $C$ of $W'$, the function $v \mapsto \det J\psi_v$ has constant sign. Since $\psi: W' \to W$ is a diffeomorphism, $\det J\psi_v \ne 0$ for every $v \in W'$ (the Jacobian is invertible at every point of a diffeomorphism). The map $v \mapsto \det J\psi_v$ is continuous (the determinant is a polynomial in the matrix entries, and the entries are continuous functions of $v$). A continuous nowhere-vanishing real-valued function on a connected set has constant sign, by the [intermediate value theorem](/theorems/180). The hypothesis of the theorem provides a global sign convention: either $\det J\psi_v > 0$ for all $v \in W'$ (orientation-preserving) or $\det J\psi_v < 0$ for all $v \in W'$ (orientation-reversing). Define $\varepsilon(\psi) \in \{+1, -1\}$ by the sign of $\det J\psi$. Then \begin{align*} \det J\psi_v \;=\; \varepsilon(\psi)\, |\det J\psi_v| \qquad \text{for every } v \in W'. \end{align*} [/step] [step:Apply the change of variables theorem for Lebesgue integrals] Equation $(\ast)$ from the previous step combined with the sign identity $\det J\psi_v = \varepsilon(\psi)\,|\det J\psi_v|$ gives \begin{align*} \int_{W'} \psi^*(\Phi^*\omega) \;=\; \varepsilon(\psi) \int_{W'} f(\psi(v))\, |\det J\psi_v|\, d\mathcal{L}^k(v). \end{align*} We now invoke [Change of Variables (general)](/theorems/22) for the [Lebesgue integral](/page/Lebesgue%20Integral). Theorem 22 requires: (i) $\psi: W' \to W$ is a $C^1$-diffeomorphism between open subsets of $\mathbb{R}^k$, and (ii) the integrand $f: W \to \mathbb{R}$ is Lebesgue-measurable and either non-negative or integrable. We verify both: (i) holds because $\psi$ is a smooth diffeomorphism by hypothesis, which is in particular $C^1$. (ii) holds because $f$ is smooth (hence measurable) and compactly supported in $W$ (Step 2), hence integrable. The theorem then yields \begin{align*} \int_W f(u)\, d\mathcal{L}^k(u) \;=\; \int_{W'} f(\psi(v))\, |\det J\psi_v|\, d\mathcal{L}^k(v). \end{align*} [guided] The classical change of variables formula in $\mathbb{R}^k$ states: for a $C^1$-diffeomorphism $\psi: W' \to W$ between open sets in $\mathbb{R}^k$ and an integrable function $f: W \to \mathbb{R}$, \begin{align*} \int_W f(u)\, d\mathcal{L}^k(u) \;=\; \int_{W'} f(\psi(v))\, |\det J\psi_v|\, d\mathcal{L}^k(v). \end{align*} The **absolute value** of the Jacobian determinant appears in this measure-theoretic formula, because Lebesgue measure is a positive measure and the change of variables tracks how positive volume transforms under $\psi$. Contrast this with the previous step, where the pullback formula for the top form $du_1 \wedge \cdots \wedge du_k$ involves the signed determinant $\det J\psi$, because differential forms carry an algebraic sign that records orientation. The two formulas differ by exactly the sign $\varepsilon(\psi)$ — this is the entire content of the theorem. We verify the hypotheses of [Change of Variables (general)](/theorems/22) explicitly. The map $\psi: W' \to W$ is smooth and a diffeomorphism, hence in particular $C^1$ with $C^1$ inverse: hypothesis (i) is satisfied. The function $f: W \to \mathbb{R}$ from Step 2 is smooth (it is the coefficient of $\Phi^*\omega$ in the standard top-form basis, and pullbacks of smooth forms have smooth coefficients) and compactly supported in $W$ (the support of $f$ equals the support of $\Phi^*\omega$, compact by hypothesis). A smooth compactly supported function is bounded and supported in a set of finite Lebesgue measure, so $f \in L^1(W)$: hypothesis (ii) is satisfied. [/guided] [/step] [step:Combine to obtain the reparametrisation identity] Chaining the equalities established above: \begin{align*} \int_{\Phi \circ \psi}\omega &\overset{\text{Step 1}}{=} \int_{W'}\psi^*(\Phi^*\omega) \\ &\overset{\text{Step 3, Step 4}}{=} \varepsilon(\psi)\int_{W'} f(\psi(v))\, |\det J\psi_v|\, d\mathcal{L}^k(v) \\ &\overset{\text{Step 5}}{=} \varepsilon(\psi)\int_W f(u)\, d\mathcal{L}^k(u) \\ &\overset{\text{Step 2}}{=} \varepsilon(\psi)\int_\Phi \omega. \end{align*} If $\psi$ is orientation-preserving, $\varepsilon(\psi) = +1$ and $\int_{\Phi \circ \psi}\omega = \int_\Phi \omega$. If $\psi$ is orientation-reversing, $\varepsilon(\psi) = -1$ and $\int_{\Phi \circ \psi}\omega = -\int_\Phi \omega$. This completes the proof. [/step]