[proofplan] The argument is a direct application of the change-of-variables formula for top-degree forms. Because $\phi^{-1} \circ \psi = g$, we have $\psi = \phi \circ g$, and functoriality of the pullback gives $\psi^* \omega = g^*(\phi^* \omega)$. Since $\phi^* \omega$ is a top-degree form on the [open set](/page/Open%20Set) $A \subseteq \mathbb{R}^k$, it admits a unique representation $\phi^* \omega = f \, dx_1 \wedge \cdots \wedge dx_k$ with $f: A \to \mathbb{R}$ smooth. The pullback under $g$ then introduces the factor $\det Jg$ in front of the volume form on $B$. Comparing both integrals to the Lebesgue change-of-variables formula — which carries an absolute value $|\det Jg|$ — produces the claimed equality with a sign determined by the sign of $\det Jg$. [/proofplan] [step:Express $\psi$ as $\phi \circ g$ and pull $\omega$ back via functoriality] By hypothesis $g = \phi^{-1} \circ \psi : B \to A$, so post-composing with $\phi$ gives \begin{align*} \phi \circ g = \phi \circ \phi^{-1} \circ \psi = \psi \end{align*} as maps $B \to \mathbb{R}^n$ (the inverse $\phi^{-1}$ is defined on $P = \psi(B)$, which is exactly the image of $\psi$, so the composition makes sense pointwise). Both $\phi: A \to \mathbb{R}^n$ and $g: B \to A$ are smooth, hence so is $\psi = \phi \circ g$. Applying the [Functoriality of the Pullback of Differential Forms](/theorems/3568) to the composition $\psi = \phi \circ g$ — whose hypotheses (smoothness of both factors with compatible domains and codomains) we verified above — we obtain the identity of $k$-forms on $B$: \begin{align*} \psi^* \omega = (\phi \circ g)^* \omega = g^*(\phi^* \omega). \end{align*} [guided] The first observation is that the transition map $g$ relates $\phi$ and $\psi$ by $\psi = \phi \circ g$. Indeed, $g$ is defined precisely so that $\phi(g(y)) = \psi(y)$ for every $y \in B$: starting from $g(y) = \phi^{-1}(\psi(y))$ and applying $\phi$ to both sides yields $\phi(g(y)) = \psi(y)$. The inverse $\phi^{-1}$ is defined on $P = \phi(A) = \psi(B)$, so the composition is well-defined pointwise on $B$. Smoothness of $\psi$ now follows because both $\phi$ and $g$ are smooth. Now we want to convert the integral $\int_B \psi^* \omega$ into something involving $\phi^* \omega$, because the whole statement is about comparing the two pullbacks. Pullback is contravariantly functorial: for smooth maps $X \xrightarrow{u} Y \xrightarrow{v} Z$ and a $k$-form $\eta$ on $Z$, one has $(v \circ u)^* \eta = u^*(v^* \eta)$. We apply this with $u = g$, $v = \phi$, $\eta = \omega$. Smoothness of both $g$ and $\phi$ is the only hypothesis required, and we have it. The [Functoriality of the Pullback of Differential Forms](/theorems/3568) gives \begin{align*} \psi^* \omega = (\phi \circ g)^* \omega = g^*(\phi^* \omega). \end{align*} This is the key reduction: the integral $\int_B \psi^* \omega$ is now an integral of a $k$-form on $B$ obtained by pulling back another $k$-form on $A$ via $g$. The problem has been reduced to comparing the integrals of $\phi^*\omega$ over $A$ and of $g^*(\phi^*\omega)$ over $B$ — i.e. to the change-of-variables theorem on open subsets of $\mathbb{R}^k$. [/guided] [/step] [step:Write $\phi^*\omega$ in the canonical top-form basis on $A$] Since $A \subseteq \mathbb{R}^k$ is open and $\phi^* \omega$ is a smooth $k$-form on $A$, the [Coordinate Basis for Differential Forms on an Open Subset of $\mathbb{R}^k$](/theorems/3562) gives a unique smooth function \begin{align*} f : A &\to \mathbb{R} \\ x &\mapsto f(x) \end{align*} such that \begin{align*} \phi^* \omega = f \, dx_1 \wedge \cdots \wedge dx_k \qquad \text{on } A, \end{align*} where $(x_1, \dots, x_k)$ are the standard coordinates on $\mathbb{R}^k$. By the definition of integration of a top-degree form on an open subset of $\mathbb{R}^k$, this means \begin{align*} \int_A \phi^* \omega = \int_A f(x) \, d\mathcal{L}^k(x). \end{align*} [/step] [step:Compute $g^*(\phi^* \omega)$ via the pullback of the top-degree volume form] The transition map $g: B \to A$ is a diffeomorphism between open subsets of $\mathbb{R}^k$. By the [Pullback of the Euclidean Volume Form](/theorems/3571), for any smooth map $h: B \to A \subseteq \mathbb{R}^k$ one has \begin{align*} h^*(dx_1 \wedge \cdots \wedge dx_k) = (\det Jh) \, dy_1 \wedge \cdots \wedge dy_k \end{align*} as $k$-forms on $B$, where $(y_1, \dots, y_k)$ are the standard coordinates on $\mathbb{R}^k \supseteq B$ and $Jh_y \in \mathbb{R}^{k \times k}$ is the Jacobian matrix of $h$ at $y$. Applying this with $h = g$ and using the $C^\infty(B)$-linearity of pullback in the function factor: \begin{align*} g^*(\phi^* \omega) = g^*\bigl( f \, dx_1 \wedge \cdots \wedge dx_k \bigr) = (f \circ g)(y) \, \det Jg_y \, dy_1 \wedge \cdots \wedge dy_k. \end{align*} Combining with Step 1, $\psi^* \omega = (f \circ g) \, \det Jg \, dy_1 \wedge \cdots \wedge dy_k$, and so \begin{align*} \int_B \psi^* \omega = \int_B (f \circ g)(y) \, \det Jg_y \, d\mathcal{L}^k(y). \end{align*} [guided] We need an explicit formula for the pullback under $g$ of the canonical top form on $A$. The general fact is that for a smooth map $h$ between open subsets of $\mathbb{R}^k$, the pullback of $dx_1 \wedge \cdots \wedge dx_k$ picks up the Jacobian determinant of $h$ as a scalar factor. This is the content of the [Pullback of the Euclidean Volume Form](/theorems/3571). Concretely, if $h: B \to A \subseteq \mathbb{R}^k$ is smooth, then at each $y \in B$, \begin{align*} \bigl(h^*(dx_1 \wedge \cdots \wedge dx_k)\bigr)_y = \det Jh_y \cdot (dy_1 \wedge \cdots \wedge dy_k)_y, \end{align*} where $Jh_y$ is the $k \times k$ Jacobian matrix with entries $(Jh_y)_{ij} = \partial_{y_j} h_i(y)$. The determinant appears because of the multilinear-alternating defining property of the top exterior power: pulling back $k$ covectors and wedging them produces the determinant of the matrix of partial derivatives. We apply this with $h = g$. Pullback is $C^\infty$-linear in the function multiplying a form, so we may pull the factor $f$ outside in the form of $f \circ g$: \begin{align*} g^*(f \, dx_1 \wedge \cdots \wedge dx_k) = (f \circ g) \cdot g^*(dx_1 \wedge \cdots \wedge dx_k) = (f \circ g) \, \det Jg \, dy_1 \wedge \cdots \wedge dy_k. \end{align*} Together with the identity $\psi^*\omega = g^*(\phi^*\omega)$ from Step 1, this means \begin{align*} \psi^* \omega = (f \circ g) \, \det Jg \, dy_1 \wedge \cdots \wedge dy_k \qquad \text{on } B, \end{align*} and therefore by the definition of integration of a top-degree form, \begin{align*} \int_B \psi^* \omega = \int_B (f \circ g)(y) \, \det Jg_y \, d\mathcal{L}^k(y). \end{align*} Note crucially that the determinant appears here with its **sign**, not in absolute value — this is the geometric content of the orientation convention in integration of forms. [/guided] [/step] [step:Apply the Lebesgue change-of-variables formula to compare the two integrals] The map $g : B \to A$ is a $C^1$ diffeomorphism between open subsets of $\mathbb{R}^k$, so the hypotheses of the [Change of Variables (general)](/theorems/22) formula are satisfied. Applied to the integrable function $f : A \to \mathbb{R}$ from Step 2, the formula reads \begin{align*} \int_A f(x) \, d\mathcal{L}^k(x) = \int_B (f \circ g)(y) \, |\det Jg_y| \, d\mathcal{L}^k(y). \end{align*} The existence of either integral (assumed in the theorem statement) is equivalent to the existence of the other under this transformation, since $|\det Jg|$ is continuous and strictly positive on the [open set](/page/Open%20Set) $B$. [guided] We have two integral expressions on the table: \begin{align*} \int_A \phi^* \omega &= \int_A f(x) \, d\mathcal{L}^k(x), & \int_B \psi^* \omega &= \int_B (f \circ g)(y) \, \det Jg_y \, d\mathcal{L}^k(y). \end{align*} The classical Lebesgue change-of-variables formula relates them, but with the **absolute value** of the Jacobian determinant rather than the signed determinant — because $d\mathcal{L}^k$ is an unsigned measure, insensitive to orientation. The [Change of Variables (general)](/theorems/22) theorem requires $g: B \to A$ to be a $C^1$ diffeomorphism between open subsets of $\mathbb{R}^k$, which is exactly our hypothesis. It then asserts \begin{align*} \int_A f(x) \, d\mathcal{L}^k(x) = \int_B (f \circ g)(y) \, |\det Jg_y| \, d\mathcal{L}^k(y). \end{align*} The integrability hypothesis transfers cleanly across this identity because $|\det Jg|$ is continuous on $B$ and never vanishes (a diffeomorphism has nonsingular Jacobian at every point), so existence of one side implies existence of the other. The remaining step is to relate the signed integrand $(f \circ g) \, \det Jg$ to the unsigned integrand $(f \circ g)\,|\det Jg|$, which is where the sign hypothesis on $\det Jg$ enters. [/guided] [/step] [step:Conclude by the sign of $\det Jg$] Two cases, according to the sign hypothesis. **Case 1: $\det Jg_y > 0$ for all $y \in B$.** Then $|\det Jg_y| = \det Jg_y$ on $B$, so the identity of Step 4 reads \begin{align*} \int_A \phi^* \omega = \int_A f \, d\mathcal{L}^k = \int_B (f \circ g) \, |\det Jg| \, d\mathcal{L}^k = \int_B (f \circ g) \, \det Jg \, d\mathcal{L}^k = \int_B \psi^* \omega, \end{align*} using Step 2 for the first equality, the Lebesgue change-of-variables identity for the third, the sign hypothesis for the fourth, and Step 3 for the last. **Case 2: $\det Jg_y < 0$ for all $y \in B$.** Then $|\det Jg_y| = -\det Jg_y$ on $B$, so \begin{align*} \int_A \phi^* \omega = \int_B (f \circ g)(y) \, |\det Jg_y| \, d\mathcal{L}^k(y) = -\int_B (f \circ g)(y) \, \det Jg_y \, d\mathcal{L}^k(y) = -\int_B \psi^* \omega. \end{align*} Equivalently, $\int_B \psi^* \omega = -\int_A \phi^* \omega$, as claimed. [guided] We now combine all the pieces. In **Case 1** ($\det Jg > 0$ on $B$), the signed and unsigned Jacobian agree: $|\det Jg| = \det Jg$. The chain of equalities is \begin{align*} \int_A \phi^* \omega &\stackrel{\text{Step 2}}{=} \int_A f \, d\mathcal{L}^k \\ &\stackrel{\text{Step 4}}{=} \int_B (f \circ g) \, |\det Jg| \, d\mathcal{L}^k \\ &\stackrel{\det Jg > 0}{=} \int_B (f \circ g) \, \det Jg \, d\mathcal{L}^k \\ &\stackrel{\text{Step 3}}{=} \int_B \psi^* \omega. \end{align*} This is the orientation-preserving identity. In **Case 2** ($\det Jg < 0$ on $B$), the absolute value flips the sign: $|\det Jg| = -\det Jg$. The same chain becomes \begin{align*} \int_A \phi^* \omega &= \int_B (f \circ g) \, |\det Jg| \, d\mathcal{L}^k = -\int_B (f \circ g) \, \det Jg \, d\mathcal{L}^k = -\int_B \psi^* \omega, \end{align*} which rearranges to $\int_B \psi^* \omega = -\int_A \phi^* \omega$. The minus sign is the geometric record of the fact that $g$ reverses orientation. The dichotomy is exhaustive on each connected component of $B$ (since $\det Jg$ is continuous and nonvanishing on $B$, it has constant sign on each component); the hypothesis that the sign is the same at every $y \in B$ rules out the mixed case in which $g$ preserves orientation on some components and reverses it on others. This completes the proof. [/guided] [/step]