[proofplan] Fix a basepoint $x_0 \in U$ and define a candidate potential $f: U \to \mathbb{R}$ by integrating $\omega$ along any piecewise smooth path from $x_0$ to a given point. The strategy reduces to showing that this definition does not depend on the path, equivalently that $\int_\beta \omega = 0$ for every piecewise smooth loop $\beta$ based at $x_0$. Simple connectivity provides a continuous null-homotopy of $\beta$ which, after smoothing rel boundary, can be combined with [Naturality of the Pullback](/theorems/3574) and [Stokes' Theorem](/theorems/1530) on the unit square to produce the vanishing. Once $f$ is well-defined, the [Fundamental Theorem of Calculus](/theorems/632) applied along coordinate segments yields $\partial_{x_i} f = \omega_i$, so $df = \omega$ and bootstrapping regularity from $\omega \in C^\infty$ gives $f \in C^\infty(U)$. The vector-field corollary follows by translating between $F$ and $\omega_F$, with the closedness condition $d\omega_F = 0$ being the symmetry $\partial_{x_i} F_j = \partial_{x_j} F_i$ for the components. [/proofplan] [step:Reduce exactness to vanishing of line integrals over all piecewise smooth loops] Write $\omega = \sum_{i=1}^n \omega_i\, dx_i$ with $\omega_i \in C^\infty(U)$. For any piecewise $C^1$ path $\gamma: [a, b] \to U$, the line integral is defined as \begin{align*} \int_\gamma \omega := \int_a^b \sum_{i=1}^n \omega_i(\gamma(t))\, \dot\gamma_i(t)\, d\mathcal{L}^1(t). \end{align*} Fix any basepoint $x_0 \in U$. By path-connectedness of $U$, for every $x \in U$ there exists a piecewise smooth path $\gamma_x: [0, 1] \to U$ with $\gamma_x(0) = x_0$ and $\gamma_x(1) = x$. We claim the following equivalence: if \begin{align*} \int_\beta \omega = 0 \qquad \text{for every piecewise smooth loop } \beta: [0,1] \to U \text{ based at } x_0, \tag{$\star$} \end{align*} then the map \begin{align*} f: U &\to \mathbb{R} \\ x &\mapsto \int_{\gamma_x} \omega \end{align*} is independent of the choice of $\gamma_x$. Indeed, if $\gamma, \tilde\gamma$ are two such piecewise smooth paths from $x_0$ to $x$, the concatenation $\beta := \gamma * \tilde\gamma^{-1}$ (traverse $\gamma$, then $\tilde\gamma$ in reverse) is a piecewise smooth loop at $x_0$, and the line integral is additive over concatenations and antisymmetric under reversal, so \begin{align*} \int_\gamma \omega - \int_{\tilde\gamma} \omega = \int_\beta \omega = 0 \end{align*} by $(\star)$. Thus $f(x)$ is unambiguous. It therefore suffices to establish $(\star)$. [guided] We must produce a smooth function $f$ with $df = \omega$. The classical construction defines $f(x)$ as the value of the line integral of $\omega$ from a basepoint $x_0$ to $x$, mimicking how an antiderivative is constructed in one variable. The obstruction is that there are many paths from $x_0$ to $x$, and a priori the integral depends on which one is chosen. The choice-independence question reduces cleanly to a loop question: two paths $\gamma, \tilde\gamma$ from $x_0$ to $x$ produce the same integral if and only if the concatenated loop $\gamma * \tilde\gamma^{-1}$ has zero integral. So the entire proof of well-definedness is reduced to property $(\star)$ — vanishing of $\omega$ around every piecewise smooth loop based at $x_0$. Write $\omega = \sum_{i=1}^n \omega_i\, dx_i$ with $\omega_i \in C^\infty(U)$. The line integral of $\omega$ along a piecewise $C^1$ path $\gamma: [a, b] \to U$ is defined as \begin{align*} \int_\gamma \omega := \int_a^b \sum_{i=1}^n \omega_i(\gamma(t))\, \dot\gamma_i(t)\, d\mathcal{L}^1(t). \end{align*} Fix once and for all a basepoint $x_0 \in U$. By path-connectedness, every $x \in U$ is reachable by a piecewise smooth path $\gamma_x: [0,1] \to U$ from $x_0$ — and we may freely upgrade "continuous" to "piecewise smooth" in an open subset of $\mathbb{R}^n$ because we can replace any continuous path by a piecewise linear approximation along a sufficiently fine subdivision and connecting line segments stay in $U$ provided they lie in a ball $B(x_k, r_k) \subseteq U$ around each chosen point. Once $(\star)$ holds, choose paths $\gamma_x$ for each $x \in U$ and set $f(x) := \int_{\gamma_x} \omega$. The earlier computation shows the choice of $\gamma_x$ is immaterial. The remaining steps will produce $(\star)$ and verify $df = \omega$. [/guided] [/step] [step:Smooth the continuous null-homotopy of $\beta$ to a smooth null-homotopy with fixed boundary behaviour] Let $\beta: [0, 1] \to U$ be a piecewise smooth loop at $x_0$. Choose a finite partition $0=s_0<s_1<\cdots<s_m=1$ such that $\beta$ is smooth on each interval $[s_{k-1},s_k]$. Choose a smooth nondecreasing surjection \begin{align*} \theta: [0,1] &\to [0,1] \end{align*} which is constant near $0$ and $1$, maps a finite partition of $[0,1]$ onto the partition $(s_k)_{k=0}^m$, and is flat at the preimages of the breakpoints $s_1,\dots,s_{m-1}$. Such a map is obtained by integrating a nonnegative $C^\infty$ function that vanishes to infinite order at the prescribed subdivision points and is positive elsewhere. Define \begin{align*} \hat\beta: [0,1] &\to U \\ \tau &\mapsto \beta(\theta(\tau)). \end{align*} Then $\hat\beta$ is smooth and has sitting instants at its endpoints, meaning that $\hat\beta(\tau)=x_0$ for $\tau$ in a neighbourhood of $0$ and for $\tau$ in a neighbourhood of $1$. Applying the one-dimensional change-of-variables formula on each interval of smoothness gives $\int_{\hat\beta}\omega=\int_\beta\omega$. Hence it suffices to prove the vanishing for smooth loops with sitting instants. Replacing $\beta$ by $\hat\beta$, assume from now on that $\beta\in C^\infty([0,1];U)$ and that $\beta$ has sitting instants. By the [Characterisation of Simply Connected Spaces](/theorems/1883), simple connectivity of $U$ gives a continuous homotopy $H_0: [0, 1]^2 \to U$ with \begin{align*} H_0(s, 0) = \beta(s), \quad H_0(s, 1) = x_0, \quad H_0(0, t) = H_0(1, t) = x_0 \end{align*} for all $s, t \in [0, 1]$. We claim there exists a smooth map $H: [0, 1]^2 \to U$ satisfying the same four boundary identities. [claim:Smoothing of the null-homotopy rel boundary] There exists a smooth map $H: [0, 1]^2 \to U$ with $H(s, 0) = \beta(s)$, $H(s, 1) = x_0$, and $H(0, t) = H(1, t) = x_0$ for all $s, t \in [0, 1]$. [proof] Set $Q := [0,1]^2$. The image $K := H_0(Q)$ is compact and contained in the [open set](/page/Open%20Set) $U$. Choose a number $\delta_* > 0$ such that every point of $\mathbb{R}^n$ whose Euclidean distance from $K$ is less than $\delta_*$ lies in $U$. Because $\beta$ has sitting instants, choose $a \in (0,1/8)$ such that $\beta(s)=x_0$ for every $s \in [0,4a]\cup[1-4a,1]$. Decrease $a$ if necessary so that [uniform continuity](/page/Uniform%20Continuity) of $H_0$ on $Q$ gives \begin{align*} |H_0(y)-H_0(y')| < \frac{\delta_*}{4} \end{align*} whenever $y,y' \in Q$ and $|y-y'|<2a$. Define the collar neighbourhood \begin{align*} N_a := \{(s,t)\in Q : s<2a \text{ or } 1-s<2a \text{ or } t<2a \text{ or } 1-t<2a\}. \end{align*} Define a smooth map $B: N_a \to U$ as follows: on the bottom strip $\{(s,t)\in Q:t<2a\}$ set $B(s,t):=\beta(s)$, and on the side and top strips set $B(s,t):=x_0$. These definitions agree on overlaps because $a<1/8$ prevents the top strip from meeting the bottom strip, and because $\beta(s)=x_0$ when $s<2a$ or $1-s<2a$. Thus $B$ is smooth on $N_a$ and agrees with the prescribed boundary values. Moreover, for every $y\in N_a$ there is a boundary point $y_\partial\in\partial Q$ on the same collar strip with $|y-y_\partial|<2a$ and $B(y)=H_0(y_\partial)$, hence \begin{align*} |B(y)-H_0(y)| < \frac{\delta_*}{4}. \end{align*} Let $\pi: \mathbb{R}^2 \to Q$ be the nearest-point projection onto the convex square $Q$, and define $\tilde H_0: \mathbb{R}^2 \to \mathbb{R}^n$ by $\tilde H_0(z):=H_0(\pi(z))$. Let $\eta \in C_c^\infty(B(0,1)\subseteq\mathbb{R}^2)$ be a [Standard Mollifier](/page/Standard%20Mollifier) with $\int_{\mathbb{R}^2}\eta(z)\,d\mathcal{L}^2(z)=1$, and set $\eta_\varepsilon(z):=\varepsilon^{-2}\eta(z/\varepsilon)$. Define the componentwise mollification \begin{align*} M_\varepsilon: \mathbb{R}^2 &\to \mathbb{R}^n \\ z &\mapsto \int_{\mathbb{R}^2}\eta_\varepsilon(z-y)\,\tilde H_0(y)\,d\mathcal{L}^2(y). \end{align*} [Uniform convergence](/page/Uniform%20Convergence) of mollifications on compact sets gives an $\varepsilon_0\in(0,a/4)$ such that \begin{align*} \|M_{\varepsilon_0}-H_0\|_{L^\infty(Q)} < \frac{\delta_*}{4}. \end{align*} Choose $\rho\in C^\infty([0,1];[0,1])$ with $\rho\equiv0$ on $[0,a/2]\cup[1-a/2,1]$ and $\rho\equiv1$ on $[a,1-a]$, and define \begin{align*} \psi: Q &\to [0,1] \\ (s,t) &\mapsto \rho(s)\rho(t). \end{align*} Now define $H: Q\to\mathbb{R}^n$ by \begin{align*} H(y):= \begin{cases} \psi(y)M_{\varepsilon_0}(y)+(1-\psi(y))B(y), & y\in N_a,\\ M_{\varepsilon_0}(y), & y\in Q\setminus N_a. \end{cases} \end{align*} The two formulas agree smoothly near the interface because $\psi\equiv1$ on a neighbourhood of $Q\setminus N_a$. Therefore $H$ is smooth on $Q$ in the sense that it is the restriction of a smooth map on a neighbourhood of $Q$. If $y\in Q\setminus N_a$, then $|H(y)-H_0(y)|<\delta_*/4$. If $y\in N_a$, the preceding estimates and $0\le\psi\le1$ give \begin{align*} |H(y)-H_0(y)| \leq \psi(y)|M_{\varepsilon_0}(y)-H_0(y)|+(1-\psi(y))|B(y)-H_0(y)| < \frac{\delta_*}{4}. \end{align*} Thus $H(Q)\subseteq U$. Finally, if $y\in\partial Q$, then $\psi(y)=0$ and $H(y)=B(y)$, so $H(s,0)=\beta(s)$, $H(s,1)=x_0$, and $H(0,t)=H(1,t)=x_0$ for all $s,t\in[0,1]$. [/proof] [/claim] Henceforth $H \in C^\infty([0, 1]^2; U)$ denotes a smooth null-homotopy of the smooth loop $\beta$ rel boundary, satisfying the four identities above. [/step] [step:Apply Stokes' Theorem to $H^*\omega$ on the unit square to conclude $\int_\beta \omega = 0$] The pullback $H^*\omega$ is a smooth $1$-form on $[0,1]^2$; cf. [Pullback of a Differential Form Is a Smooth Differential Form](/theorems/3567). By [Naturality of the Pullback](/theorems/3574), [exterior derivative](/theorems/1525) commutes with pullback: \begin{align*} d(H^*\omega) = H^*(d\omega) = H^*(0) = 0. \end{align*} Apply [Stokes' Theorem](/theorems/1530) to the smooth $1$-form $H^*\omega$ on the compact oriented manifold-with-corners $[0,1]^2$ (corners do not obstruct Stokes once one verifies the standard orientation on $\partial[0,1]^2$): \begin{align*} \int_{[0,1]^2} d(H^*\omega) = \int_{\partial [0,1]^2} H^*\omega. \end{align*} The left-hand side is $0$ by the previous display. Decompose the boundary of $[0,1]^2$ (traversed with the induced orientation) into the four edges \begin{align*} e_b(s) &= (s, 0), \ s \in [0, 1]; & e_r(t) &= (1, t), \ t \in [0, 1]; \\ e_t(s) &= (1-s, 1), \ s \in [0, 1]; & e_\ell(t) &= (0, 1 - t), \ t \in [0, 1]. \end{align*} On the right edge $H(e_r(t)) = H(1, t) = x_0$, so $H \circ e_r$ is the constant map $x_0$; its velocity is identically zero and $\int_{e_r} H^*\omega = 0$. The same holds for $e_t$ (since $H(\cdot, 1) \equiv x_0$) and for $e_\ell$ (since $H(0, \cdot) \equiv x_0$). On the bottom edge $H(e_b(s)) = \beta(s)$, so $H \circ e_b = \beta$, and by definition of pullback line integrals \begin{align*} \int_{e_b} H^*\omega = \int_\beta \omega. \end{align*} Combining, \begin{align*} 0 = \int_{\partial[0,1]^2} H^*\omega = \int_\beta \omega + 0 + 0 + 0, \end{align*} which proves $(\star)$ for smooth loops with sitting instants. The reparametrisation reduction in the smoothing step preserves the line integral, so $(\star)$ also holds for the original piecewise smooth loop. [guided] This is the heart of the proof. We have a smooth null-homotopy $H: [0,1]^2 \to U$ of $\beta$ rel its constant endpoint, and a closed $1$-form $\omega$ on $U$. The two combine via pullback to produce a $1$-form on the square whose [exterior derivative](/theorems/1525) we can compute, and whose boundary integral we want to recognise as $\int_\beta \omega$. **Computing $d(H^*\omega)$.** [Naturality of the Pullback](/theorems/3574) is the identity $d \circ H^* = H^* \circ d$ for smooth maps $H$ between smooth manifolds (here, $H: [0,1]^2 \to U$, both regarded as open subsets of Euclidean space). Applied to $\omega$: \begin{align*} d(H^*\omega) = H^*(d\omega) = 0, \end{align*} where the last equality uses the standing hypothesis $d\omega = 0$. This is the only place the closedness hypothesis is used. **Applying Stokes.** [Stokes' Theorem](/theorems/1530) for a smooth $1$-form $\eta$ on a compact oriented $2$-manifold-with-boundary $D$ states $\int_D d\eta = \int_{\partial D} \eta$. We take $D = [0,1]^2$ with the standard orientation from $\mathbb{R}^2$. Strictly speaking $[0,1]^2$ has corners, not boundary, but Stokes extends to manifolds-with-corners essentially without change because the corners have measure zero. The four edges of $\partial[0,1]^2$ traversed with the induced (counterclockwise) orientation are as listed in the exact version. **Computing the boundary integrals.** Three of the four boundary edges are mapped by $H$ to the single point $x_0$: \begin{align*} H(1, t) &= x_0 \quad (\text{right edge: top boundary identity for } s = 1), \\ H(1 - s, 1) &= x_0 \quad (\text{top edge: by } H(\cdot, 1) \equiv x_0), \\ H(0, 1 - t) &= x_0 \quad (\text{left edge: bottom boundary identity for } s = 0). \end{align*} The pullback $H^*\omega$ restricted to each such edge is the pullback of $\omega$ along a constant map, which is the zero $1$-form. So those three integrals vanish. Only the bottom edge contributes: $H \circ e_b(s) = H(s, 0) = \beta(s)$, and by the change-of-variables formula for line integrals (or directly from the definition of pullback), \begin{align*} \int_{e_b} H^*\omega = \int_0^1 \sum_i \omega_i(\beta(s))\, \dot\beta_i(s)\, d\mathcal{L}^1(s) = \int_\beta \omega. \end{align*} **Conclusion.** Combining $\int_D d(H^*\omega) = 0$ (from closedness) with the boundary decomposition (from the homotopy) yields $\int_\beta \omega = 0$, which is $(\star)$. **Why piecewise smooth is enough.** If $\beta$ has finitely many corners, smooth across each corner using a one-dimensional mollification within $U$ — this works because each corner is interior to $U$, hence sits in a small ball $B(\beta(s_j), r_j) \subseteq U$, and convex linear combinations stay in $U$. The integral $\int_\beta \omega$ is continuous in $\beta$ under $C^1$ convergence on each smooth piece, so passing to the smoothed loop and back recovers $(\star)$ for the original piecewise smooth $\beta$. [/guided] [/step] [step:Define the potential $f$ and prove $\partial_{x_i} f = \omega_i$ via the fundamental theorem of calculus] Equation $(\star)$ holds for every piecewise smooth loop based at $x_0$, so by the reduction of the first step, the function \begin{align*} f: U &\to \mathbb{R} \\ x &\mapsto \int_{\gamma_x} \omega \end{align*} is well-defined, where $\gamma_x$ is any piecewise smooth path from $x_0$ to $x$ in $U$. In particular $f(x_0) = 0$ (taking the constant path). Fix $x \in U$ and $i \in \{1, \dots, n\}$. Since $U$ is open, there exists $r > 0$ with $B(x, r) \subseteq U$. For $h \in (-r, r)$, let $\sigma_h: [0, 1] \to U$ be the line segment $\sigma_h(\tau) := x + \tau h e_i$, where $e_i$ is the $i$-th standard basis vector; this segment lies in $B(x, r) \subseteq U$ since $B(x, r)$ is convex. The concatenation $\gamma_x * \sigma_h$ is a piecewise smooth path from $x_0$ to $x + h e_i$, so \begin{align*} f(x + h e_i) - f(x) &= \int_{\sigma_h} \omega \\ &= \int_0^1 \sum_{j=1}^n \omega_j(x + \tau h e_i)\, \dot\sigma_{h,j}(\tau)\, d\mathcal{L}^1(\tau) \\ &= \int_0^1 \omega_i(x + \tau h e_i)\, h\, d\mathcal{L}^1(\tau), \end{align*} where in the last line only the $j = i$ term survives because $\dot\sigma_h(\tau) = h e_i$. Substituting $u = \tau h$ gives $du = h\, d\mathcal{L}^1(\tau)$ and \begin{align*} f(x + h e_i) - f(x) = \int_0^h \omega_i(x + u e_i)\, d\mathcal{L}^1(u). \end{align*} The integrand $u \mapsto \omega_i(x + u e_i)$ is continuous on $(-r, r)$ since $\omega_i \in C^\infty(U)$. By the [Fundamental Theorem of Calculus](/theorems/632), \begin{align*} \partial_{x_i} f(x) = \lim_{h \to 0} \frac{f(x + h e_i) - f(x)}{h} = \omega_i(x). \end{align*} This holds for every $x \in U$ and every $i$, so $\partial_{x_i} f = \omega_i$ on $U$. Since each $\omega_i \in C^\infty(U)$, $f$ admits continuous first partial derivatives, hence $f \in C^1(U)$. Iterating, $\partial_{x_i} f = \omega_i$ is itself $C^\infty$, so all higher partial derivatives of $f$ exist and are continuous: $f \in C^\infty(U)$. Finally, \begin{align*} df = \sum_{i=1}^n \partial_{x_i} f \, dx_i = \sum_{i=1}^n \omega_i \, dx_i = \omega. \end{align*} **Uniqueness up to a constant.** If $g \in C^\infty(U)$ also satisfies $dg = \omega$, then $d(f - g) = 0$, i.e., all partial derivatives of $f - g$ vanish on the path-connected [open set](/page/Open%20Set) $U$. For any $x, y \in U$, choose a piecewise smooth path $\gamma$ from $x$ to $y$; then $(f-g)(y) - (f-g)(x) = \int_\gamma d(f - g) = 0$. Hence $f - g$ is constant on $U$. [/step] [step:Deduce the conservative vector field equivalence] Let $F: U \to \mathbb{R}^n$ be smooth with components $F_i = F \cdot e_i \in C^\infty(U)$, and define $\omega_F = \sum_{i=1}^n F_i\, dx_i \in \Omega^1(U)$. *Forward direction.* If $F$ is conservative, say $F = \nabla f$ for some $f \in C^\infty(U)$, then $\omega_F = \sum_i \partial_{x_i} f\, dx_i = df$. By [Exact Forms Are Closed](/theorems/3565), $d\omega_F = d(df) = 0$. *Reverse direction.* If $d\omega_F = 0$, apply the result already proved to obtain $f \in C^\infty(U)$ with $\omega_F = df$. Reading off components yields $F_i = \partial_{x_i} f$ for each $i$, i.e., $F = \nabla f$, so $F$ is conservative. Combining both directions gives the stated equivalence and completes the proof. $\blacksquare$ [/step]