No. "Every closed form is exact" is equivalent to the claim that the first de Rham cohomology $H^1_{dR}(U, \mathbb{R})$ vanishes. This means, equivalently, that there are no nontrivial homomorphisms from the fundamental group $\pi_1(U)$ to $\mathbb{R}$. But it does not imply that the fundamental group is trivial (which is equivalent to simply connected).
In fact any reasonable space (e.g. a manifold) is homotopy equivalent to an open subset of some $\mathbb{R}^n$ (embed it into $\mathbb{R}^n$ in a nice way, then take a tubular neighborhood), and lots of reasonable spaces have the property that their fundamental group is nontrivial but admits no nontrivial maps to $\mathbb{R}$. Maybe the simplest such space is the real projective plane $\mathbb{RP}^2$, which embeds nicely into $\mathbb{R}^4$ (I think) and has fundamental group $\mathbb{Z}_2$.
If you really want an explicit open subset of some $\mathbb{R}^n$, you can take $GL_3^{+}(\mathbb{R})$, the space of $3 \times 3$ matrices with positive determinant, which is homotopy equivalent to $SO(3)$, which is in turn the real projective space $\mathbb{RP}^3$, and which in particular again has fundamental group $\mathbb{Z}_2$. This is a connected open subset of $\mathbb{R}^9$.
In general, homomorphisms $\pi_1(U) \to \mathbb{R}$ correspond to homomorphisms $H_1(U, \mathbb{Z}) \to \mathbb{R}$, where $H_1$ is the first singular homology. If the fundamental group $\pi_1(U)$ is finitely generated (which is again true in all reasonable cases), every such homomorphism is trivial iff $H_1(U, \mathbb{Z})$ is torsion iff it is finite.
This proof is not quite right. You need a form defined on all of $M$ in order to apply Stokes's Theorem. By your logic, any time you integrate a form on the boundary you would get $0$. So what is the easy fix here? How do you get an $n$-form on all of $M$?
Best Answer
This is true but as far as I know it is a hard theorem without any elementary proof. By the de Rham theorem, $\omega$ is exact iff its integral over every smooth $k$-cycle is $0$. So, your question is equivalent to asking whether the homology group $H_k(U;\mathbb{R})$ is generated by the fundamental classes of closed oriented submanifolds $M\subseteq U$. This (stated more generally for arbitrary manifolds) is Corollary II.30 of Thom's famous paper
which introduced cobordism theory and earned him his Fields medal.
To very briefly sketch the argument, Thom uses transversality theory to show that if $f:U\to MSO(m)$ is any continuous map that approaches the basepoint as you approach the boundary of $U$, and $u\in H^m(MSO(m))$ is the Thom class, then the Poincaré dual of $f^*(u)\in H_c^m(U)$ is the fundamental class of a closed oriented submanifold of $U$ (namely, the inverse image of the zero section in $MSO(m)$ when you homotope $f$ to be transverse to it). So, it suffices to show that $H_c^m(U;\mathbb{R})$ is generated by classes obtained by pulling back the Thom class along such maps $U\to MSO(m)$. Every class in $H_c^m(U)$ is classified by a map from $U$ to the $n$-skeleton of $K(\mathbb{Z},m)$. By induction on $n$, you can show that the universal $m$th cohomology class on the $n$-skeleton of $K(\mathbb{Z},m)$ has a multiple that is pulled back from the Thom class via a map to $MSO(m)$ (since the relevant obstruction group for extending from the $n$-skeleton to the $(n+1)$-skeleton is torsion). Thus every class in $H_c^m(U)$ has a multiple which is Poincaré dual to the fundamental class of a closed oriented submanifold, and so $H_c^m(U;\mathbb{R})$ is generated by such classes.