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$U$ is the union of two disjoint open simply connected sets. $\mathbf{F}:U\to\mathbb{R}^3$ is $C^1$. Then is it true that if $\nabla\times \mathbf{F}=\mathbf{0}$, then $\mathbf{F}$ is conservative?

The calculus book I am reading allows this type of $U$.

I think it is true. If $\gamma:[a,b]\to U\subset\mathbb{R}^3$ is a piecewise $C^1$ closed path, $\gamma([a,b])$ is in one of the two simply connected sets since $\gamma$ is continuous. Then using Stoke's Theorem, $\int_{\gamma}\mathbf{F}=0$, which implies $\mathbf{F}$ is conservative.

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  • $\begingroup$ This is trivially true. The Poincare lemma gives us a potential $\Phi$ whose gradient is $F$ in each of the simply connected components of $U\,.$ $\endgroup$
    – Kurt G.
    Commented Apr 13 at 8:09
  • $\begingroup$ So if $U=U_1\cup U_2$, $\Phi$ such that $\Phi=\Phi_1$ on $U_1$ and $\Phi=\Phi_2$ on $U_2$? $\endgroup$
    – user1099762
    Commented Apr 13 at 8:17
  • $\begingroup$ Yes. You don't even have to use the notation $\Phi_1,\Phi_2\,.$ The potential is by (Poincare's lemma) defined on each $U_i$ and therefore trivially defined on the disjoint union $U=U_1\cup U_2\,.$ $\endgroup$
    – Kurt G.
    Commented Apr 13 at 10:13
  • $\begingroup$ Wikipedia and a book I read have Poincare's lemma for a simply connected open set. Would you mind referring me to a source with the lemma for $U=U_1\cup U_2$? $\endgroup$
    – user1099762
    Commented Apr 13 at 16:46
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    $\begingroup$ Looks right. $\phantom{.}$ $\endgroup$
    – Kurt G.
    Commented Apr 13 at 19:01

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Let $U=U_1\sqcup U_2$ where $U_1,U_2$ are open simply connected. $F\vert_{U_i}$ is $C^1$. Then since $\nabla\times F\vert_{U_i}=0$, by Poincare's lemma, we have $\Phi_i$ on $U_i$ such that $\nabla \Phi_i=F\vert_{U_i}$.Defining $\Phi=\Phi_i$ on $U_i$, $\nabla \Phi=F$.

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