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    $\begingroup$ 1. Penultimate sentence should reach "since this holds for all closed paths, this means ...". 2. You have proved that if ${\bf A} = \nabla \psi$ then $\nabla \times {\bf A} = 0$ but not the converse. The question is whether $\nabla \times {\bf A}=0$ is enough to guarantee the existence of a $\psi$ of which $\bf A$ is the gradient. $\endgroup$
    – Andrew Steane
    Commented Oct 5, 2023 at 11:38
  • $\begingroup$ @AndrewSteane Thank you for the first comment. For the second bit, since we already see that $\nabla \times \mathbf{A} = 0$, then via Helmholtz decomposition (en.wikipedia.org/wiki/…) doesn't it then follow that $\mathbf{A} = \nabla \psi$? In particular, I have shown that if $\nabla \times \mathbf{A} = 0$ then $\mathbf{A} = \nabla \psi$. $\endgroup$
    – S.G
    Commented Oct 5, 2023 at 20:53