There are two questions here at MSE that are very similar to mine, namely Does zero curl imply a conservative field? and Why does $\vec\nabla\times\vec A=\vec0$ imply $\vec A=\vec\nabla B$? But they don't fully answer my question.
The accepted answer for the first linked question claims that $\operatorname{curl}(\mathbf a)=\mathbf0$ does not imply that $\mathbf a$ is conservative but the point that confuses me is that once you extend from the field defined over a simply connected subspace
to the "non conservative" version over a space that is not simply connected
then there is no continuous scalar field whose gradient can produce such a vector field. There would be a discontinuity at the line where they join.
With this, can we say that if $\mathbf a\colon U\subset\mathbb R^m\to\mathbf{R}^m$ is a vector field (with $m=2, 3$) then $\operatorname{curl}(\mathbf a)=\mathbf 0\nRightarrow\mathbf{a}=\boldsymbol{\nabla}b$ unless $U$ is simply connected?
Also, if $\mathbf{a}=\boldsymbol{\nabla}b$ can we say that $\mathbf{a}$ is conservative even if $U$ is not simply connected?