We do know that
For a vector field $f=(f_1, f_2)\in C^1$ on simply connected set $S \in \mathbb{R}^n$ satisfying $$D_1 f_2 = D_2 f_1$$ is a gradient of some potential function $\phi$.
, and I have also seen examples when the set is not simply connected, but only connected, the above theorem is not sufficient for $f$ to be a gradient.
However, I do not understand why do we need the set to be simply connected. In other words, when the set is only connected, what might cause a problem for sufficiency, and when it is simply connected, how does simpleness solve that problem.
Edit:
I mean I have read the proof, (from the book Multivariable Calculus by Apostle), but the author assumes a convex set instead of a simply connected set, and uses the convexity to construct a potential function while taking the line integral along the direct line segment joining the fixed point $a$ and the point $x$, but this theorem holds even when the set is only simply connected, and with Apostol's sketch of the proof, we do not need simpleness, but only path connectedness, so I do not understand why do we need simpleness in here ?
Further:
Is this condition (simply connecteness) is the "smallest" condition that this theorem holds ?
You can find $\phi$ by integrating along a path. However, in order for this to be well-defined, the result must not depend on the path taken. Simply put, if $S$ is simply connected then any tow different paths from $a$ to $b$ can be transformed into one another continuously, and the condition on $f$ guarantees that infinitesimal changes to the path do not change the integral.