Restriction of a differential form along an inclusion (example)?

Question

I am trying to figure out if the restriction of a differential form on a manifold $M$ onto an open subspace of $M$ is the same as the original form?

Take, for example, the $1$-form $\alpha=-ydx+xdy$ on $S^1$ and the open $U = S^1 \setminus \left\{(0,1)\right\}$ of $S^1$. We know that $\alpha$ is not exact on $S^1$, but its restriction to $U$ is exact according to Poincare lemma. Therefore, we can write $$\alpha = df$$ for some function $f$ on $U$. I cannot understand this as it seems contradictory to me. I mean why we cannot do this on $S^1$ as well?

In general, is the restriction of a differential form $\alpha$ on $M$ to open $U$ of $M$ have the same formula?

Answer 1

It is true that the restriction to an open subset is given by the same formula, but this does not lead to any contradiction. For the example you use, the function $f$ is just the angular coordinate in polar coordinates, for example you can take $f(x,y)$ for $(x,y)\in S^1\setminus\{(-1,0)\}$ to be the unique $t\in (-\frac{3\pi}2,\frac{\pi}2)$ such that $x=\cos t$ and $y=\sin t$. This is a smooth function on $S^1\setminus\{(-1,0)\}$ such that $df=\alpha$ on this subset. (Indeed, this is the unique solution of $df=\alpha$ up to an additive constant.) But obviously $f$ does not extend to $S^1$ (not even continuously), so this does not say anything about $\alpha$ on $S^1$.