Question Regarding a form that is closed but not exact, and smooth angle function on the punctured plane In Jack Lee's introduction to smooth manifold. Example 11.48 in Chapter 11.
It discusses that the 1 form $\omega = \frac{x dy - y dx}{x^2+y^2}$ is a closed 1 form but not exact (because it is not conservative).
Then Jack goes on to discuss why this form cannot be exact intuitively, here's what he says
"On the other hand, if we restrict the domain to the right
half-plane U = {(x,y) | x>0}, a computation shows that $\omega = d (\tan^{-1}\frac{y}{x})$
.
This can be seen more clearly in polar coordinates, where $\omega = d(\theta)$. The problem,
of course, is that there is no smooth (or even continuous) angle function on all of $\mathbb{R}^2 \backslash  (0,0)$, which is a consequence of the “hole” in the center."
What is a smooth angle function defined on a smooth manifold, and why is the hole in the center a problem, shouldn't the removal of the origin actually takes care of the problem of not having an angle defined at $(0,0)$?
I am at the beginning of the De Rham Cohomology Chapter, and I am trying to understand the relationship between the shape of a smooth manifold and what it has to do with differential forms not being exact, and this first example rather baffles me.
Thank you for your help. 
 A: Identify $\mathbb R^2$ with $\mathbb C$.  The trick is to define the angle function implicitly: an angle function on a subset $U \subset S^1$ is a continuous function $\theta : U \to \mathbb R$ such that $e^{iθ(p)} = p,$ for all $p \in U.$ This definition makes sense intutively because if $e^{iθ(p)}=\cos \theta(p)+i\sin \theta (p)=p$, then $\theta$ must map $p$ to the angle that the line segment from the origin to $p$ makes with the positive $x$ axis. Of course, there is the question of well-definedness and continuity of $\theta$, which is addressed by the following standard result:
There exists an angle function on an open subset $U\subset  S^1$ if and only if $U\neq S^1$.
Suppose $\theta :U\to \mathbb R$ is an angle function and $U = S^1
\theta$ is continuous and $S^1$ is compact and connected, so $\theta(S^1)$ is an interval closed interval. And since $\theta$ is injective (why?), it is in fact, a homeomorphism from $S^1$ onto its $\theta (S^1)$. But, removing one point from $\theta (S^1)$ disconnects it, whereas removing one point from $S^1$ does not. Contradiction.
On the other hand,  Suppose $U \nsubseteq S^1$ is open in $S^1$. Define $\varphi : [0, 2\pi]\to S^1$ by $\varphi(t) = e^{it}$
.
Continuity of $\varphi$ implies that $V =\varphi^{-1}(U)$ is open in $[0, 2\pi].$ It follows that $\varphi$ is injective on $V$ (why?). So the restriction of $\varphi$ to $V$ is a continuous bijection  $:V\to U$. Now. define $\theta = (\varphi|_V )^{-1}$. Then $e^{iθ(p)} = p$, for all $p \in U$ and if $I\subset V$ is an open interval then $\theta^{-1}(I)=\varphi (I)$ so $\theta$ is continuous. 
Now that we have an angle function, we can relate it to the $1$-form $\omega = \frac{x dy - y dx}{x^2+y^2}$:
Without loss of generality, $(0,1)\notin U$. Then, if $p=(x,y)\in S^1$, we may take $\theta(x,y)=\tan^{-1}(y/x)$ 
and then because 
$d\theta \left (\frac{\partial}{\partial x}\right)=\frac{\partial \theta}{\partial x}=\frac{-y}{y^2+x^2}=\omega\left (\frac{\partial}{\partial x}\right)$ 
and 
$d\theta \left (\frac{\partial}{\partial y}\right)=\frac{\partial \theta}{\partial y}=\frac{x}{y^2+x^2}=\omega\left (\frac{\partial}{\partial y}\right)$ 
it follows that  $\omega=d\theta.$
