A proof of exactness of closed 1-forms on the two-sphere Remark. I'm aware that here the same problem is solved. I'm in trouble with the proof I'm going to quote.
Consider the following

Claim. Every closed 1-form $\beta$ on $S^2$ is exact.

This is an exercise from Schutz, Geometrical methods of mathematical physics, ex. 4.24(c). The idea is to construct a function $f$ such that $\beta = \mbox{d} f$. The solutions is as follows. 

Solution. We are given $\beta$ defined everywhere on $S^2$ with $\mbox{d} \beta = 0$. Integrate $\mbox{d}\beta$ over any region of $S^2$ bounded by a single closed curve $C$ to find $\oint_C \beta = 0$ for any $C$. This can only be true if $\beta = \mbox{d}f$ for some $f$: otherwise some $C$ can be found on which $\beta$ has nonvanishing integral. In fact $f$ can be constructed by choosing an arbitrary value $f_0$ at a point $P$ and integrating $\beta$ on a curve from $P$ to any point $Q$ definining 
  $$f(Q) = f_0 + \int \beta.$$
  The condition $\oint \beta = $ guarantees that $f(Q)$ is independent of the path from $P$ to $Q$. 

I can't understand statements emphasized in italics: for the first one, assigned arbitrarly a closed curve $C$ on $S^2$, $C$ is the boundary of some region $U$. Here $\mbox{d}\beta = 0$, hence by Stokes' theorem $\int_U \mbox{d}\beta = \oint_c \beta = 0$.
So the $C$ to which the author refers can't be closed, so what's the point?
For the second one, choosing the curve connecting $P$ and $Q$ to be closed, the integral must vanish, hence $f(Q) = f_0$ for any $Q$, being $Q$ arbitrary. Again, I can't see the point here.
Probably, I'm missing something. Might anyone elucidate these arguments to me? Thanks in advance.
 A: What you (and the textbook you are using) are missing is Stokes' theorem for non-simple closed curves $C$ in $S^2$: $\int_C \omega=0$ provided that $\omega$ is a closed 1-form on $S^2$. The curves $C$ appear as concatenations $C= C_1 * C_2$, where $C_1$ is a curve from $P$ to $Q$ and $C_2$ is a curve from $Q$ to $P$. If $\bar{C}_2$ denotes the reverse of the curve $C_2$ then you get:
$$
0=\int_{C} \omega= \int_{C_1}\omega + \int_{C_2}\omega= \int_{C_1}\omega - \int_{\bar{C}_2}\omega. 
$$
Hence,
$$
\int_{C_1}\omega = \int_{\bar{C}_2}\omega. 
$$
In particular, the integral of $\omega$ is independent of the choice of a curve connecting $P$ to $Q$ (but does depend, of course, on $P$ and $Q$). Note that the curves $C_1, C_2$ could intersect (and in infinitely many point!) which makes $C$ potentially non-simple. 
You can reduce this more general Stokes formula for curves with finitely many self-intersections to the case of simple curves by cut-and-paste procedure. However, this will become quite nasty if set of self-intersection points is "large" (e.g. a Cantor set).  
Proving the general Stokes formula for non-simple curves hinges on the following property (called simple connectivity): For every closed curve $C$ in $S^2$, regarded as a map $\gamma$ from the unit circle  $S^1$ to $S^2$, there exists an extension of $\gamma$ to a map of the unit disk $\Gamma: D^2\to S^2$. (There is an issue of the degree of differentiability of the maps involved, which I am ignoring.) Then the (more general) Stokes theorem states:
$$
\int_C \omega:= \int_\gamma \omega= \int_\Gamma d\omega:= \int_{D^2} \Gamma^*(d\omega). 
$$
If $\omega$ is closed, this integral vanishes:
$$
\int_C \omega=0. 
$$ 
You can avoid using non-simple curves by restricting integration to arcs of great circles in the 2-sphere.  
