1 -form on Compact Manifold This question was asked in my quiz on smooth manifolds and I couldn't solve it during the exam. I tried this problem again but still couldn't solve it. I am not really good in solving problems related to differential forms despite reading course notes 3 times. So, I am posting it here looking for hints. I have been following introduction to smooth manifolds by John Lee along with my course notes.

Question:(a) Let $ M$ be a compact manifold of dimension $n>1$. Show that a nowhere vanishing $1$-form on $M$ cannot be exact.

(b) Let $w=fdx + gdy$ be a closed $1$-form on $\mathbb{R}^2$. Show that $w$ is exact.
Attempt: (a) Let on the contrary the $1$-form be exact.  That means if a was $1$-form then it is exterior derivative of another $1$-form $b$. But unfortunately, I am not sure which result to use to proceed towards a contradiction. Can you please give some hints?
(b) form $w$ is closed implies that $dw =0$, which implies that $dw= df dx + dg dy+ f d^2x + g d^2 y$. I have to show that $w$ is exact. But again I am not sure how should I proceed.
Kindly give a couple of hints. I shall be really thankful.
 A: For (a), if $w = db$ for $b$ a zero-form (smooth function $M \to \mathbb{R}$), then by compactness this smooth function achieves a global maximum at some $p \in M$. Choosing a local coordinate system $x^1, \ldots, x^n$ around $p$, the differential $d$ is given by
$$
   db = \sum_i \frac{\partial b}{\partial x^i} dx^i,
$$
which is zero at the maximum since $b$ is smooth.
For (b), there are several ways to prove this. Since we are only working in $\mathbb{R}^2$, the most down-to-earth would be something like the following: the condition that $\omega = f dx + g dy$ is closed implies that $$\frac{\partial f}{\partial y} = \frac{\partial g}{\partial x}.$$ (Why?) This is a good sign: we are after all looking for some function $F$ with the property that its $x$-partial derivative is $f$ and its $y$-partial derivative is $g$, and then this equality is just saying
$$\frac{\partial^2 F}{\partial x \partial y} = \frac{\partial^2 F}{\partial y \partial x},$$
which is true. To reverse-engineer what $F$ should be, we start at one point and then integrate along paths to get the result at any other point. Start by defining $F(0) = 0$ (this is arbitrary, because shifting $F$ up or down by a constant won't affect any of these partial derivatives). Then for any other point $p = (x_0, y_0)$, define
$$
   F(x_0, y_0) = \int_{0}^{x_0} f(x, 0) dx + \int_{0}^{y_0} g(x_0, y) dy.
$$
Is it clear what these integrals are doing? They are traversing from $(0,0)$ to $(x_0, y_0)$ by going first horizontally along the $x$-axis, then vertically up to $(x_0, y_0)$. As they progress, they are measuring how much $F$ is supposed to be increasing, via its partial derivatives in the appropriate direction. One can then do some analysis of partial derivatives and use the FTOC to show that the function $F$ we have defined this way is smooth, and that $dF = f dx + g dy$.
This can all be done much more generally. The path we took to get to $(x_0, y_0)$ was arbitrary; one could use a straight line, or another curve instead. For a lot of this worked out carefully on manifolds, see Lee's Smooth Manifolds, Chapter 8, I think.
I'd be remiss if I didn't mention for posterity that this is all subsumed by the theorem that de Rham cohomology is homotopy-invariant. De Rham cohomology is an algebraic gadget measuring how far apart the properties of being closed and exact are. Homotopy invariance means that we can squish the whole plane down to a point, check that every closed 1-form is exact on a point (which is trivial because there aren't any since a point is zero-dimensional), and conclude that the same is true for $\mathbb{R}^n$ for any $n$. Very powerful! See for example Lee, Chapter 17.
