Let $U := \mathbb R^2 \setminus \{0\}$ be the punctured plane and $\gamma$ the counter clockwise once traversed unit circle. Show that the map $$\tag{1} [α] \mapsto \int^{2 \pi}_{0} \alpha_{\gamma (t)}(\gamma' (t))dt $$ between the first deRham cohomology group $H^1_{dR}(U, R)$ and $\mathbb R$ is a linear isomorphism.
Remark: In a separate question I asked if the mapping $(1)$ is well-defined.
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So, the first cohomology group is the closed forms in $\Omega^{1}_{dR}$ modulo the exact forms.
And we know that $\gamma$ is the once traversed unit circle (which may be parameterized by $cos^{2}(\theta) + sin^{2}(\theta) = 1$)
Kyle, here is an outline of what you need to do:
(1) Find a (reasonably famous) closed $1$-form $\omega$ on $U$ whose integral around the unit circle is not $0$ (it may well be $2\pi$). How does using $\omega$ show that the map is surjective?
(2) Say $C$ is the unit circle oriented counterclockwise. Suppose $\displaystyle\int_C\alpha = 0$. You must show that $\alpha$ is exact. The idea is to define $f(x) = \displaystyle\int_{(1,0)}^x \alpha$ and show that this is well-defined, independent of path from $(1,0)$ to $x$ [you may need a little bit of hand-waving here to see that any two paths in $U$ differ by a path that is "essentially" some number of times of going around the unit circle — see if you can make this more precise]. Then you need to know or check that $df = \alpha$. Perhaps you did this in class.