Let $U := \mathbb R^2 \setminus \{0\}$ be the punctured plane and $\gamma$ the counter clockwise once traversed unit circle. Consider $$\tag{1} [α] \mapsto \int^{2 \pi}_{0} \alpha_{\gamma (t)}(\gamma' (t))dt, $$ where $[\alpha]$ is a class in the first deRham cohomology group $H^1_{dR}(U, R)$.
Question: Show that this map is well-defined, i.e., independent of the choice of representative of the coset $[\alpha]$. In other words, if $\alpha = df$ then $$\int^{2 \pi}_{0} \alpha_{\gamma (t)}(\gamma' (t))dt = 0 $$
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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 $cos^{2}(\theta) + sin^{2}(\theta) = 1$)
We know $\alpha$ is a $1$-form (since it's the exterior derivative of a function) and that $[\alpha ]$ is the coset of $\alpha$ and is equivalent to multiplying by $\gamma'$ then integrating from 0 to $2\pi$, but would $\alpha$ just be the exterior derivative of $\gamma$? Or is it the exterior derivative of just some arbitrary function, $f$?
You are asked to prove that the mapping defined in $(1)$ is well defined: That is, if $[\alpha] = [\beta]$, is it true that
$$ \tag{2}\int^{2 \pi}_{0} \alpha_{\gamma (t)}(\gamma' (t))dt = \int^{2 \pi}_{0} \beta_{\gamma (t)}(\gamma' (t))dt?$$
By definition of de Rham cohomology, if $[\alpha] = [\beta]$, then $\alpha$ and $\beta$ differs by an exact form: that is,
$$\beta -\alpha =df, \ \ \text{or } \beta = \alpha + df$$
for some function $f: U\to \mathbb R$. So showing $(2)$ is the same as showing
$$\int^{2 \pi}_{0} df_{\gamma (t)}(\gamma' (t))dt =0.$$
But showing $(3)$ is indeed very easy: since
$$df_{\gamma (t)}(\gamma' (t)) = (f\circ \gamma)'(t)$$
by Chain rule (check!). Since $\gamma(0) = \gamma(2\pi)$, we have
$$\int^{2 \pi}_{0} df_{\gamma (t)}(\gamma' (t))dt =f(\gamma(2\pi)) - f(\gamma(0)) = 0$$
and we are done checking $(2)$.