I'm reading from Bott-Tu's Differential forms in algebraic topology, and came across the following exercise:
(p52 Exercise 5.16(b)) Let $S$ be the unit circle in the plane, and $M := \mathbb{R}^2-\{0\}$. Show that the closed Poincare dual of the unit circle $i : S\hookrightarrow M$ in $H^1(M)$ is 0, but the compact Poincare dual is the nontrivial generator $\rho(r)dr$ in $H^1_c(M)$, where $\rho(r)$ is a bump function with total integral 1 (By a bump function we mean a smooth function whose support is contained in some disc and whose graph looks like a "bump").
(My question is near the bottom)
To compute the closed dual, we must check that there is a closed form $\eta = \eta_r dr + \eta_\theta d\theta$ on $M$ such that for every closed compactly supported form $\omega = \omega_rdr + \omega_\theta d\theta$ on $M$, we have $$\int_S i^*\omega = \int_M\omega\wedge\eta$$ However, since $\omega$ is compactly supported and closed, realizing $S$ as the boundary of the punctured unit disk, Stoke's theorem tells us that $\int_S i^*\omega = 0$, so we may pick $\eta = 0$.
To compute the compact dual, we must now find a closed compactly supported form $\eta' = \eta_r' dr + \eta_\theta' d\theta$ on $M$ such that for every closed (not necessarily compactly supported) form $\omega = \omega_rdr + \omega_\theta d\theta$, we have again:
$$\int_S i^*\omega = \int_M\omega\wedge\eta$$
If we assume $\eta' = \rho(r)dr$ as in the exercise, we wish to check: $$\int_S i^*\omega = \int_0^{2\pi}\omega_\theta(1,\theta)d\theta = \int_\epsilon^{\epsilon'}\rho(r)\left(\int_0^{2\pi}\omega_\theta(r,\theta)d\theta \right)dr$$ where the support of $\rho(r)$ is contained in the annuli of radii $\epsilon,\epsilon'$. Again, applying Stoke's theorem to the closed annulus with radii $R_1,R_2$, we find that the integral $\int_{0}^{2\pi}\omega_\theta(r,\theta)d\theta$ is independent of the choice of $r$. Thus, $\rho(r)dr$ has total integral 1, the equality is checked as desired.
My question is: Suppose we didn't know to guess $\eta' = \rho(r)dr$. How would we have deduced it? Ie, let us just assume that $\eta' = \eta'_rdr + \eta'_\theta d\theta$, with support on the annulus of radii $\epsilon,\epsilon'$. Thus we want to find $\eta'_r,\eta'_\theta$ such that: $$\int_0^{2\pi}\omega_\theta(1,\theta)d\theta = \int_0^{2\pi}\int_\epsilon^{\epsilon'}\omega_r\eta'_\theta drd\theta - \int_0^{2\pi}\int_\epsilon^{\epsilon'}\omega_\theta\eta'_r drd\theta$$ The fact that $\omega,\eta$ are closed is the same as saying that $\frac{\partial\omega_r}{\partial\theta} = \frac{\partial\omega_\theta}{\partial r}$ and $\frac{\partial\eta'_r}{\partial\theta} = \frac{\partial\eta'_\theta}{\partial r}$. How can we deduce that we may pick $\eta'$ to be of the form $\eta' = \rho(r)dr$?
We already have abstractly that $$\int \iota^* \omega = \int \omega \wedge \eta,$$ for some $\eta$, so all we need is to identify it. As you point out, the left-hand side is independent of $\omega_r$, so the first term in $$\int_0^{2\pi} \int_0^\infty \omega_r \eta_\theta drd\theta - \int_0^{2\pi} \int_0^\infty\omega_\theta \eta_r drd\theta$$
had better be $0$, which necessitates $\eta_\theta = 0$.
Then your closedness equation gives $\frac{\partial \eta_r}{\partial \theta} = 0$, so $\eta = \eta_r(r)dr$. The normalization is clear, I think.