Let $M = \mathbb{R}^2 \setminus \{0\}$. We have a global coordinates $(x, y)$ on $M$. We want to compute the cohomology of the complex
$$0 \to \Omega^0(M)\ \stackrel{d}{\to}\ \Omega^1(M)\ \stackrel{d}{\to}\ \Omega^2(M)\ \to 0$$
where $\Omega^k(M)$ is the space of smooth $k$-forms on $M$. So we want to compute $H^k(M) = Z^k(M)/B^k(M)$ where $Z^k(M) = \{\alpha \in \Omega^k(M),\, d\alpha = 0\}$ (closed $k$-forms) and $B^k(M) = \{d\beta,\, \beta \in \Omega^{k-1}(M)\}$ (exact $k$-forms). In this situation,
- Elements of $\Omega^0(M)$ are just smooth functions $f(x,y)$ on $M$
- Elements of $\Omega^1(M)$ can be written $u(x,y) dx + v(x,y) dy$ (where $u$ and $v$ are smooth functions on $M$)
- Elements of $\Omega^2(M)$ can be written $g(x,y) dx \wedge dy$ (where $g$ is a smooth function on $M$)
Compute $H^0(M)$:
$B^0(M) = \{0\}$ and $Z^0(M)$ consists of functions $f(x,y)$ such that $df = 0$. Since $M$ is connected, this implies that $f$ is constant. It follows that $H^0(M)$ is isomorphic to $\mathbb{R}$.
Compute $H^1(M)$:
This is the where the all the fun happens :)
Let $\alpha \in Z^1(M)$, this means that $\alpha = u(x,y) dx + v(x,y) dy$ with $\frac{\partial v}{\partial x} - \frac{\partial u }{\partial y} = 0$.
Lemma 1: Let $R$ be a closed rectangle in $\mathbb{R}^2$ which does not contain the origin. Then $\int_{\partial R} \alpha = 0$.
Proof: This is just Green's theorem (or Stokes' theorem) (in its most simple setting, where it's not hard to show directly).
Lemma 2: Let $R$ and $R'$ be closed rectangles in $\mathbb{R}^2$ whose interiors contain the origin. Then $\int_{\partial R} \alpha = \int_{\partial R'} \alpha$.
Proof: This is a consequence of Lemma 1. It might be a bit tedious to write down (several cases need to be addressed, according to the configuration of the two rectangles), but it's fairly easy. You need to cut and rearrange integrals along a bunch of rectangles so that the two initial integrals agree up to integrals along rectangles who do not contain the origin.
Let's denote by $\lambda(\alpha)$ the common value of all integrals $\int_{\partial R} \alpha$ when $R$ is a closed rectangle whose interior contains the origin.
Lemma 3: If $\alpha$ is exact iff $\lambda(\alpha) = 0$.
Proof: "$\Rightarrow$" is trivial1, let's prove the converse. Fix permanently some point $m_0 \in M$, whichever you like best. For any $m\in M$, consider a rectangle $R$ in $\mathbb{R}^2$ whose boundary contains $m_0$ and $m$ but avoids the origin. Let $\gamma$ be one of the two paths joining $m_0$ and $m$ along $\partial R$. Let $f(m) = \int_\gamma \alpha$. Since $\lambda(\alpha) = 0$, this definition does not depend on the choice of the rectangle or the path. In other words $f: M \rightarrow \mathbb{R}$ is well defined. Let's see that $df = \alpha$. Check that $\frac{f(x+h, y) - f(x,y)}{h} = \frac{1}{h}\int_x^{x+h} u(t,y) dt$, so that taking the limit when $h\rightarrow 0$ yields $\frac{\partial f}{\partial x} = u$. Same for $v$.
Finally let's consider the $1$-form $d\theta = \frac{-ydx + xdy}{x^2 + y^2}$. NB: Be well aware that $d\theta$ is a misleading (but standard) notation: it is not an exact form.
Lemma 4: $d\theta$ is a closed $1$-form and $\lambda(d\theta) = 2\pi$.
Proof: This is a direct computation.
Now're done:
Proposition: $H^1(M)$ is the one-dimensional vector space spanned by $[d\theta]$.
where $[d\theta]$ denotes the class of $d\theta$ in $H^1(M)$. Note that $[d\theta] \neq 0$: see Lemma 4 and 3.
Proof: Let $\alpha$ be a closed $1$-form. Consider $\beta = \alpha - \frac{\lambda(\alpha)}{2 \pi} d\theta$. We have $\lambda(\beta) = 0$ so $\beta$ is exact by Lemma 3. This proves that $[\alpha] = \frac{\lambda(\alpha)}{2 \pi} [d\theta]$.
Compute $H^2(M)$:
Let's show that $H^2(M) = 0$, in other words every closed $2$-form on $M$ is exact. This solution is taken from Ted Shifrin in the comments below.
Here's the idea: in polar coordinates, a $2$-form $\omega$ can be written $\omega = f(r,\theta) dr \wedge d\theta$. Then $\eta = (\int_1^r f(\rho, \theta) d\rho)\, d\theta$ is a primitive of $\omega$.
Although it's not very insightful, this can be checked by a direct computation without refering to a change of variables.
Let $\omega(x,y) = g(x,y) dx \wedge dy$. Define $$h(x,y) = \int_1^{\sqrt{x^2+y^2}} t\, g\left(\frac{tx}{\sqrt{x^2+y^2}}, \frac{ty}{\sqrt{x^2+y^2}}\right) \,dt$$
Check that $h(x,y) d\theta$ is a primitive of $\omega$.
NB: Any course / book / notes on de Rham cohomology will show that if $M$ is a connected compact orientable manifold, $H^n(M) \approx \mathbb{R}$. However, I don't think I've read anywhere that when $M$ is not compact, $H^n(M) = 0$. I wonder if there is an "elementary" proof.
1 I'll explain this by request of OP. It is a general fact that if $\gamma : [a, b] \rightarrow M$ is a ${\cal C}^1$ path and $\alpha$ is a smooth exact one-form i.e. $\alpha = df$ where $f$ is a smooth function, then $\int_\gamma \alpha = f(\gamma(b)) - f(\gamma(a))$. This is because $\int_\gamma \alpha = \int_a^b \alpha(\gamma(t))(\gamma'(t))\,dt$ (by definition) and here $\alpha(\gamma(t))(\gamma'(t)) = df(\gamma(t))(\gamma'(t)) = f'(\gamma(t)) \gamma'(t) = \frac{d}{dt}\left(f(\gamma(t))\right)$.
This fact extends to piecewise ${\cal C}^1$ maps by cutting the integral. In particular, is $\gamma$ is a closed piecewise ${\cal C}^1$ path and $\alpha$ is exact, then $\int_\gamma \alpha = 0$.
In fact, it is useful (e.g. for Cauchy theory in complex analysis) to know that
- A one-form is closed iff its integral along any homotopically trivial loop is zero (any boundary of a rectangle contained in the open set is enough).
- A one-form is exact iff its integral along any loop is zero.
Note that the "smooth" condition can be weakened.
This is not a conclusive answer, but here's how I would approach this.
Let $\Delta_n(M)$ be the Abelian group of singular $n$-chains and $\Delta^n(M;\mathbb{R})=\mathrm{Hom}_\mathbb{Z}(\Delta_n(M),\mathbb{R})$ the $\mathbb R$-valued signular $n$-cochains. There are two subcomplexes that are relevant to the question: the complex $\Delta_*^\infty(M)$ of smooth singular chains (as explained in Bredon's book, for example), and the complex $\Delta_c^*(M)$ of compactly supported cochains, i.e. singular cochains that vanish on all chains with image outside of a compact set (which depends on the cochain).
Now, without having a reference or a proof, I would bet some money that the inclusion $\Delta_*^\infty(M)\hookrightarrow\Delta_*(M)$ is a chain homotopy equivalence. This should, in turn, dualize to chain homotopy equivalences
$$\Delta^*(M;\mathbb R)\to\Delta^*_\infty(M;\mathbb R) \quad\text{and}\quad \Delta^*_c(M;\mathbb R)\to \Delta^*_{\infty,c}(M)$$
where $\Delta^n_\infty = \mathrm{Hom}(\Delta_n^\infty,\mathbb R)$ and $\Delta^n_{\infty,c}$ is the compactly supported analogue.
Next, integration gives rise to chain maps
$$\Psi\colon \Omega^*(M)\to \Delta_\infty^*(M) \quad\text{and}\quad \Psi_c\colon \Omega^*_c(M)\to \Delta_{\infty,c}^*(M).$$
Bredon proves that $\Psi$ induces an isomorphism on cohomology and it should be possible to adapt the proof to show the same for $\Psi_c$.
Finally, the cohomology of $\Delta_c^*(M)$ is known a singular cohomology with compact supports, denoted by $H^*_c(M;\mathbb R)$. If $M$ has an orientation, then Poincaré duality gives isomorphisms
$$H^{n-i}_c(M;\mathbb R)\cong H_{i}(M;\mathbb R)$$
with singular homology on the right hand side. And if everything above goes through as claimed, then the left hand side is isomorphic to compactly supported de Rham cohomology $H^{n-i}_{dR,c}(M)$.
I'll try to find some references later. Other duties are calling.
Best Answer
Here are some remarks related to the questions you raise:
First of all, you mention homotopy groups, rather than homology groups. The relationship between homotopy and homology with $\mathbb Z$-coefficients (to the extent that there is one) is given by the Hurewicz theorem. In any event, in discussing Poincaré duality, it is homology and cohomology groups that are directly relevant, rather than homotopy groups.
The relationship between singular homology with $\mathbb Z$ and $\mathbb R$ coefficients is given by the universal coefficient theorem. It states that $H_i(X,\mathbb R) = H_i(X,\mathbb Z)\otimes_{\mathbb Z} \mathbb R.$ (In general, there would be a Tor contribution as well, but that vanishes here, since $\mathbb R$ is torsion free.)
The de Rham Theorem states that the $k$th de Rham cohomology of a smooth manifold is isomorphic to the $k$th singular cohomology of the manifold with $\mathbb R$-coefficients, or, equivalently (by universal coefficients for cohomology), is dual to the $k$th singular homology with $\mathbb R$-coefficients.
Poincaré duality itself, when phrased in terms of de Rham cohomology, states that for a closed, connected, and orientable $n$-dimensional smooth manifold, the $n$th de Rham cohomology group $H^n$ is one-dimensional over $\mathbb R$, and the cup product from $H^k \times H^{n-k}$ to $H^n$ (which in de Rham cohomology is induced by wedge product on forms) is a perfect pairing.
Taking into account the relationship between de Rham and singular theory stated above, this can also be phrased as an isomorphism between $H^{n-k}$ and $H_k$ with $\mathbb R$-coefficients. But note that the homology under consideration has $\mathbb R$-coefficients, not $\mathbb Z$-coefficients! Note also that Poincaré duality is a statement for closed manifolds (and so does not apply, at least in the naive form that you have stated it, to a non-compact manifold such as the punctured plane).
There is a version of Poincaré duality with $\mathbb Z$-coefficients (as well as generalizations to non-closed and/or non-orientable manifolds); see the wikipedia page. Note though that if you want to consider homology with $\mathbb Z$-coefficients, then you will also need to consider cohomology with $\mathbb Z$-coefficients, and de Rham theory doesn't do this. (You will need to use some other form of cohomology, such as singular, Cech, or sheaf cohomology.)