Meaning of Cauchy integral theorem - the (co)homology viewpoint

I'm not sure what follows is not just a complicated way to deduce a blatant triviality, or if is even correct. Let's try.

In the elementary theory of analytic functions of $1$ complex variable, one proves the classical result:

Theorem 1(Cauchy) Let $U\subseteq \mathbb{C}$ be a domain (i.e. open connected), $\omega:=fdz\in\Omega^1(U)$ a holomorphic 1-form, and $\gamma\in B_1(U)\subseteq Z_1(U)$ a closed cycle which is a boundary, i.e. it's homologous to 0 in $U$. Then $\langle \omega, \gamma \rangle :=\int_\gamma\omega=0$.

And also a kind of inverse (which I will state in less generality than usual):

Theorem 2 (Morera) Let $\omega:=fdz\in \mathcal{A}^{1,0}(U)$ a smooth 1-form of type $(1,0)$. Suppose that for every homologous to zero cycle $\gamma \in B_1(U)$ we have $\langle\omega,\gamma\rangle=0$. Then $\omega\in\Omega^1(U)$, i.e. it's holomorphic (which, in this case, is the same of saying that $\omega$ is $d$-closed where $d=\partial + \bar{\partial}$ is the de Rham differential).

[Edit:] The goal of this question is to understand the (co)homological interpretation of the classical Cauchy's and Morera's integral theorems, separating their "topological" content from their "complex analytic" content, perhaps disregarding the "Real Analysis" content (i.e. Morera's theorem could be stated for any continuous $f$, not just smooth).

Essentially, Cauchy's theorem states that the 'integral' pairing

$\int: \Omega^1(U)\times Z_1(U)\to \mathbb{C}$

where $\Omega^k$ denotes holomorphic differential k-forms, descends to a pairing involving homology:

$\int:\Omega^1(U)\times H_1(U)\to \mathbb{C}$.

Let $Z^1_{hol}(U)$ and $B^1_{hol}(U)$ be the $\partial$-closed and $\partial$-exact holomorphic 1-forms. One can see that $Z^1_{hol}(U)=\Omega^1(U)$ and $B^1_{hol}(U)=\Omega^1(U)\cap\mathrm{d}\mathcal{C}^{\infty}(U)$. Since exact forms have zero integral on closed loops, the above pairing descends further to:

$H^1(\Gamma(U,\Omega^{\bullet}_U))\times H_1(U)\to\mathbb{C}$.

Now, since $U$ is a Stein manifold, if I'm not mistaken, we have $H^i(\Gamma(U,\Omega^{\bullet}_U))\cong \mathbb{H}^i(\Omega^{\bullet}_{U})$ where the latter is holomorphic de Rham cohomology. And now I think that by Dolbeault/Poincaré lemma $\Omega^{\bullet}_{U}$ is a resolution of the constant sheaf $\mathbb{C}$, so the above hypercohomology is just sheaf cohomology of the sheaf $\mathbb{C}$, which is dual to $H_1(U,\mathbb{C})$, which, by P.D. on non-compact manifolds is in turn dual to de Rham with compact supports (and complex coefficients) $H^1_{dR,c}(U)$ [edit: added the compact supports thing]. So we get a pairing:

$H^1_{dR\, c}(U)\times H_1(U) \to \mathbb{C}$

that is a map

$PD: H^1_{dR\, c}(U)\to H_1(U,\mathbb{C})^*$

which, were $U$ compact, would just be the Poincaré duality [edit: which should be nothing but Poincaré duality for non-compact manifolds].

So my questions:

0) Are the above passages correct?

1) Is it possible to recover Cauchy's theorem (Theorem 1) from some version of Poincaré duality by just doing the above passages in the inverse order? [edit: so, it seems the answer is "yes"]

2) Can we say the map $PD$ defined above is an isomorphism? [edit: given 1), the answer is "yes"]

3) What about Theorem 2? Does it have a cohomological interpretation? Maybe something like: $\forall R, 0=\langle \partial R, \omega\rangle=\langle R, d \omega\rangle$ then $d\omega=0$ ?

• I hope you don't mind; I fixed some of your TeX. – Daniel Litt Mar 21 '11 at 1:42
• ....arrgh!! It should be Stokes', not Stoke's...I keep making that mistake!!! – Zen Harper Mar 21 '11 at 9:17
• @Zen Harper: On the other hand, as you remark, Morera's theorem is usually stated in greater generality, i.e. assuming $f$ to be just continuous (rather than smooth). But my goal in this question was to give a cohomological interpretation of Cauchy's and Morera's theorems, separating their "topological" content from their "complex-analytic" content. If my argument is reliable, Cauchy's theorem boils down to local exactness of holomorphic differential forms (Dolbeault's lemma) and Poincaré duality. – Qfwfq Mar 21 '11 at 11:51
• Its worth noting that for a complex one dimensional manifold $U$, $\Omega^{1,1}(U)=\Omega^2(U)$. That is every two form is of type $(1,1)$. Hence requiring $\overline{\del}(\alpha)=0$ for a $(1,0)$-form is equivalent to $d\alpha=0$. So in fact your version of Moreira's theorem is Poincare duality decorated with the Dolbeaut decomposition of the De Rham complex. – Charlie Frohman Mar 21 '11 at 13:04
• Yes, of course, I'm an Analyst rather than a Topologist, so my mathematical tastes might well be quite different to yours! As I say, it's just a matter of opinion. Actually I think it's interesting that two mathematicians can have such different attitudes towards the same theorem; one mathematician's "beautiful theorem" is another mathematician's "tedious technical details", maybe?! (Measure theory also springs to mind...!) [Note: this comment is for humour purposes only] – Zen Harper Mar 22 '11 at 4:53