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This is probably a fairly basic question:

Poincaré Duality states the following: Given an $n$-manifold, the $k^{th}$ homology is isomorphic to the $(n-k)^{th}$ cohomology.

So I was curious is there some certain relation you get when dealing with de Rham cohomology? For example lets take the punctured plane, then the 1st cohomology group is simply $\mathbb{R}$. However, when looking at the first homotopy group of the punctured plane, we get $\mathbb{Z}$.

But, these two groups are not isomorphic. What am I missing?

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    Dear Benjamin, The relationship between homotopy and homology with $\mathbb Z$-coefficients (to the extent that there is one) is given by the Hurewicz theorem. The relationship between homology with $\mathbb Z$ and $\mathbb R$ coefficients is given by universal coefficients. Also, Poincare duality holds 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). Regards,2011-11-14
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    @Matt E: Why not make that an answer?2011-11-14
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    @Brandon: Dear Brandon, I thought at first that the OP may have been looking for something more substantive. I will post a version as answer, though. Regards,2011-11-14
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    When you allow the manifold to have boundary, frequently people call the appropriately-generalized version of Poincare Duality that applies to that situation *Alexander Duality*. It's also frequently phrased as $H_i(M,X) \simeq H^{m-i}(M,Y)$ where $X \sqcup Y = \partial M$.2011-11-14
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    There is a version Poincare duality between de Rham cohomology and compactly-supported de Rham cohomology. You can read about this in Section I of Bott & Tu's beautiful book, "Differential Forms in Algebraic Topology".2011-11-14

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