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Question: Calculate the de Rham cohomology groups of: $U=\mathbb{R}^3 - (L \cup C)$ and $V= \mathbb{R}^3 - (L' \cup C)$, where $L' = \{x = 2, y = 0\}$, $L = \{x = y = 0\}$ and $C=\{ x^2 + y^2 = 1, z=0\}$. Then conclude that they are not diffeomorphic.

I did the calculation of the Cohomology groups and they are the same. But I'm not seeing why $U$ and $V$ are not diffeomorphic.

I found $H^0(U) = \mathbb{R}$, $H^1(U) = \mathbb{R}\times \mathbb{R}$, $H^2(U) = \mathbb{R}$ and $H^r(U) = 0$ for all $r\geq 3$.

Jude
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    I'm guessing the fundamental groups are different. – Matt Samuel Apr 17 '16 at 21:22
  • I didn't use this to find the cohomoly groups but I think that $U$ has the same homotopy type of the torus, so the fundamental group would be easy, but in the case of $V$ how can I do the calculation? – Jude Apr 17 '16 at 21:29
  • I would've known how to do it years ago but I've forgotten. – Matt Samuel Apr 17 '16 at 21:30
  • There aren't that many tools for computing fundamental groups - what tools do you have available? –  Apr 17 '16 at 21:37
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    You can use whatever works =) – Jude Apr 17 '16 at 22:29
  • Have you computed what the wedge map (i.e. the cup product) does to cohomology? In other words in each of your examples for essential closed one forms $\alpha$ and $\beta$ what is $[\alpha \wedge \beta]$? – PVAL-inactive Apr 18 '16 at 23:30

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